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[PDF] Application of Derivativespmd - NCERT
In this chapter, we will study applications of the derivative in various disciplines, e g , surface area increasing when the length of an edge is 12 cm?
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16223_2leep206.pdf 6.1 Overview
6.1.1Rate of change of quantities
For the functiony =f (x),d
dx(f (x)) represents the rate of change ofy with respect tox. Thus if 's" represents the distance and 't" the time, thends dtrepresents the rate of change of distance with respect to time.6.1.2Tangents and normals
A line touching a curve y=f (x) at a point (x1,y1) is called the tangent to the curve at that point and its equation is given1 11( , ) 1( - )x ydyy yx xdxae ö- =ç ÷è ø. The normal to the curve is the line perpendicular to the tangent at the point of contact, and its equation is given as: y -y1 =1 11 ( , )-1 ( ) x yx xdy dx-ae öç ÷è øThe angle of intersection between two curves is the angle between the tangents to the
curves at the point of intersection.6.1.3Approximation
Sincef¢(x) =0( )- ( )lim
xf x x f x xD ®+D
D, we can say thatf¢(x) is approximately equal
to( )- ( )f x x f x x+D DÞ approximate value off (x +D x) =f(x) +Dx .f¢ (x).Chapter 6APPLICATION OF DERIVATIVES
118 MATHEMATICS6.1.4Increasing/decreasing functions
A continuous function in an interval (a,b) is :
(i) strictly increasing if for allx1,x2Î (a,b),x16.1.5Theorem:Letf be a continuous function on [a,b] and differentiable in (a,b) then
(i)f is increasing in [a,b] iff¢ (x) > 0 for each xÎ (a,b) (ii)f is decreasing in [a,b] iff¢ (x) < 0 for eachxÎ (a,b) (iii)f is a constant function in [a, b] iff¢ (x) = 0 for eachxÎ (a,b).6.1.6Maxima and minima
Local Maximum/Local Minimum for a real valued functionf A pointc in the interior of the domain off, is called (i) local maxima, if there exists an h> 0 , such thatf(c) > f (x), for allx in (c -h,c +h). The valuef (c) is called the local maximum value of f . (ii) local minima if there exists anh > 0 such thatf (c)APPLICATION OF DERIVATIVES 119(ii) Iff¢ (x) changes sign from negative to positive asx increases through
c, thenc is a point of local minima, andf(c) is local minimum value. (iii) Iff¢ (x) does not change sign as x increases throughc, thencis neither a point of local minima nor a point of local maxima. Such a point is called a point of inflection. (b)Second Derivative test: Letf be a function defined on an interval I and cÎ I. Letf be twice differentiable atc. Then (i)x =c is a point of local maxima iff¢(c) = 0 andf²(c) < 0. In this case f (c) is then the local maximum value. (ii)x =c is a point of local minima iff¢ (c) = 0 andf²(c) > 0. In this case f (c) is the local minimum value. (iii) The test fails iff¢(c) = 0 andf² (c) = 0. In this case, we go back to first derivative test.6.1.8Working rule for finding absolute maxima and or absolute minima :
Step 1 : Find all the critical points off in the given interval. Step 2 : At all these points and at the end points of the interval, calculate the values off. Step 3 : Identify the maximum and minimum values off out of the values calculated in step 2. The maximum value will be the absolute maximum value off and the minimum value will be the absolute minimum value of f.6.2 Solved Examples
Short Answer Type (S.A.)
Example 1For the curvey = 5x - 2x3, ifx increases at the rate of 2 units/sec, then how fast is the slope of curve changing whenx = 3?SolutionSlopeof curve =
dy dx = 5 - 6x2Þd dy
dt dxae öç ÷è ø = -12x.dx dt120 MATHEMATICS= -12 . (3) . (2)
= -72 units/sec. Thus, slope of curve is decreasing at the rate of 72 units/sec whenx is increasing at the rate of 2 units/sec. Example 2 Water is dripping out from a conical funnel of semi-vertical angle4pat the uniform rate of 2 cm2/sec in the surface area, through a tiny hole at the vertex of the
bottom. When the slant height of cone is 4 cm, find the rate of decrease of the slant height of water.SolutionIfs represents the surface area, thend s
dt= 2cm2 /sec s =p r.l =pl .sin4p.l =22lpTherefore,ds
dt=2.2dlldtp=2 .dlldtpwhenl =4 cm,1 1 2.2 cm/s42 .4 2 2dl dt= = =pp p. Example 3Find the angle of intersection of the curvesy2=xandx2=y. SolutionSolving the given equations, we havey2=xandx2=yÞx4 =x or x4 -x = 0Þ x (x3- 1) = 0Þx = 0,x = 1
Therefore,y = 0, y = 1
i.e. points of intersection are (0, 0) and (1, 1)Furthery2 =xÞ2ydy
dx = 1Þdy dx =12yandx2 =yÞdy
dx = 2x.APPLICATION OF DERIVATIVES 121At (0, 0), the slope of the tangent to the curvey2 =x is parallel toy-axis and the
tangent to the curvex2 =y is parallel tox-axis. Þ angle of intersection =2pAt (1, 1), slope of the tangent to the curvey2 =x is equal to12 and that ofx2= y is 2.
tanq =12-21 1+ =3
4.Þ q = tan-13
4ae öç ÷è øExample 4Prove that the functionf (x) = tanx - 4x is strictly decreasing on-,3 3p p
ae öç ÷è ø.Solutionf (x) = tanx- 4xÞf¢(x) = sec2x - 4
When-3p Therefore, 1 < sec
2x < 4Þ -3 < (sec2x - 4) < 0
Thus for-
3p Hencef is strictly decreasing on-,3 3p p
ae öç ÷è ø. Example 5Determine for which values ofx, the functiony =x4 -34 3xis increasing
and for which values, it is decreasing. Solutiony =x4 -34
3xÞdy
dx= 4x3 - 4x2= 4x2 (x- 1) 122 MATHEMATICSNow,dy
dx= 0Þx = 0,x = 1. Sincef¢ (x) < 0x" Î(-¥, 0)È(0, 1) andf is continuous in (-¥, 0] and [0, 1]. Thereforef is decreasing in (-¥, 1] andf is increasing in [1,¥). Note: Heref is strictly decreasing in (-¥, 0)È(0, 1) and is strictly increasing in (1,¥). Example 6Show that the functionf (x) = 4x3 - 18x2 + 27x - 7 has neither maxima nor minima. Solutionf (x) = 4x3 - 18x2 + 27x - 7
f¢ (x) = 12x2 - 36x + 27 = 3 (4x2 - 12x + 9) = 3 (2x - 3)2 f¢ (x) = 0Þx =3 2(critical point)
Sincef¢ (x) > 0 for allx3
2< and for allx >3
2Hencex =3
2is a point of inflexion i.e., neither a point of maxima nor a point of minima.
x =3 2is the only critical point, andf has neither maxima nor minima.
Example 7Using differentials, find the approximate value of0.082Solution Letf (x) =xUsingf (x +Dx)f (x) +Dx . f¢(x), takingx = .09 andDx= - 0.008,
we getf(0.09 - 0.008) = f(0.09) + (- 0.008) f¢ (0.09) Þ0.082 =0.09 - 0.008 .1
2 0.09ae öç ÷è ø = 0.3 -0.008
0.6= 0.3 - 0.0133 = 0.2867.
APPLICATION OF DERIVATIVES 123Example 8 Find the condition for the curves2 2 2 2 -x y a b= 1;xy =c2 to intersect orthogonally. Solution Let the curves intersect at (x1,y1). Therefore,2 2 2 2 -x y a b= 1Þ2 22 2-x y dy dx a b= 0Þ 2 2dy b x
dx a y=Þ slope of tangent at the point of intersection (m1) =2 1 2 1b x a yAgain xy =c2Þdyx ydx+= 0Þ-dy y dx x=Þ m2 =1 1y x- . For orthoganality,m1 ×m2= - 1Þ 2
2b a= 1 ora2 -b2 = 0. Example 9Find all the points of local maxima and local minima of the function f (x) =4 3 23 45- - 8 - 1054 2x x x+. Solutionf¢ (x) = -3x3- 24x2 - 45x
=- 3x (x2 + 8x + 15) = - 3x (x + 5) (x + 3) f¢ (x) = 0Þx = -5,x = -3, x = 0 f²(x) = -9x2 - 48x - 45 = -3 (3x2 + 16x + 15) f²(0) = - 45 < 0. Therefore,x = 0 is point of local maxima f²(-3) = 18 > 0. Therefore,x = -3 is point of local minima f²(-5) = -30 < 0. Thereforex = -5 is point of local maxima. 124 MATHEMATICSExample 10 Show that the local maximum value of1xx+is less than local minimum
value. SolutionLety=1xx+Þdy
dx= 1 -21 x,dy dx= 0Þx2= 1Þx = ± 1.2 2d y dx= +32 x, therefore2 2d y dx(atx = 1) > 0 and2 2d y dx(atx = -1) < 0. Hence local maximum value ofy is atx = -1 and the local maximum value = - 2. Local minimum value ofy is atx = 1 and local minimum value = 2. Therefore, local maximum value (-2) is less than local minimum value 2. Long Answer Type (L.A.)
Example 11Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is6p. SolutionGiven thatdv
dt = 1 cm3/s, wherev is the volume of water in the conical vessel. From the Fig.6.2,l = 4cm,h =l cos6p =3
2l andr =l sin6p=2l.
Therefore,v =1
3pr2h =233 3
3 4 2 24ll lpp=.
APPLICATION OF DERIVATIVES 12523
8dv dlldt dtp=Therefore, 1 =316.8dl
dtpÞ1 2 3dl dt=pcm/s. Therefore, the rate of decrease of slant height =1 2 3pcm/s.
Example 12 Find the equation of all the tangents to the curvey = cos (x +y), -2 p £x£ 2p, that are parallel to the line x + 2y = 0. SolutionGiven that y = cos (x +y)Þdy
dx= - sin (x +y)1dy dxé ù+ê úë û...(i) ordy dx = -( ) ( )sin 1 sinx y
x y+ + +Since tangent is parallel tox + 2y = 0, therefore slope of tangent =1-2Therefore,( ) ( )sin-1 sinx y x y+ + +=1-2Þ sin (x +y) = 1.... (ii) Since cos (x +y) =y and sin (x +y)=1Þ cos2 (x +y)+ sin2(x +y) =y2 + 1 Þ1 =y2 + 1 ory = 0.
Therefore, cosx = 0.
Therefore, x = (2n + 1)2p, n= 0, ± 1, ± 2... 126 MATHEMATICSThus,x =3,2 2p p± ±, butx =2p,x =-3
2psatisfy equation (ii)
Hence, the points are, 02p
ae öç ÷è ø,-3,02p ae öç ÷è ø. Therefore, equation of tangent at, 02p
ae öç ÷è ø isy =1-2-2xpae öç ÷è ø or 2x + 4y -p = 0, and equation of tangent at-3,02p ae öç ÷è øisy =1-23 2xpae ö+ç ÷è øor 2x + 4y + 3p = 0.
Example 13Find the angle of intersection of the curvesy2 = 4ax andx2 = 4by. SolutionGiven thaty2 = 4ax...(i) andx2 = 4by...(ii). Solving (i) and (ii), we get22 4x bae öç ÷è ø= 4axÞ x4 = 64ab2x orx(x3 - 64ab2) = 0 Þ x = 0,1 2 3 34x a b=Therefore, the points of intersection are (0, 0) and1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø.
Again,y2 = 4axÞ4 2
2dy a a
dx y y= =andx2 = 4byÞ 2 4 2dy x x
dx b b= =Therefore, at (0, 0) the tangent to the curvey2 = 4ax is parallel toy-axis and tangent to the curvex2= 4by is parallel tox-axis. Þ Angle between curves =2pAt1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø,m1(slope of the tangent to the curve (i)) =1
32a
bae öç ÷è ø=1 3 2 1 3 32 1
24a a
ba bae ö =ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21 3 33422a b a
b bae ö=ç ÷è ø APPLICATION OF DERIVATIVES 127Therefore, tanq =2 1 1 2- 1m m m m+ =11 33
1 1 3 31 2 -2 1 1 22a a
b b a a b bae ö ae ö ç ÷ ç ÷è ø è ø
ae ö ae ö +ç ÷ ç ÷è ø è ø =1 1 3 3 2 2 3 33 .
2a b a b ae ö+ç ÷ç ÷è ø Hence,q = tan-11 1
3 3 2 2 3 33 .
2a b a baeöç÷ç÷ç÷ae öç÷+ç ÷ç÷ç ÷è øèøExample 14Show that the equation of normal at any point on the curve
x = 3cosq - cos3q,y= 3sinq - sin3q is 4 (ycos3q -x sin3q) = 3 sin 4q. SolutionWe havex = 3cosq - cos3q
Therefore,dx
dq = -3sinq + 3cos2q sinq = - 3sinq (1 - cos2q) = -3sin3q .dy dq = 3cosq - 3sin2q cosq = 3cosq (1 - sin2q) = 3cos3q3 3cos-sindy
dxq=q. Therefore, slope of normal =3 3sin cosq+qHence the equation of normal is y - (3sinq - sin3q) =3 3sin cosq q[x - (3cosq - cos3q)] Þ ycos3q - 3sinq cos3q + sin3q cos3q =xsin3q - 3sin3q cosq + sin3q cos3q Þ ycos3q -xsin3q = 3sinq cosq (cos2q - sin2q)
128 MATHEMATICS =3
2sin2q . cos2q
=3 4sin4q
or 4 (ycos3q -xsin3q) = 3 sin4q. Example 15Find the maximum and minimum values of
f (x) = secx + log cos2x,0 Solution f(x) = secx + 2 log cosx Therefore, f
(x) = secx tanx - 2 tanx = tanx (secx -2) f (x) = 0Þ tanx = 0 or secx = 2 or cosx =1 2Therefore, possible values ofx arex = 0, orx=p and
x =3porx =5 3pAgain,f¢¢ (x) = sec2x (secx -2) + tanx (secx tanx)
= sec 3x + secx tan2x - 2sec2x
=secx (sec2x + tan2x - 2secx). We note thatf¢¢(0) = 1 (1 + 0 - 2) = -1 < 0. Therefore,x = 0 is a point of maxima.f¢¢(p) = -1 (1 + 0 + 2) = -3 < 0. Therefore,x =p is a point of maxima.f¢¢3p
ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =3p is a point of minima.f¢¢5 3p ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =5 3p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1 Maximum Value ofy atx =p is-1 + 0 = -1
Minimum Value ofy atx =3pis 2 + 2 log1
2 = 2 (1 - log2)
Minimum Value ofy atx =5
3pis 2 + 2 log1
2 = 2 (1 - log2)
Example 16 Find the area of greatest rectangle that can be inscribed in an ellipse2 2 2 2 1x y a b+ =. Solution Let ABCD be the rectangle of maximum area with sides AB = 2xand BC =2y,where C (x,y) is a point on the ellipse2 2
2 2 1x y a b+ = as shown in the Fig.6.3. The area A of the rectangle is 4xyi.e. A = 4xywhich gives A2 = 16x2y2 =s (say) Therefore,s = 16x222
21- .xbaae öç ÷è ø =2
216b
a (a2x2 -x4) Þ2 216ds b
dx a=. [2a2x - 4x3]. Again,ds
dx = 0Þ x =and2 2a by=Now,2 2 2 216d s b
dx a=[2a2 - 12x2] At2 222 22
2 221616,[2 6 ] ( 4 ) 02a d s bbxa aadx aa= = - = - <
130 MATHEMATICSThus atx =2a, y =2b, s is maximum and hence the area A is maximum.
Maximum area = 4.x.y =4 .2a.2b= 2ab sq units.
Example 17Find the difference between the greatest and least values of the functionf (x) = sin2x -x, on- ,2 2p p é ùê úë û.
Solutionf (x) = sin2x -x
Þf¢(x) = 2 cos2x - 1
Therefore,f¢(x) = 0Þ cos2x =1
2Þ 2xisor3 3- p pÞx =- or6 6p p-2fpae öç ÷è ø =sin (- p) +2 2p p=-6fpae öç ÷è ø =2sin -6 6p p
ae ö+ç ÷è ø=3-2 6p+6fpae öç ÷è ø =2sin -6 6p p ae öç ÷è ø=3-2 6p2fpae öç ÷è ø =( )sin -2pp=-2pClearly,2pis the greatest value and-2pis the least.
Therefore, difference =2p +2p=p
APPLICATION OF DERIVATIVES 131Example 18An isosceles triangle of vertical angle 2q is inscribed in a circle of radius
a . Show that the area of triangle is maximum whenq =6p. SolutionLet ABC be an isosceles triangle inscribed in the circle with radiusa such that AB = AC. AD = AO + OD =
a +a cos2qand BC = 2BD = 2a sin2q (see fig. 16.4) Therefore, area of the triangle ABC i.e.D =1
2BC . AD
=1 22asin2q . (a +a cos2q)
= a2sin2q (1 + cos2q) Þ D= a2sin2q +1
2a2sin4q
Therefore,d
dD q= 2a2cos2q + 2a2cos4q = 2a2(cos2q + cos4q)d dD q= 0Þ cos2q = -cos4q = cos (p - 4q) Therefore, 2q =p - 4q Þ q =6p2
2d dD q = 2a2 (-2sin2q - 4sin4q) < 0 (atq =6p). Therefore, Area of triangle is maximum whenq =6p.
132 MATHEMATICSObjective Type Questions
Choose the correct answer from the given four options in each of the following Examples 19 to 23.
Example 19The abscissa of the point on the curve 3y = 6x - 5x3, the normal at which passes through origin is: (A) 1(B)1 3(C) 2(D)1
2SolutionLet (x1,y1) be the point on the given curve 3y = 6x - 5x3 at which the normal
passes through the origin. Then we have1 12 1 ( , )2-5 x ydy xdxae ö=ç ÷è ø. Again the equation of the normal at (x1,y1) passing through the origin gives2112 11- -32-56-5xxyx= =. Sincex1 = 1 satisfies the equation, therefore, Correct answer is (A). Example 20The two curvesx3 - 3xy2 + 2 = 0 and 3x2y -y3 = 2 (A) touch each other(B) cut at right angle (C) cut at an angle3p(D) cut at an angle4pSolutionFrom first equation of the curve, we have 3x2 - 3y2 - 6xydy
dx = 0 Þdy
dx =2 2- 2x y xy= (m1) say and second equation of the curve gives 6 xy + 3x2dy dx- 3y2dy dx= 0Þdy dx=2 2-2 -xy x y= (m2) say Since m1 . m2= -1. Therefore, correct answer is (B). APPLICATION OF DERIVATIVES 133Example 21The tangent to the curve given byx =et . cost,y =et . sint att=4pmakes
withx-axis an angle: (A) 0(B)4p(C)3p(D)2pSolutiondx dt=- et . sint+etcost,dy dt=etcost+etsint Therefore,4tdy
dx p=ae öç ÷è ø=cos sin cos - sint t t t+=2 0 and hence the correct answer is (D).
Example 22The equation of the normal to the curvey = sinx at (0, 0) is: (A)x = 0(B)y = 0(C)x +y = 0 (D)x -y = 0 Solutiondy
dx= cosx.Therefore, slope of normal =0-1 cos xx=ae öç ÷è ø= -1. Hence the equation of normal isy - 0 = -1(x - 0) orx +y = 0 Therefore, correct answer is (C).
Example 23The point on the curvey2 =x, where the tangent makes an angle of4pwithx-axis is (A)1,2 41 ae öç ÷è ø(B)1,4 21 ae öç ÷è ø(C) (4, 2)(D) (1, 1) Solution1
2dy dx y== tan4p= 1Þy =1 2Þx =41Therefore, correct answer is B.
134 MATHEMATICSFill in the blanks in each of the following Examples 24 to 29.
Example 24The values ofa for whichy =x2 +ax + 25 touches the axis ofx are______. Solution0 2 0dyx adx= Þ + =i.e.x =2a-,
Therefore,2
25 04 2a aaae ö+ - + =ç ÷è øÞa =±10
Hence, the values ofa are ± 10.
Example 25If f(x) =21
4 2 1x x+ +, then its maximum value is _______.
Solution Forf to be maximum, 4x2 + 2x+ 1 should be minimum i.e. 4 x 2 + 2x + 1 = 4 (x +1
4)2 +114ae ö-ç ÷è øgiving the minimum value of 4x2+ 2x + 1 =3
4. Hence maximum value off =4
3. Example 26Letfhave second deriative atc such thatf¢(c) = 0 and f ²(c) > 0, thenc is a point of ______. Solution Local minima.
Example 27Minimum value of f iff (x) = sinx in-,2 2p p é ùê úë ûis _____.
Solution -1
Example 28The maximum value of sinx + cosx is _____. Solution2.
APPLICATION OF DERIVATIVES 135Example 29The rate of change of volume of a sphere with respect to its surface
area, when the radius is 2 cm, is______. Solution1 cm3/cm2
v =32443dvr rdrp Þ = p,s =24dsrdrp Þ =82dv rrdsp Þ == 1 atr = 2. 6.3 EXERCISE
Short Answer (S.A.)
1.A spherical ball of salt is dissolving in water in such a manner that the rate of
decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate. 2.If the area of a circle increases at a uniform rate, then prove that perimetervaries inversely as the radius.
3.A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is10 m/s, how fast is the string being let out; when the kite is 250 m away fromthe boy who is flying the kite? The height of boy is 1.5 m.
4.Two men A and B start with velocitiesv at the same time from the junction of
two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated.. 5.Find an angleq, 0 , which increases twice as fast as its sine. 6.Find the approximate value of (1.999)5.
7.Find the approximate volume of metal in a hollow spherical shell whose internaland external radii are 3 cm and 3.0005 cm, respectively.
8.A man, 2m tall, walks at the rate of213m/s towards a street light which is153m above the ground. At what rate is the tip of his shadow moving? At what
136 MATHEMATICSrate is the length of the shadow changing when he is133m from the base of
the light? 9.A swimming pool is to be drained for cleaning. If L represents the number of
litres of water in the poolt seconds after the pool has been plugged off to drain and L = 200 (10 -t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds? 10.The volume of a cube increases at a constant rate. Prove that the increase inits surface area varies inversely as the length of the side.
11.x andyare the sides of two squares such thaty =x -x2. Find the rate of
change of the area of second square with respect to the area of first square. 12.Find the condition that the curves 2x =y2 and 2xy =k intersect orthogonally.
13.Prove that the curvesxy = 4 andx2 +y2 = 8 touch each other.
14.Find the co-ordinates of the point on the curve+x y= 4 at which tangent
is equally inclined to the axes. 15.Find the angle of intersection of the curvesy = 4 -x2 andy =x2.
16.Prove that the curvesy2 = 4x andx2 +y2 - 6x + 1 = 0 touch each other at the
point (1, 2). 17.Find the equation of the normal lines to the curve 3x2 -y2 = 8 which are
parallel to the linex + 3y = 4. 18.At what points on the curvex2 +y2 - 2x - 4y + 1 = 0, the tangents are parallel
to they-axis? 19.Show that the line+x y
a b= 1, touches the curvey =b .-x aeat the point where the curve intersects the axis ofy. 20.Show thatf (x) = 2x + cot-1x +log()21x x+ -is increasing inR.
APPLICATION OF DERIVATIVES 13721.Show that fora³1,f (x) =3sinx -cosx -2ax + bis decreasing inR.
22.Show thatf (x) = tan-1(sinx + cosx) is an increasing function in0,4
ae öç ÷è ø. 23.At what point, the slope of the curvey = -x3 + 3x2 + 9x - 27 is maximum?
Also find the maximum slope.
24.Prove thatf (x) = sinx +3cosxhas maximum value atx =6
. Long Answer (L.A.)
25.If the sum of the lengths of the hypotenuse and a side of a right angled triangle
is given, show that the area of the triangle is maximum when the angle between them is3 . 26.Find the points of local maxima, local minima and the points of inflection of the
functionf (x) =x5 - 5x4 + 5x3 - 1. Also find the corresponding local maximum and local minimum values. 27.A telephone company in a town has 500 subscribers on its list and collectsfixed charges of Rs 300/- per subscriber per year. The company proposes toincrease the annual subscription and it is believed that for every increase ofRe 1/- one subscriber will discontinue the service. Find what increase willbring maximum profit?
28.If the straight linex cosa +y sina =p touches the curve2 2
2 2 +x y a b= 1, then prove thata2 cos2a +b2 sin2a =p2. 29.An open box with square base is to be made of a given quantity of card board
of areac2. Show that the maximum volume of the box is3 6 3c cubic units.
30.Find the dimensions of the rectangle of perimeter 36 cm which will sweep out
a volume as large as possible, when revolved about one of its sides. Also find the maximum volume. 138 MATHEMATICS31.If the sum of the surface areas of cube and a sphere is constant, what is the
ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum? 32.AB is a diameter of a circle and C is any point on the circle. Show that thearea ofD ABC is maximum, when it is isosceles.
33.A metal box with a square base and vertical sides is to contain 1024 cm3. The
material for the top and bottom costs Rs 5/cm 2 and the material for the sides
costs Rs 2.50/cm 2 . Find the least cost of the box.
34.The sum of the surface areas of a rectangular parallelopiped with sidesx, 2x
and3xand a sphere is given to be constant. Prove that the sum of their volumes is minimum, ifx is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes. Objective Type Questions
Choose the correct answer from the given four options in each of the following questions 35 to 39:
35.The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The
rate at which the area increases, when side is 10 cm is: (A) 10 cm 2/s (B)3cm2/s (C) 103cm2/s (D)10
3cm2/s
36.A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical
wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A)1 10radian/sec (B)1
20radian/sec (C) 20 radian/sec
(D) 10 radian/sec 37.The curvey =1
5xhas at (0, 0)
APPLICATION OF DERIVATIVES 139(A) a vertical tangent (parallel toy-axis) (B) a horizontal tangent (parallel tox-axis) (C) an oblique tangent (D) no tangent 38.The equation of normal to the curve 3x2 -y2 = 8 which is parallel to the line
x + 3y = 8 is (A) 3x -y = 8(B) 3x +y + 8 = 0 (C)x + 3y± 8 = 0(D)x + 3y = 0 39.If the curveay +x2 = 7 andx3 =y, cut orthogonally at (1, 1), then the value of
a is: (A) 1 (B) 0(C) - 6(D) .6 40.Ify =x4 - 10 and ifx changes from 2 to 1.99, what is the change iny
(A) .32 (B) .032 (C) 5.68 (D) 5.968 41.The equation of tangent to the curvey (1 +x2) = 2 -x, where it crossesx-axis
is: (A) x + 5y = 2(B) x - 5y = 2 (C) 5 x -y = 2(D) 5x +y = 2 42.The points at which the tangents to the curvey =x3 - 12x + 18 are parallel to
x-axis are: (A) (2, -2), (-2, -34) (B) (2, 34), (-2, 0) (C) (0, 34), (-2, 0)(D) (2, 2), (-2, 34) 43.The tangent to the curvey =e2x at the point (0, 1) meetsx-axis at:
(A) (0, 1) (B)1- ,02ae öç ÷è ø(C) (2, 0)(D) (0, 2) 44.The slope of tangent to the curvex =t2 + 3t - 8,y= 2t2 - 2t - 5 at the point
(2, -1) is: 140 MATHEMATICS(A)22
7(B)6 7(C)-6
7(D) - 6
45.The two curvesx3 - 3xy2+ 2 = 0 and 3x2y -y3 - 2 = 0 intersect at an angle of
(A)4 (B)3 (C)2 (D)6 46.The interval on which the functionf (x) = 2x3 + 9x2 + 12x - 1 is decreasing is: (A) [-1,¥)(B) [-2, -1](C)(-¥, -2] (D) [-1, 1] 47.Let thef :R®R be defined byf(x) = 2x + cosx, then f:
(A) has a minimum atx =p(B) has a maximum, atx = 0 (C) is a decreasing function(D) is an increasing function 48.y =x (x - 3)2decreases for the values ofx given by :
(A) 1 0 (D) 0 < x <3 249.The functionf (x) = 4 sin3x - 6 sin2x + 12 sinx + 100 is strictly
(A) increasing in3ππ,2ae öç ÷è ø(B) decreasing inπ,π2ae öç ÷è ø(C) decreasing in-,2 2 é ùê úë û(D) decreasing in0,2
é ùê úë û50.Which of the following functions is decreasing on0,2 ae öç ÷è ø(A) sin2x(B) tanx(C) cosx(D) cos 3x 51.The functionf (x) = tanx -x
(A) always increases (B) always decreases (C) never increases(D) sometimes increases and sometimes decreases. APPLICATION OF DERIVATIVES 14152.Ifx is real, the minimum value ofx2 - 8x + 17 is (A) -1(B) 0 (C) 1 (D) 2 53.The smallest value of the polynomialx3 - 18x2 + 96x in [0, 9] is
(A) 126 (B) 0 (C) 135 (D) 160 54.The functionf (x) = 2x3 - 3x2 - 12x + 4, has
(A) two points of local maximum (B) two points of local minimum (C) one maxima and one minima (D) no maxima or minima 55.The maximum value of sinx . cosx is
(A)1 4(B)1 2(C)2(D)2 256.Atx =5
6,f (x) = 2 sin3x + 3 cos3x is:
(A) maximum(B) minimum (C) zero(D) neither maximum nor minimum. 57.Maximum slope of the curvey = -x3 + 3x2 + 9x - 27 is:
(A) 0 (B) 12 (C) 16 (D) 32 58.f (x) =xxhas a stationary point at
(A)x =e(B)x=1 e(C)x = 1 (D)x =e59.The maximum value of1ae öç ÷è øx xis: (A)e(B)ee(C)1 ee(D)11ae öç ÷è øe e 142 MATHEMATICSFill in the blanks in each of the following Exercises 60 to 64:
60.The curvesy = 4x2 + 2x - 8 and y =x3 -x + 13 touch each other at the
point_____. 61.The equation of normal to the curvey = tanx at (0, 0) is ________.
62.The values ofa for which the function f (x) = sinx -ax +b increases onRare
______. 63.The functionf (x) =2
42 -1x
x,x > 0, decreases in the interval _______. 64.The least value of the functionf (x) =ax +b
x (a > 0,b > 0,x > 0) is ______.
Therefore, 1 < sec
2x < 4Þ -3 < (sec2x - 4) < 0
Thus for-
3p Hencef is strictly decreasing on-,3 3p p
ae öç ÷è ø. Example 5Determine for which values ofx, the functiony =x4 -34 3xis increasing
and for which values, it is decreasing. Solutiony =x4 -34
3xÞdy
dx= 4x3 - 4x2= 4x2 (x- 1) 122 MATHEMATICSNow,dy
dx= 0Þx = 0,x = 1. Sincef¢ (x) < 0x" Î(-¥, 0)È(0, 1) andf is continuous in (-¥, 0] and [0, 1]. Thereforef is decreasing in (-¥, 1] andf is increasing in [1,¥). Note: Heref is strictly decreasing in (-¥, 0)È(0, 1) and is strictly increasing in (1,¥). Example 6Show that the functionf (x) = 4x3 - 18x2 + 27x - 7 has neither maxima nor minima. Solutionf (x) = 4x3 - 18x2 + 27x - 7
f¢ (x) = 12x2 - 36x + 27 = 3 (4x2 - 12x + 9) = 3 (2x - 3)2 f¢ (x) = 0Þx =3 2(critical point)
Sincef¢ (x) > 0 for allx3
2< and for allx >3
2Hencex =3
2is a point of inflexion i.e., neither a point of maxima nor a point of minima.
x =3 2is the only critical point, andf has neither maxima nor minima.
Example 7Using differentials, find the approximate value of0.082Solution Letf (x) =xUsingf (x +Dx)f (x) +Dx . f¢(x), takingx = .09 andDx= - 0.008,
we getf(0.09 - 0.008) = f(0.09) + (- 0.008) f¢ (0.09) Þ0.082 =0.09 - 0.008 .1
2 0.09ae öç ÷è ø = 0.3 -0.008
0.6= 0.3 - 0.0133 = 0.2867.
APPLICATION OF DERIVATIVES 123Example 8 Find the condition for the curves2 2 2 2 -x y a b= 1;xy =c2 to intersect orthogonally. Solution Let the curves intersect at (x1,y1). Therefore,2 2 2 2 -x y a b= 1Þ2 22 2-x y dy dx a b= 0Þ 2 2dy b x
dx a y=Þ slope of tangent at the point of intersection (m1) =2 1 2 1b x a yAgain xy =c2Þdyx ydx+= 0Þ-dy y dx x=Þ m2 =1 1y x- . For orthoganality,m1 ×m2= - 1Þ 2
2b a= 1 ora2 -b2 = 0. Example 9Find all the points of local maxima and local minima of the function f (x) =4 3 23 45- - 8 - 1054 2x x x+. Solutionf¢ (x) = -3x3- 24x2 - 45x
=- 3x (x2 + 8x + 15) = - 3x (x + 5) (x + 3) f¢ (x) = 0Þx = -5,x = -3, x = 0 f²(x) = -9x2 - 48x - 45 = -3 (3x2 + 16x + 15) f²(0) = - 45 < 0. Therefore,x = 0 is point of local maxima f²(-3) = 18 > 0. Therefore,x = -3 is point of local minima f²(-5) = -30 < 0. Thereforex = -5 is point of local maxima. 124 MATHEMATICSExample 10 Show that the local maximum value of1xx+is less than local minimum
value. SolutionLety=1xx+Þdy
dx= 1 -21 x,dy dx= 0Þx2= 1Þx = ± 1.2 2d y dx= +32 x, therefore2 2d y dx(atx = 1) > 0 and2 2d y dx(atx = -1) < 0. Hence local maximum value ofy is atx = -1 and the local maximum value = - 2. Local minimum value ofy is atx = 1 and local minimum value = 2. Therefore, local maximum value (-2) is less than local minimum value 2. Long Answer Type (L.A.)
Example 11Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is6p. SolutionGiven thatdv
dt = 1 cm3/s, wherev is the volume of water in the conical vessel. From the Fig.6.2,l = 4cm,h =l cos6p =3
2l andr =l sin6p=2l.
Therefore,v =1
3pr2h =233 3
3 4 2 24ll lpp=.
APPLICATION OF DERIVATIVES 12523
8dv dlldt dtp=Therefore, 1 =316.8dl
dtpÞ1 2 3dl dt=pcm/s. Therefore, the rate of decrease of slant height =1 2 3pcm/s.
Example 12 Find the equation of all the tangents to the curvey = cos (x +y), -2 p £x£ 2p, that are parallel to the line x + 2y = 0. SolutionGiven that y = cos (x +y)Þdy
dx= - sin (x +y)1dy dxé ù+ê úë û...(i) ordy dx = -( ) ( )sin 1 sinx y
x y+ + +Since tangent is parallel tox + 2y = 0, therefore slope of tangent =1-2Therefore,( ) ( )sin-1 sinx y x y+ + +=1-2Þ sin (x +y) = 1.... (ii) Since cos (x +y) =y and sin (x +y)=1Þ cos2 (x +y)+ sin2(x +y) =y2 + 1 Þ1 =y2 + 1 ory = 0.
Therefore, cosx = 0.
Therefore, x = (2n + 1)2p, n= 0, ± 1, ± 2... 126 MATHEMATICSThus,x =3,2 2p p± ±, butx =2p,x =-3
2psatisfy equation (ii)
Hence, the points are, 02p
ae öç ÷è ø,-3,02p ae öç ÷è ø. Therefore, equation of tangent at, 02p
ae öç ÷è ø isy =1-2-2xpae öç ÷è ø or 2x + 4y -p = 0, and equation of tangent at-3,02p ae öç ÷è øisy =1-23 2xpae ö+ç ÷è øor 2x + 4y + 3p = 0.
Example 13Find the angle of intersection of the curvesy2 = 4ax andx2 = 4by. SolutionGiven thaty2 = 4ax...(i) andx2 = 4by...(ii). Solving (i) and (ii), we get22 4x bae öç ÷è ø= 4axÞ x4 = 64ab2x orx(x3 - 64ab2) = 0 Þ x = 0,1 2 3 34x a b=Therefore, the points of intersection are (0, 0) and1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø.
Again,y2 = 4axÞ4 2
2dy a a
dx y y= =andx2 = 4byÞ 2 4 2dy x x
dx b b= =Therefore, at (0, 0) the tangent to the curvey2 = 4ax is parallel toy-axis and tangent to the curvex2= 4by is parallel tox-axis. Þ Angle between curves =2pAt1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø,m1(slope of the tangent to the curve (i)) =1
32a
bae öç ÷è ø=1 3 2 1 3 32 1
24a a
ba bae ö =ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21 3 33422a b a
b bae ö=ç ÷è ø APPLICATION OF DERIVATIVES 127Therefore, tanq =2 1 1 2- 1m m m m+ =11 33
1 1 3 31 2 -2 1 1 22a a
b b a a b bae ö ae ö ç ÷ ç ÷è ø è ø
ae ö ae ö +ç ÷ ç ÷è ø è ø =1 1 3 3 2 2 3 33 .
2a b a b ae ö+ç ÷ç ÷è ø Hence,q = tan-11 1
3 3 2 2 3 33 .
2a b a baeöç÷ç÷ç÷ae öç÷+ç ÷ç÷ç ÷è øèøExample 14Show that the equation of normal at any point on the curve
x = 3cosq - cos3q,y= 3sinq - sin3q is 4 (ycos3q -x sin3q) = 3 sin 4q. SolutionWe havex = 3cosq - cos3q
Therefore,dx
dq = -3sinq + 3cos2q sinq = - 3sinq (1 - cos2q) = -3sin3q .dy dq = 3cosq - 3sin2q cosq = 3cosq (1 - sin2q) = 3cos3q3 3cos-sindy
dxq=q. Therefore, slope of normal =3 3sin cosq+qHence the equation of normal is y - (3sinq - sin3q) =3 3sin cosq q[x - (3cosq - cos3q)] Þ ycos3q - 3sinq cos3q + sin3q cos3q =xsin3q - 3sin3q cosq + sin3q cos3q Þ ycos3q -xsin3q = 3sinq cosq (cos2q - sin2q)
128 MATHEMATICS =3
2sin2q . cos2q
=3 4sin4q
or 4 (ycos3q -xsin3q) = 3 sin4q. Example 15Find the maximum and minimum values of
f (x) = secx + log cos2x,0 Solution f(x) = secx + 2 log cosx Therefore, f
(x) = secx tanx - 2 tanx = tanx (secx -2) f (x) = 0Þ tanx = 0 or secx = 2 or cosx =1 2Therefore, possible values ofx arex = 0, orx=p and
x =3porx =5 3pAgain,f¢¢ (x) = sec2x (secx -2) + tanx (secx tanx)
= sec 3x + secx tan2x - 2sec2x
=secx (sec2x + tan2x - 2secx). We note thatf¢¢(0) = 1 (1 + 0 - 2) = -1 < 0. Therefore,x = 0 is a point of maxima.f¢¢(p) = -1 (1 + 0 + 2) = -3 < 0. Therefore,x =p is a point of maxima.f¢¢3p
ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =3p is a point of minima.f¢¢5 3p ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =5 3p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1 Maximum Value ofy atx =p is-1 + 0 = -1
Minimum Value ofy atx =3pis 2 + 2 log1
2 = 2 (1 - log2)
Minimum Value ofy atx =5
3pis 2 + 2 log1
2 = 2 (1 - log2)
Example 16 Find the area of greatest rectangle that can be inscribed in an ellipse2 2 2 2 1x y a b+ =. Solution Let ABCD be the rectangle of maximum area with sides AB = 2xand BC =2y,where C (x,y) is a point on the ellipse2 2
2 2 1x y a b+ = as shown in the Fig.6.3. The area A of the rectangle is 4xyi.e. A = 4xywhich gives A2 = 16x2y2 =s (say) Therefore,s = 16x222
21- .xbaae öç ÷è ø =2
216b
a (a2x2 -x4) Þ2 216ds b
dx a=. [2a2x - 4x3]. Again,ds
dx = 0Þ x =and2 2a by=Now,2 2 2 216d s b
dx a=[2a2 - 12x2] At2 222 22
2 221616,[2 6 ] ( 4 ) 02a d s bbxa aadx aa= = - = - <
130 MATHEMATICSThus atx =2a, y =2b, s is maximum and hence the area A is maximum.
Maximum area = 4.x.y =4 .2a.2b= 2ab sq units.
Example 17Find the difference between the greatest and least values of the functionf (x) = sin2x -x, on- ,2 2p p é ùê úë û.
Solutionf (x) = sin2x -x
Þf¢(x) = 2 cos2x - 1
Therefore,f¢(x) = 0Þ cos2x =1
2Þ 2xisor3 3- p pÞx =- or6 6p p-2fpae öç ÷è ø =sin (- p) +2 2p p=-6fpae öç ÷è ø =2sin -6 6p p
ae ö+ç ÷è ø=3-2 6p+6fpae öç ÷è ø =2sin -6 6p p ae öç ÷è ø=3-2 6p2fpae öç ÷è ø =( )sin -2pp=-2pClearly,2pis the greatest value and-2pis the least.
Therefore, difference =2p +2p=p
APPLICATION OF DERIVATIVES 131Example 18An isosceles triangle of vertical angle 2q is inscribed in a circle of radius
a . Show that the area of triangle is maximum whenq =6p. SolutionLet ABC be an isosceles triangle inscribed in the circle with radiusa such that AB = AC. AD = AO + OD =
a +a cos2qand BC = 2BD = 2a sin2q (see fig. 16.4) Therefore, area of the triangle ABC i.e.D =1
2BC . AD
=1 22asin2q . (a +a cos2q)
= a2sin2q (1 + cos2q) Þ D= a2sin2q +1
2a2sin4q
Therefore,d
dD q= 2a2cos2q + 2a2cos4q = 2a2(cos2q + cos4q)d dD q= 0Þ cos2q = -cos4q = cos (p - 4q) Therefore, 2q =p - 4q Þ q =6p2
2d dD q = 2a2 (-2sin2q - 4sin4q) < 0 (atq =6p). Therefore, Area of triangle is maximum whenq =6p.
132 MATHEMATICSObjective Type Questions
Choose the correct answer from the given four options in each of the following Examples 19 to 23.
Example 19The abscissa of the point on the curve 3y = 6x - 5x3, the normal at which passes through origin is: (A) 1(B)1 3(C) 2(D)1
2SolutionLet (x1,y1) be the point on the given curve 3y = 6x - 5x3 at which the normal
passes through the origin. Then we have1 12 1 ( , )2-5 x ydy xdxae ö=ç ÷è ø. Again the equation of the normal at (x1,y1) passing through the origin gives2112 11- -32-56-5xxyx= =. Sincex1 = 1 satisfies the equation, therefore, Correct answer is (A). Example 20The two curvesx3 - 3xy2 + 2 = 0 and 3x2y -y3 = 2 (A) touch each other(B) cut at right angle (C) cut at an angle3p(D) cut at an angle4pSolutionFrom first equation of the curve, we have 3x2 - 3y2 - 6xydy
dx = 0 Þdy
dx =2 2- 2x y xy= (m1) say and second equation of the curve gives 6 xy + 3x2dy dx- 3y2dy dx= 0Þdy dx=2 2-2 -xy x y= (m2) say Since m1 . m2= -1. Therefore, correct answer is (B). APPLICATION OF DERIVATIVES 133Example 21The tangent to the curve given byx =et . cost,y =et . sint att=4pmakes
withx-axis an angle: (A) 0(B)4p(C)3p(D)2pSolutiondx dt=- et . sint+etcost,dy dt=etcost+etsint Therefore,4tdy
dx p=ae öç ÷è ø=cos sin cos - sint t t t+=2 0 and hence the correct answer is (D).
Example 22The equation of the normal to the curvey = sinx at (0, 0) is: (A)x = 0(B)y = 0(C)x +y = 0 (D)x -y = 0 Solutiondy
dx= cosx.Therefore, slope of normal =0-1 cos xx=ae öç ÷è ø= -1. Hence the equation of normal isy - 0 = -1(x - 0) orx +y = 0 Therefore, correct answer is (C).
Example 23The point on the curvey2 =x, where the tangent makes an angle of4pwithx-axis is (A)1,2 41 ae öç ÷è ø(B)1,4 21 ae öç ÷è ø(C) (4, 2)(D) (1, 1) Solution1
2dy dx y== tan4p= 1Þy =1 2Þx =41Therefore, correct answer is B.
134 MATHEMATICSFill in the blanks in each of the following Examples 24 to 29.
Example 24The values ofa for whichy =x2 +ax + 25 touches the axis ofx are______. Solution0 2 0dyx adx= Þ + =i.e.x =2a-,
Therefore,2
25 04 2a aaae ö+ - + =ç ÷è øÞa =±10
Hence, the values ofa are ± 10.
Example 25If f(x) =21
4 2 1x x+ +, then its maximum value is _______.
Solution Forf to be maximum, 4x2 + 2x+ 1 should be minimum i.e. 4 x 2 + 2x + 1 = 4 (x +1
4)2 +114ae ö-ç ÷è øgiving the minimum value of 4x2+ 2x + 1 =3
4. Hence maximum value off =4
3. Example 26Letfhave second deriative atc such thatf¢(c) = 0 and f ²(c) > 0, thenc is a point of ______. Solution Local minima.
Example 27Minimum value of f iff (x) = sinx in-,2 2p p é ùê úë ûis _____.
Solution -1
Example 28The maximum value of sinx + cosx is _____. Solution2.
APPLICATION OF DERIVATIVES 135Example 29The rate of change of volume of a sphere with respect to its surface
area, when the radius is 2 cm, is______. Solution1 cm3/cm2
v =32443dvr rdrp Þ = p,s =24dsrdrp Þ =82dv rrdsp Þ == 1 atr = 2. 6.3 EXERCISE
Short Answer (S.A.)
1.A spherical ball of salt is dissolving in water in such a manner that the rate of
decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate. 2.If the area of a circle increases at a uniform rate, then prove that perimetervaries inversely as the radius.
3.A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is10 m/s, how fast is the string being let out; when the kite is 250 m away fromthe boy who is flying the kite? The height of boy is 1.5 m.
4.Two men A and B start with velocitiesv at the same time from the junction of
two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated.. 5.Find an angleq, 0 , which increases twice as fast as its sine. 6.Find the approximate value of (1.999)5.
7.Find the approximate volume of metal in a hollow spherical shell whose internaland external radii are 3 cm and 3.0005 cm, respectively.
8.A man, 2m tall, walks at the rate of213m/s towards a street light which is153m above the ground. At what rate is the tip of his shadow moving? At what
136 MATHEMATICSrate is the length of the shadow changing when he is133m from the base of
the light? 9.A swimming pool is to be drained for cleaning. If L represents the number of
litres of water in the poolt seconds after the pool has been plugged off to drain and L = 200 (10 -t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds? 10.The volume of a cube increases at a constant rate. Prove that the increase inits surface area varies inversely as the length of the side.
11.x andyare the sides of two squares such thaty =x -x2. Find the rate of
change of the area of second square with respect to the area of first square. 12.Find the condition that the curves 2x =y2 and 2xy =k intersect orthogonally.
13.Prove that the curvesxy = 4 andx2 +y2 = 8 touch each other.
14.Find the co-ordinates of the point on the curve+x y= 4 at which tangent
is equally inclined to the axes. 15.Find the angle of intersection of the curvesy = 4 -x2 andy =x2.
16.Prove that the curvesy2 = 4x andx2 +y2 - 6x + 1 = 0 touch each other at the
point (1, 2). 17.Find the equation of the normal lines to the curve 3x2 -y2 = 8 which are
parallel to the linex + 3y = 4. 18.At what points on the curvex2 +y2 - 2x - 4y + 1 = 0, the tangents are parallel
to they-axis? 19.Show that the line+x y
a b= 1, touches the curvey =b .-x aeat the point where the curve intersects the axis ofy. 20.Show thatf (x) = 2x + cot-1x +log()21x x+ -is increasing inR.
APPLICATION OF DERIVATIVES 13721.Show that fora³1,f (x) =3sinx -cosx -2ax + bis decreasing inR.
22.Show thatf (x) = tan-1(sinx + cosx) is an increasing function in0,4
ae öç ÷è ø. 23.At what point, the slope of the curvey = -x3 + 3x2 + 9x - 27 is maximum?
Also find the maximum slope.
24.Prove thatf (x) = sinx +3cosxhas maximum value atx =6
. Long Answer (L.A.)
25.If the sum of the lengths of the hypotenuse and a side of a right angled triangle
is given, show that the area of the triangle is maximum when the angle between them is3 . 26.Find the points of local maxima, local minima and the points of inflection of the
functionf (x) =x5 - 5x4 + 5x3 - 1. Also find the corresponding local maximum and local minimum values. 27.A telephone company in a town has 500 subscribers on its list and collectsfixed charges of Rs 300/- per subscriber per year. The company proposes toincrease the annual subscription and it is believed that for every increase ofRe 1/- one subscriber will discontinue the service. Find what increase willbring maximum profit?
28.If the straight linex cosa +y sina =p touches the curve2 2
2 2 +x y a b= 1, then prove thata2 cos2a +b2 sin2a =p2. 29.An open box with square base is to be made of a given quantity of card board
of areac2. Show that the maximum volume of the box is3 6 3c cubic units.
30.Find the dimensions of the rectangle of perimeter 36 cm which will sweep out
a volume as large as possible, when revolved about one of its sides. Also find the maximum volume. 138 MATHEMATICS31.If the sum of the surface areas of cube and a sphere is constant, what is the
ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum? 32.AB is a diameter of a circle and C is any point on the circle. Show that thearea ofD ABC is maximum, when it is isosceles.
33.A metal box with a square base and vertical sides is to contain 1024 cm3. The
material for the top and bottom costs Rs 5/cm 2 and the material for the sides
costs Rs 2.50/cm 2 . Find the least cost of the box.
34.The sum of the surface areas of a rectangular parallelopiped with sidesx, 2x
and3xand a sphere is given to be constant. Prove that the sum of their volumes is minimum, ifx is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes. Objective Type Questions
Choose the correct answer from the given four options in each of the following questions 35 to 39:
35.The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The
rate at which the area increases, when side is 10 cm is: (A) 10 cm 2/s (B)3cm2/s (C) 103cm2/s (D)10
3cm2/s
36.A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical
wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A)1 10radian/sec (B)1
20radian/sec (C) 20 radian/sec
(D) 10 radian/sec 37.The curvey =1
5xhas at (0, 0)
APPLICATION OF DERIVATIVES 139(A) a vertical tangent (parallel toy-axis) (B) a horizontal tangent (parallel tox-axis) (C) an oblique tangent (D) no tangent 38.The equation of normal to the curve 3x2 -y2 = 8 which is parallel to the line
x + 3y = 8 is (A) 3x -y = 8(B) 3x +y + 8 = 0 (C)x + 3y± 8 = 0(D)x + 3y = 0 39.If the curveay +x2 = 7 andx3 =y, cut orthogonally at (1, 1), then the value of
a is: (A) 1 (B) 0(C) - 6(D) .6 40.Ify =x4 - 10 and ifx changes from 2 to 1.99, what is the change iny
(A) .32 (B) .032 (C) 5.68 (D) 5.968 41.The equation of tangent to the curvey (1 +x2) = 2 -x, where it crossesx-axis
is: (A) x + 5y = 2(B) x - 5y = 2 (C) 5 x -y = 2(D) 5x +y = 2 42.The points at which the tangents to the curvey =x3 - 12x + 18 are parallel to
x-axis are: (A) (2, -2), (-2, -34) (B) (2, 34), (-2, 0) (C) (0, 34), (-2, 0)(D) (2, 2), (-2, 34) 43.The tangent to the curvey =e2x at the point (0, 1) meetsx-axis at:
(A) (0, 1) (B)1- ,02ae öç ÷è ø(C) (2, 0)(D) (0, 2) 44.The slope of tangent to the curvex =t2 + 3t - 8,y= 2t2 - 2t - 5 at the point
(2, -1) is: 140 MATHEMATICS(A)22
7(B)6 7(C)-6
7(D) - 6
45.The two curvesx3 - 3xy2+ 2 = 0 and 3x2y -y3 - 2 = 0 intersect at an angle of
(A)4 (B)3 (C)2 (D)6 46.The interval on which the functionf (x) = 2x3 + 9x2 + 12x - 1 is decreasing is: (A) [-1,¥)(B) [-2, -1](C)(-¥, -2] (D) [-1, 1] 47.Let thef :R®R be defined byf(x) = 2x + cosx, then f:
(A) has a minimum atx =p(B) has a maximum, atx = 0 (C) is a decreasing function(D) is an increasing function 48.y =x (x - 3)2decreases for the values ofx given by :
(A) 1 0 (D) 0 < x <3 249.The functionf (x) = 4 sin3x - 6 sin2x + 12 sinx + 100 is strictly
(A) increasing in3ππ,2ae öç ÷è ø(B) decreasing inπ,π2ae öç ÷è ø(C) decreasing in-,2 2 é ùê úë û(D) decreasing in0,2
é ùê úë û50.Which of the following functions is decreasing on0,2 ae öç ÷è ø(A) sin2x(B) tanx(C) cosx(D) cos 3x 51.The functionf (x) = tanx -x
(A) always increases (B) always decreases (C) never increases(D) sometimes increases and sometimes decreases. APPLICATION OF DERIVATIVES 14152.Ifx is real, the minimum value ofx2 - 8x + 17 is (A) -1(B) 0 (C) 1 (D) 2 53.The smallest value of the polynomialx3 - 18x2 + 96x in [0, 9] is
(A) 126 (B) 0 (C) 135 (D) 160 54.The functionf (x) = 2x3 - 3x2 - 12x + 4, has
(A) two points of local maximum (B) two points of local minimum (C) one maxima and one minima (D) no maxima or minima 55.The maximum value of sinx . cosx is
(A)1 4(B)1 2(C)2(D)2 256.Atx =5
6,f (x) = 2 sin3x + 3 cos3x is:
(A) maximum(B) minimum (C) zero(D) neither maximum nor minimum. 57.Maximum slope of the curvey = -x3 + 3x2 + 9x - 27 is:
(A) 0 (B) 12 (C) 16 (D) 32 58.f (x) =xxhas a stationary point at
(A)x =e(B)x=1 e(C)x = 1 (D)x =e59.The maximum value of1ae öç ÷è øx xis: (A)e(B)ee(C)1 ee(D)11ae öç ÷è øe e 142 MATHEMATICSFill in the blanks in each of the following Exercises 60 to 64:
60.The curvesy = 4x2 + 2x - 8 and y =x3 -x + 13 touch each other at the
point_____. 61.The equation of normal to the curvey = tanx at (0, 0) is ________.
62.The values ofa for which the function f (x) = sinx -ax +b increases onRare
______. 63.The functionf (x) =2
42 -1x
x,x > 0, decreases in the interval _______. 64.The least value of the functionf (x) =ax +b
x (a > 0,b > 0,x > 0) is ______.
Hencef is strictly decreasing on-,3 3p p
ae öç ÷è ø. Example 5Determine for which values ofx, the functiony =x4 -343xis increasing
and for which values, it is decreasing.Solutiony =x4 -34
3xÞdy
dx= 4x3 - 4x2= 4x2 (x- 1)122 MATHEMATICSNow,dy
dx= 0Þx = 0,x = 1. Sincef¢ (x) < 0x" Î(-¥, 0)È(0, 1) andf is continuous in (-¥, 0] and [0, 1]. Thereforef is decreasing in (-¥, 1] andf is increasing in [1,¥). Note: Heref is strictly decreasing in (-¥, 0)È(0, 1) and is strictly increasing in (1,¥). Example 6Show that the functionf (x) = 4x3 - 18x2 + 27x - 7 has neither maxima nor minima.Solutionf (x) = 4x3 - 18x2 + 27x - 7
f¢ (x) = 12x2 - 36x + 27 = 3 (4x2 - 12x + 9) = 3 (2x - 3)2 f¢ (x) = 0Þx =32(critical point)
Sincef¢ (x) > 0 for allx3
2< and for allx >3
2Hencex =3
2is a point of inflexion i.e., neither a point of maxima nor a point of minima.
x =32is the only critical point, andf has neither maxima nor minima.
Example 7Using differentials, find the approximate value of0.082Solution Letf (x) =xUsingf (x +Dx)f (x) +Dx . f¢(x), takingx = .09 andDx= - 0.008,
we getf(0.09 - 0.008) = f(0.09) + (- 0.008) f¢ (0.09)Þ0.082 =0.09 - 0.008 .1
2 0.09ae öç ÷è ø = 0.3 -0.008
0.6= 0.3 - 0.0133 = 0.2867.
APPLICATION OF DERIVATIVES 123Example 8 Find the condition for the curves2 2 2 2 -x y a b= 1;xy =c2 to intersect orthogonally. Solution Let the curves intersect at (x1,y1). Therefore,2 2 2 2 -x y a b= 1Þ2 22 2-x y dy dx a b= 0Þ 22dy b x
dx a y=Þ slope of tangent at the point of intersection (m1) =2 1 2 1b x a yAgain xy =c2Þdyx ydx+= 0Þ-dy y dx x=Þ m2 =1 1y x- .For orthoganality,m1 ×m2= - 1Þ 2
2b a= 1 ora2 -b2 = 0. Example 9Find all the points of local maxima and local minima of the function f (x) =4 3 23 45- - 8 - 1054 2x x x+.Solutionf¢ (x) = -3x3- 24x2 - 45x
=- 3x (x2 + 8x + 15) = - 3x (x + 5) (x + 3) f¢ (x) = 0Þx = -5,x = -3, x = 0 f²(x) = -9x2 - 48x - 45 = -3 (3x2 + 16x + 15) f²(0) = - 45 < 0. Therefore,x = 0 is point of local maxima f²(-3) = 18 > 0. Therefore,x = -3 is point of local minima f²(-5) = -30 < 0. Thereforex = -5 is point of local maxima.124 MATHEMATICSExample 10 Show that the local maximum value of1xx+is less than local minimum
value.SolutionLety=1xx+Þdy
dx= 1 -21 x,dy dx= 0Þx2= 1Þx = ± 1.2 2d y dx= +32 x, therefore2 2d y dx(atx = 1) > 0 and2 2d y dx(atx = -1) < 0. Hence local maximum value ofy is atx = -1 and the local maximum value = - 2. Local minimum value ofy is atx = 1 and local minimum value = 2. Therefore, local maximum value (-2) is less than local minimum value 2.Long Answer Type (L.A.)
Example 11Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is6p.SolutionGiven thatdv
dt = 1 cm3/s, wherev is the volume of water in the conical vessel.From the Fig.6.2,l = 4cm,h =l cos6p =3
2l andr =l sin6p=2l.
Therefore,v =1
3pr2h =233 3
3 4 2 24ll lpp=.
APPLICATION OF DERIVATIVES 12523
8dv dlldt dtp=Therefore, 1 =316.8dl
dtpÞ1 2 3dl dt=pcm/s. Therefore, the rate of decrease of slant height =12 3pcm/s.
Example 12 Find the equation of all the tangents to the curvey = cos (x +y), -2 p £x£ 2p, that are parallel to the line x + 2y = 0.SolutionGiven that y = cos (x +y)Þdy
dx= - sin (x +y)1dy dxé ù+ê úë û...(i) ordy dx = -( ) ( )sin1 sinx y
x y+ + +Since tangent is parallel tox + 2y = 0, therefore slope of tangent =1-2Therefore,( ) ( )sin-1 sinx y x y+ + +=1-2Þ sin (x +y) = 1.... (ii) Since cos (x +y) =y and sin (x +y)=1Þ cos2 (x +y)+ sin2(x +y) =y2 + 1Þ1 =y2 + 1 ory = 0.
Therefore, cosx = 0.
Therefore, x = (2n + 1)2p, n= 0, ± 1, ± 2...126 MATHEMATICSThus,x =3,2 2p p± ±, butx =2p,x =-3
2psatisfy equation (ii)
Hence, the points are, 02p
ae öç ÷è ø,-3,02p ae öç ÷è ø.Therefore, equation of tangent at, 02p
ae öç ÷è ø isy =1-2-2xpae öç ÷è ø or 2x + 4y -p = 0, and equation of tangent at-3,02p ae öç ÷è øisy =1-232xpae ö+ç ÷è øor 2x + 4y + 3p = 0.
Example 13Find the angle of intersection of the curvesy2 = 4ax andx2 = 4by. SolutionGiven thaty2 = 4ax...(i) andx2 = 4by...(ii). Solving (i) and (ii), we get22 4x bae öç ÷è ø= 4axÞ x4 = 64ab2x orx(x3 - 64ab2) = 0 Þ x = 0,1 23 34x a b=Therefore, the points of intersection are (0, 0) and1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø.
Again,y2 = 4axÞ4 2
2dy a a
dx y y= =andx2 = 4byÞ 24 2dy x x
dx b b= =Therefore, at (0, 0) the tangent to the curvey2 = 4ax is parallel toy-axis and tangent to the curvex2= 4by is parallel tox-axis.Þ Angle between curves =2pAt1 2 2 1
3 3 3 34 ,4a b a baeöç÷ç÷èø,m1(slope of the tangent to the curve (i)) =1
32abae öç ÷è ø=1 3 2 1
3 32 1
24a aba bae ö =ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21
3 33422a b a
b bae ö=ç ÷è ø APPLICATION OF DERIVATIVES 127Therefore, tanq =2 1 1 2- 1m m m m+ =11 331 1 3 31 2 -2 1
1 22a a
b b a a b bae ö ae öç ÷ ç ÷è ø è ø
ae ö ae ö +ç ÷ ç ÷è ø è ø =1 1 3 3 2 23 33 .
2a b a b ae ö+ç ÷ç ÷è øHence,q = tan-11 1
3 3 2 23 33 .
2a ba baeöç÷ç÷ç÷ae öç÷+ç ÷ç÷ç ÷è øèøExample 14Show that the equation of normal at any point on the curve
x = 3cosq - cos3q,y= 3sinq - sin3q is 4 (ycos3q -x sin3q) = 3 sin 4q.SolutionWe havex = 3cosq - cos3q
Therefore,dx
dq = -3sinq + 3cos2q sinq = - 3sinq (1 - cos2q) = -3sin3q .dy dq = 3cosq - 3sin2q cosq = 3cosq (1 - sin2q) = 3cos3q33cos-sindy
dxq=q. Therefore, slope of normal =3 3sin cosq+qHence the equation of normal is y - (3sinq - sin3q) =3 3sin cosq q[x - (3cosq - cos3q)] Þ ycos3q - 3sinq cos3q + sin3q cos3q =xsin3q - 3sin3q cosq + sin3q cos3qÞ ycos3q -xsin3q = 3sinq cosq (cos2q - sin2q)
128 MATHEMATICS =3
2sin2q . cos2q
=34sin4q
or 4 (ycos3q -xsin3q) = 3 sin4q.Example 15Find the maximum and minimum values of
f (x) = secx + log cos2x,0Therefore, f
(x) = secx tanx - 2 tanx = tanx (secx -2) f (x) = 0Þ tanx = 0 or secx = 2 or cosx =12Therefore, possible values ofx arex = 0, orx=p and
x =3porx =53pAgain,f¢¢ (x) = sec2x (secx -2) + tanx (secx tanx)
= sec3x + secx tan2x - 2sec2x
=secx (sec2x + tan2x - 2secx). We note thatf¢¢(0) = 1 (1 + 0 - 2) = -1 < 0. Therefore,x = 0 is a point of maxima.f¢¢(p) = -1 (1 + 0 + 2) = -3 < 0. Therefore,x =p is a point of maxima.f¢¢3p
ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =3p is a point of minima.f¢¢5 3p ae öç ÷è ø = 2 (4 + 3 - 4) = 6 > 0. Therefore,x =53p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1Maximum Value ofy atx =p is-1 + 0 = -1
Minimum Value ofy atx =3pis 2 + 2 log1
2 = 2 (1 - log2)
Minimum Value ofy atx =5
3pis 2 + 2 log1
2 = 2 (1 - log2)
Example 16 Find the area of greatest rectangle that can be inscribed in an ellipse2 2 2 2 1x y a b+ =. Solution Let ABCD be the rectangle of maximum area with sides AB = 2xandBC =2y,where C (x,y) is a point on the ellipse2 2
2 2 1x y a b+ = as shown in the Fig.6.3. The area A of the rectangle is 4xyi.e. A = 4xywhich gives A2 = 16x2y2 =s (say)Therefore,s = 16x222
21- .xbaae öç ÷è ø =2
216ba (a2x2 -x4) Þ2
216ds b
dx a=. [2a2x - 4x3].Again,ds
dx = 0Þ x =and2 2a by=Now,2 22 216d s b
dx a=[2a2 - 12x2]At2 222 22
2 221616,[2 6 ] ( 4 ) 02a d s bbxa aadx aa= = - = - <
130 MATHEMATICSThus atx =2a, y =2b, s is maximum and hence the area A is maximum.
Maximum area = 4.x.y =4 .2a.2b= 2ab sq units.
Example 17Find the difference between the greatest and least values of the functionf (x) = sin2x -x, on- ,2 2p pé ùê úë û.
Solutionf (x) = sin2x -x
Þf¢(x) = 2 cos2x - 1
Therefore,f¢(x) = 0Þ cos2x =1
2Þ 2xisor3 3- p pÞx =- or6 6p p-2fpae öç ÷è ø =sin (- p) +2 2p p=-6fpae öç ÷è ø =2sin -6 6p p
ae ö+ç ÷è ø=3-2 6p+6fpae öç ÷è ø =2sin -6 6p pae öç ÷è ø=3-2 6p2fpae öç ÷è ø =( )sin -2pp=-2pClearly,2pis the greatest value and-2pis the least.
Therefore, difference =2p +2p=p
APPLICATION OF DERIVATIVES 131Example 18An isosceles triangle of vertical angle 2q is inscribed in a circle of radius
a . Show that the area of triangle is maximum whenq =6p. SolutionLet ABC be an isosceles triangle inscribed in the circle with radiusa such that AB = AC.AD = AO + OD =
a +a cos2qand BC = 2BD = 2a sin2q (see fig. 16.4)Therefore, area of the triangle ABC i.e.D =1
2BC . AD
=122asin2q . (a +a cos2q)
= a2sin2q (1 + cos2q)Þ D= a2sin2q +1
2a2sin4q
Therefore,d
dD q= 2a2cos2q + 2a2cos4q = 2a2(cos2q + cos4q)d dD q= 0Þ cos2q = -cos4q = cos (p - 4q)Therefore, 2q =p - 4q Þ q =6p2
2d dD q = 2a2 (-2sin2q - 4sin4q) < 0 (atq =6p).Therefore, Area of triangle is maximum whenq =6p.
132 MATHEMATICSObjective Type Questions
Choose the correct answer from the given four options in each of the following Examples19 to 23.
Example 19The abscissa of the point on the curve 3y = 6x - 5x3, the normal at which passes through origin is: (A) 1(B)13(C) 2(D)1
2SolutionLet (x1,y1) be the point on the given curve 3y = 6x - 5x3 at which the normal
passes through the origin. Then we have1 12 1 ( , )2-5 x ydy xdxae ö=ç ÷è ø. Again the equation of the normal at (x1,y1) passing through the origin gives2112 11- -32-56-5xxyx= =. Sincex1 = 1 satisfies the equation, therefore, Correct answer is (A). Example 20The two curvesx3 - 3xy2 + 2 = 0 and 3x2y -y3 = 2 (A) touch each other(B) cut at right angle(C) cut at an angle3p(D) cut at an angle4pSolutionFrom first equation of the curve, we have 3x2 - 3y2 - 6xydy
dx = 0Þdy
dx =2 2- 2x y xy= (m1) say and second equation of the curve gives 6 xy + 3x2dy dx- 3y2dy dx= 0Þdy dx=2 2-2 -xy x y= (m2) say Since m1 . m2= -1. Therefore, correct answer is (B).APPLICATION OF DERIVATIVES 133Example 21The tangent to the curve given byx =et . cost,y =et . sint att=4pmakes
withx-axis an angle: (A) 0(B)4p(C)3p(D)2pSolutiondx dt=- et . sint+etcost,dy dt=etcost+etsintTherefore,4tdy
dx p=ae öç ÷è ø=cos sin cos - sint t t t+=20 and hence the correct answer is (D).
Example 22The equation of the normal to the curvey = sinx at (0, 0) is: (A)x = 0(B)y = 0(C)x +y = 0 (D)x -y = 0Solutiondy
dx= cosx.Therefore, slope of normal =0-1 cos xx=ae öç ÷è ø= -1. Hence the equation of normal isy - 0 = -1(x - 0) orx +y = 0Therefore, correct answer is (C).
Example 23The point on the curvey2 =x, where the tangent makes an angle of4pwithx-axis is (A)1,2 41 ae öç ÷è ø(B)1,4 21 ae öç ÷è ø(C) (4, 2)(D) (1, 1)Solution1
2dy dx y== tan4p= 1Þy =12Þx =41Therefore, correct answer is B.
134 MATHEMATICSFill in the blanks in each of the following Examples 24 to 29.
Example 24The values ofa for whichy =x2 +ax + 25 touches the axis ofx are______.Solution0 2 0dyx adx= Þ + =i.e.x =2a-,
Therefore,2
25 04 2a aaae ö+ - + =ç ÷è øÞa =±10
Hence, the values ofa are ± 10.
Example 25If f(x) =21
4 2 1x x+ +, then its maximum value is _______.
Solution Forf to be maximum, 4x2 + 2x+ 1 should be minimum i.e. 4 x2 + 2x + 1 = 4 (x +1
4)2 +114ae ö-ç ÷è øgiving the minimum value of 4x2+ 2x + 1 =3
4.Hence maximum value off =4
3. Example 26Letfhave second deriative atc such thatf¢(c) = 0 and f ²(c) > 0, thenc is a point of ______.Solution Local minima.
Example 27Minimum value of f iff (x) = sinx in-,2 2p pé ùê úë ûis _____.
Solution -1
Example 28The maximum value of sinx + cosx is _____.Solution2.
APPLICATION OF DERIVATIVES 135Example 29The rate of change of volume of a sphere with respect to its surface
area, when the radius is 2 cm, is______.Solution1 cm3/cm2
v =32443dvr rdrp Þ = p,s =24dsrdrp Þ =82dv rrdsp Þ == 1 atr = 2.6.3 EXERCISE
Short Answer (S.A.)
1.A spherical ball of salt is dissolving in water in such a manner that the rate of
decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.2.If the area of a circle increases at a uniform rate, then prove that perimetervaries inversely as the radius.
3.A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is10 m/s, how fast is the string being let out; when the kite is 250 m away fromthe boy who is flying the kite? The height of boy is 1.5 m.
4.Two men A and B start with velocitiesv at the same time from the junction of
two roads inclined at 45° to each other. If they travel by different roads, find the rate at which they are being seperated..5.Find an angleq, 0 , which increases twice as fast as its sine. 6.Find the approximate value of (1.999)5.
7.Find the approximate volume of metal in a hollow spherical shell whose internaland external radii are 3 cm and 3.0005 cm, respectively.
8.A man, 2m tall, walks at the rate of213m/s towards a street light which is153m above the ground. At what rate is the tip of his shadow moving? At what
136 MATHEMATICSrate is the length of the shadow changing when he is133m from the base of
the light? 9.A swimming pool is to be drained for cleaning. If L represents the number of
litres of water in the poolt seconds after the pool has been plugged off to drain and L = 200 (10 -t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds? 10.The volume of a cube increases at a constant rate. Prove that the increase inits surface area varies inversely as the length of the side.
11.x andyare the sides of two squares such thaty =x -x2. Find the rate of
change of the area of second square with respect to the area of first square. 12.Find the condition that the curves 2x =y2 and 2xy =k intersect orthogonally.
13.Prove that the curvesxy = 4 andx2 +y2 = 8 touch each other.
14.Find the co-ordinates of the point on the curve+x y= 4 at which tangent
is equally inclined to the axes. 15.Find the angle of intersection of the curvesy = 4 -x2 andy =x2.
16.Prove that the curvesy2 = 4x andx2 +y2 - 6x + 1 = 0 touch each other at the
point (1, 2). 17.Find the equation of the normal lines to the curve 3x2 -y2 = 8 which are
parallel to the linex + 3y = 4. 18.At what points on the curvex2 +y2 - 2x - 4y + 1 = 0, the tangents are parallel
to they-axis? 19.Show that the line+x y
a b= 1, touches the curvey =b .-x aeat the point where the curve intersects the axis ofy. 20.Show thatf (x) = 2x + cot-1x +log()21x x+ -is increasing inR.
APPLICATION OF DERIVATIVES 13721.Show that fora³1,f (x) =3sinx -cosx -2ax + bis decreasing inR.
22.Show thatf (x) = tan-1(sinx + cosx) is an increasing function in0,4
ae öç ÷è ø. 23.At what point, the slope of the curvey = -x3 + 3x2 + 9x - 27 is maximum?
Also find the maximum slope.
24.Prove thatf (x) = sinx +3cosxhas maximum value atx =6
. Long Answer (L.A.)
25.If the sum of the lengths of the hypotenuse and a side of a right angled triangle
is given, show that the area of the triangle is maximum when the angle between them is3 . 26.Find the points of local maxima, local minima and the points of inflection of the
functionf (x) =x5 - 5x4 + 5x3 - 1. Also find the corresponding local maximum and local minimum values. 27.A telephone company in a town has 500 subscribers on its list and collectsfixed charges of Rs 300/- per subscriber per year. The company proposes toincrease the annual subscription and it is believed that for every increase ofRe 1/- one subscriber will discontinue the service. Find what increase willbring maximum profit?
28.If the straight linex cosa +y sina =p touches the curve2 2
2 2 +x y a b= 1, then prove thata2 cos2a +b2 sin2a =p2. 29.An open box with square base is to be made of a given quantity of card board
of areac2. Show that the maximum volume of the box is3 6 3c cubic units.
30.Find the dimensions of the rectangle of perimeter 36 cm which will sweep out
a volume as large as possible, when revolved about one of its sides. Also find the maximum volume. 138 MATHEMATICS31.If the sum of the surface areas of cube and a sphere is constant, what is the
ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum? 32.AB is a diameter of a circle and C is any point on the circle. Show that thearea ofD ABC is maximum, when it is isosceles.
33.A metal box with a square base and vertical sides is to contain 1024 cm3. The
material for the top and bottom costs Rs 5/cm 2 and the material for the sides
costs Rs 2.50/cm 2 . Find the least cost of the box.
34.The sum of the surface areas of a rectangular parallelopiped with sidesx, 2x
and3xand a sphere is given to be constant. Prove that the sum of their volumes is minimum, ifx is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes. Objective Type Questions
Choose the correct answer from the given four options in each of the following questions 35 to 39:
35.The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The
rate at which the area increases, when side is 10 cm is: (A) 10 cm 2/s (B)3cm2/s (C) 103cm2/s (D)10
3cm2/s
36.A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical
wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A)1 10radian/sec (B)1
20radian/sec (C) 20 radian/sec
(D) 10 radian/sec 37.The curvey =1
5xhas at (0, 0)
APPLICATION OF DERIVATIVES 139(A) a vertical tangent (parallel toy-axis) (B) a horizontal tangent (parallel tox-axis) (C) an oblique tangent (D) no tangent 38.The equation of normal to the curve 3x2 -y2 = 8 which is parallel to the line
x + 3y = 8 is (A) 3x -y = 8(B) 3x +y + 8 = 0 (C)x + 3y± 8 = 0(D)x + 3y = 0 39.If the curveay +x2 = 7 andx3 =y, cut orthogonally at (1, 1), then the value of
a is: (A) 1 (B) 0(C) - 6(D) .6 40.Ify =x4 - 10 and ifx changes from 2 to 1.99, what is the change iny
(A) .32 (B) .032 (C) 5.68 (D) 5.968 41.The equation of tangent to the curvey (1 +x2) = 2 -x, where it crossesx-axis
is: (A) x + 5y = 2(B) x - 5y = 2 (C) 5 x -y = 2(D) 5x +y = 2 42.The points at which the tangents to the curvey =x3 - 12x + 18 are parallel to
x-axis are: (A) (2, -2), (-2, -34) (B) (2, 34), (-2, 0) (C) (0, 34), (-2, 0)(D) (2, 2), (-2, 34) 43.The tangent to the curvey =e2x at the point (0, 1) meetsx-axis at:
(A) (0, 1) (B)1- ,02ae öç ÷è ø(C) (2, 0)(D) (0, 2) 44.The slope of tangent to the curvex =t2 + 3t - 8,y= 2t2 - 2t - 5 at the point
(2, -1) is: 140 MATHEMATICS(A)22
7(B)6 7(C)-6
7(D) - 6
45.The two curvesx3 - 3xy2+ 2 = 0 and 3x2y -y3 - 2 = 0 intersect at an angle of
(A)4 (B)3 (C)2 (D)6 46.The interval on which the functionf (x) = 2x3 + 9x2 + 12x - 1 is decreasing is: (A) [-1,¥)(B) [-2, -1](C)(-¥, -2] (D) [-1, 1] 47.Let thef :R®R be defined byf(x) = 2x + cosx, then f:
(A) has a minimum atx =p(B) has a maximum, atx = 0 (C) is a decreasing function(D) is an increasing function 48.y =x (x - 3)2decreases for the values ofx given by :
(A) 1 0 (D) 0 < x <3 249.The functionf (x) = 4 sin3x - 6 sin2x + 12 sinx + 100 is strictly
(A) increasing in3ππ,2ae öç ÷è ø(B) decreasing inπ,π2ae öç ÷è ø(C) decreasing in-,2 2 é ùê úë û(D) decreasing in0,2
é ùê úë û50.Which of the following functions is decreasing on0,2 ae öç ÷è ø(A) sin2x(B) tanx(C) cosx(D) cos 3x 51.The functionf (x) = tanx -x
(A) always increases (B) always decreases (C) never increases(D) sometimes increases and sometimes decreases. APPLICATION OF DERIVATIVES 14152.Ifx is real, the minimum value ofx2 - 8x + 17 is (A) -1(B) 0 (C) 1 (D) 2 53.The smallest value of the polynomialx3 - 18x2 + 96x in [0, 9] is
(A) 126 (B) 0 (C) 135 (D) 160 54.The functionf (x) = 2x3 - 3x2 - 12x + 4, has
(A) two points of local maximum (B) two points of local minimum (C) one maxima and one minima (D) no maxima or minima 55.The maximum value of sinx . cosx is
(A)1 4(B)1 2(C)2(D)2 256.Atx =5
6,f (x) = 2 sin3x + 3 cos3x is:
(A) maximum(B) minimum (C) zero(D) neither maximum nor minimum. 57.Maximum slope of the curvey = -x3 + 3x2 + 9x - 27 is:
(A) 0 (B) 12 (C) 16 (D) 32 58.f (x) =xxhas a stationary point at
(A)x =e(B)x=1 e(C)x = 1 (D)x =e59.The maximum value of1ae öç ÷è øx xis: (A)e(B)ee(C)1 ee(D)11ae öç ÷è øe e 142 MATHEMATICSFill in the blanks in each of the following Exercises 60 to 64:
60.The curvesy = 4x2 + 2x - 8 and y =x3 -x + 13 touch each other at the
point_____. 61.The equation of normal to the curvey = tanx at (0, 0) is ________.
62.The values ofa for which the function f (x) = sinx -ax +b increases onRare
______. 63.The functionf (x) =2
42 -1x
x,x > 0, decreases in the interval _______. 64.The least value of the functionf (x) =ax +b
x (a > 0,b > 0,x > 0) is ______.
6.Find the approximate value of (1.999)5.
7.Find the approximate volume of metal in a hollow spherical shell whose internaland external radii are 3 cm and 3.0005 cm, respectively.
8.A man, 2m tall, walks at the rate of213m/s towards a street light which is153m above the ground. At what rate is the tip of his shadow moving? At what
136 MATHEMATICSrate is the length of the shadow changing when he is133m from the base of
the light?9.A swimming pool is to be drained for cleaning. If L represents the number of
litres of water in the poolt seconds after the pool has been plugged off to drain and L = 200 (10 -t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?10.The volume of a cube increases at a constant rate. Prove that the increase inits surface area varies inversely as the length of the side.
11.x andyare the sides of two squares such thaty =x -x2. Find the rate of
change of the area of second square with respect to the area of first square.12.Find the condition that the curves 2x =y2 and 2xy =k intersect orthogonally.
13.Prove that the curvesxy = 4 andx2 +y2 = 8 touch each other.
14.Find the co-ordinates of the point on the curve+x y= 4 at which tangent
is equally inclined to the axes.15.Find the angle of intersection of the curvesy = 4 -x2 andy =x2.
16.Prove that the curvesy2 = 4x andx2 +y2 - 6x + 1 = 0 touch each other at the
point (1, 2).17.Find the equation of the normal lines to the curve 3x2 -y2 = 8 which are
parallel to the linex + 3y = 4.18.At what points on the curvex2 +y2 - 2x - 4y + 1 = 0, the tangents are parallel
to they-axis?19.Show that the line+x y
a b= 1, touches the curvey =b .-x aeat the point where the curve intersects the axis ofy.20.Show thatf (x) = 2x + cot-1x +log()21x x+ -is increasing inR.
APPLICATION OF DERIVATIVES 13721.Show that fora³1,f (x) =3sinx -cosx -2ax + bis decreasing inR.
22.Show thatf (x) = tan-1(sinx + cosx) is an increasing function in0,4
ae öç ÷è ø.23.At what point, the slope of the curvey = -x3 + 3x2 + 9x - 27 is maximum?
Also find the maximum slope.
24.Prove thatf (x) = sinx +3cosxhas maximum value atx =6
.Long Answer (L.A.)
25.If the sum of the lengths of the hypotenuse and a side of a right angled triangle
is given, show that the area of the triangle is maximum when the angle between them is3 .26.Find the points of local maxima, local minima and the points of inflection of the
functionf (x) =x5 - 5x4 + 5x3 - 1. Also find the corresponding local maximum and local minimum values.27.A telephone company in a town has 500 subscribers on its list and collectsfixed charges of Rs 300/- per subscriber per year. The company proposes toincrease the annual subscription and it is believed that for every increase ofRe 1/- one subscriber will discontinue the service. Find what increase willbring maximum profit?
28.If the straight linex cosa +y sina =p touches the curve2 2
2 2 +x y a b= 1, then prove thata2 cos2a +b2 sin2a =p2.29.An open box with square base is to be made of a given quantity of card board
of areac2. Show that the maximum volume of the box is36 3c cubic units.
30.Find the dimensions of the rectangle of perimeter 36 cm which will sweep out
a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.138 MATHEMATICS31.If the sum of the surface areas of cube and a sphere is constant, what is the
ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?32.AB is a diameter of a circle and C is any point on the circle. Show that thearea ofD ABC is maximum, when it is isosceles.
33.A metal box with a square base and vertical sides is to contain 1024 cm3. The
material for the top and bottom costs Rs 5/cm2 and the material for the sides
costs Rs 2.50/cm2 . Find the least cost of the box.
34.The sum of the surface areas of a rectangular parallelopiped with sidesx, 2x
and3xand a sphere is given to be constant. Prove that the sum of their volumes is minimum, ifx is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.Objective Type Questions
Choose the correct answer from the given four options in each of the following questions35 to 39:
35.The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The
rate at which the area increases, when side is 10 cm is: (A) 10 cm2/s (B)3cm2/s (C) 103cm2/s (D)10
3cm2/s
36.A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical
wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A)110radian/sec (B)1
20radian/sec (C) 20 radian/sec
(D) 10 radian/sec37.The curvey =1
5xhas at (0, 0)
APPLICATION OF DERIVATIVES 139(A) a vertical tangent (parallel toy-axis) (B) a horizontal tangent (parallel tox-axis) (C) an oblique tangent (D) no tangent38.The equation of normal to the curve 3x2 -y2 = 8 which is parallel to the line
x + 3y = 8 is (A) 3x -y = 8(B) 3x +y + 8 = 0 (C)x + 3y± 8 = 0(D)x + 3y = 039.If the curveay +x2 = 7 andx3 =y, cut orthogonally at (1, 1), then the value of
a is: (A) 1 (B) 0(C) - 6(D) .640.Ify =x4 - 10 and ifx changes from 2 to 1.99, what is the change iny
(A) .32 (B) .032 (C) 5.68 (D) 5.96841.The equation of tangent to the curvey (1 +x2) = 2 -x, where it crossesx-axis
is: (A) x + 5y = 2(B) x - 5y = 2 (C) 5 x -y = 2(D) 5x +y = 242.The points at which the tangents to the curvey =x3 - 12x + 18 are parallel to
x-axis are: (A) (2, -2), (-2, -34) (B) (2, 34), (-2, 0) (C) (0, 34), (-2, 0)(D) (2, 2), (-2, 34)43.The tangent to the curvey =e2x at the point (0, 1) meetsx-axis at:
(A) (0, 1) (B)1- ,02ae öç ÷è ø(C) (2, 0)(D) (0, 2)44.The slope of tangent to the curvex =t2 + 3t - 8,y= 2t2 - 2t - 5 at the point
(2, -1) is:140 MATHEMATICS(A)22
7(B)67(C)-6
7(D) - 6
45.The two curvesx3 - 3xy2+ 2 = 0 and 3x2y -y3 - 2 = 0 intersect at an angle of
(A)4 (B)3 (C)2 (D)6 46.The interval on which the functionf (x) = 2x3 + 9x2 + 12x - 1 is decreasing is: (A) [-1,¥)(B) [-2, -1](C)(-¥, -2] (D) [-1, 1]47.Let thef :R®R be defined byf(x) = 2x + cosx, then f:
(A) has a minimum atx =p(B) has a maximum, atx = 0 (C) is a decreasing function(D) is an increasing function48.y =x (x - 3)2decreases for the values ofx given by :
(A) 1249.The functionf (x) = 4 sin3x - 6 sin2x + 12 sinx + 100 is strictly
(A) increasing in3ππ,2ae öç ÷è ø(B) decreasing inπ,π2ae öç ÷è ø(C) decreasing in-,2 2 é ùê úë û(D) decreasing in0,2
é ùê úë û50.Which of the following functions is decreasing on0,2 ae öç ÷è ø(A) sin2x(B) tanx(C) cosx(D) cos 3x