CHAPTER 3 APPLICATIONS OF DERIVATIVES
Every textbook gives problems where algebra gets the answer - so you see how the whole method works. We start with those. A maximum can also occur at a
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CHAPTER 3 APPLICATIONS OF DERIVATIVES
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APPLICATION OF DERIVATIVES
Therefore local maximum value (–2) is less than local minimum value 2. Long Answer Type (L.A.). Example 11 Water is dripping out at a steady rate of 1 cu cm/
Applications of the Derivative
Such a problem differs in two ways from the local maximum and minimum problems we encountered when graphing functions: We are interested only in the function
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Functions Derivatives and their Applications Annuity
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 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- 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
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
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
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
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
Therefore, equation of tangent at, 02p
equation of tangent at-3,02p 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 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
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
32a
3 2 1 3 32 1
24a a
=ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21 3 33422a b a
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 +ç ÷ ç ÷è ø è ø =1 1 3 3 2 2 3 33 .
2a b a b Hence,q = tan-11 1
3 3 2 2 3 33 .
2a b 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
3p 3p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1quotesdbs_dbs21.pdfusesText_27
Therefore, 1 < sec
2x < 4Þ -3 < (sec2x - 4) < 0
Thus for-
3p Hencef is strictly decreasing on-,3 3p p
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
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
Therefore, equation of tangent at, 02p
equation of tangent at-3,02p 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 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
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
32a
3 2 1 3 32 1
24a a
=ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21 3 33422a b a
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 +ç ÷ ç ÷è ø è ø =1 1 3 3 2 2 3 33 .
2a b a b Hence,q = tan-11 1
3 3 2 2 3 33 .
2a b 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
3p 3p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1quotesdbs_dbs21.pdfusesText_27
Hencef is strictly decreasing on-,3 3p p
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
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
Therefore, equation of tangent at, 02p
equation of tangent at-3,02p 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 orx(x3 - 64ab2) = 0 Þ x = 0,1 23 34x a b=Therefore, the points of intersection are (0, 0) and1 2 2 1
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
32a3 2 1
3 32 1
24a a=ç ÷è ø,m2(slope of the tangent to the curve (ii)) =1 21
3 33422a b a
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 +ç ÷ ç ÷è ø è ø =1 1 3 3 2 23 33 .
2a b a bHence,q = tan-11 1
3 3 2 23 33 .
2a b 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
3p3p is a point of minima.
APPLICATION OF DERIVATIVES 129Maximum Value ofy atx = 0 is 1 + 0 = 1quotesdbs_dbs21.pdfusesText_27[PDF] application of fermentation
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