Rd Sharma XII Vol 1 2019 Solutions for Class 12 Commerce Math Chapter 13 Derivative As A Rate Measurer are provided here with simple step-by-step explanations. These solutions for Derivative As A Rate Measurer are extremely popular among Class 12 Commerce students for Math Derivative As A Rate Measurer Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Commerce Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

Page No 13.19:

Question 1:

The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is 8 cm long?

Answer:

Given: A=x2 and dxdt= 4 cm/min  Let x be the side of the square and A be its area at any time t. Then,A=x2 dAdt=2xdxdtdAdt=2×8×4            x=8 cm and dxdt=4 cm/mindAdt=64 cm2/min

Page No 13.19:

Question 2:

An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

Let be the side and V be the volume of the cube at any time t. Then, V=x3dVdt=3x2dxdtdVdt=3×102×3      ∵ x=10 cm and dxdt=3 cm/secdVdt=900 cm3/sec

Page No 13.19:

Question 3:

The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of the perimeter of the square.

Answer:

Let x be the side and P be the perimeter of the square at any time t. Then,  P=4xdPdt=4dxdtdPdt=4×0.2                ∵ dxdt= 0.2 cm/secdPdt=0.8 cm/sec

Page No 13.19:

Question 4:

The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?

Answer:

Let r be the radius and C be the circumference of the circle at any time t. Then, C=2πrdCdt=2πdrdtdCdt=2π×0.7               ∵ drdt=0.7 cm/secdCdt=1.4π cm/sec

Page No 13.19:

Question 5:

The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.

Answer:

Let r be the radius and S be the surface area of the spherical ball at any time t. ThenS=4πr2dSdt=8πrdrdtdSdt=8π×7×0.2             r=7 cm and drdt=0.2 cm/secdSdt=11.2π cm2/sec

Page No 13.19:

Question 6:

A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.

Answer:

Let r be the radius and V be the volume of the spherical balloon at any time t. Then, V=43πr3⇒ dVdt=4πr2drdtdrdt=14πr2dVdtdrdt=9004π152                       r=15 cm and dVdt=900 cm3/secdrdt=900900πdrdt=1πcm/sec

Page No 13.19:

Question 7:

The radius of an air bubble is increasing at the rate of 0.5 cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

Let r be the radius and V be the volume of the air bubble at any time t. Then,V=43πr3⇒ dVdt=4πr2drdtdVdt=4π12× 0.5            r=1 cm and drdt=0.5 cm/secdVdt=2π cm3/sec

Page No 13.19:

Question 8:

A man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.


Since triangles ABE and CDE are similar,ABCD=AECE62=x+yy3y=x+yx=2ydxdt=2dydtdydt=12dxdtdydt=125 dxdt=5dydt=52 km/hr

Page No 13.19:

Question 9:

A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Answer:

Let r be the radius and A be the area of the circle at any time t. Then, Ar2dAdt=2πrdrdtdAdt=2π×4×10           r=4 cm and drdt=10 cm/secdAdt=80π cm2/sec

Page No 13.19:

Question 10:

A man 160 cm tall, walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1 m/sec. How fast is the length of his shadow increasing when he is 1 m away from the pole?

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD is at a distance of x km from the lamp post and y m is the length of his shadow CE.


Since triangles ABE and CDE are similar,ABCD=AECE61.6=x+yyxy=61.6-1xy=4.41.6y=1644xdydt=1644dxdtdydt=1644×1.1      dxdt=1.1dydt=0.4 m/sec

Page No 13.19:

Question 11:

A man 180 cm tall walks at a rate of 2 m/sec. away, from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is 3 m away from the base of light?

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD is at a distance x km from the lamp post and y m is the length of his shadow CE.



Since triangles ABE and CDE are similar,ABCD=AECE

91.8=x+yyxy=91.8-1xy=7.21.8x=4ydydt=14dxdtdydt=14×2              dxdt=2dydt=0.5 m/sec



Page No 13.20:

Question 12:

A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall.

Answer:

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.


Then,

tan θ=yx and x2+y2=132   x21+tan2 θ=169sec2 θ=169x22sec2 θ tan θdθdt=169 -2x3dxdtdθdt=-338×1.51232sec2 θ tan θ                          ...1When x=12, y=169-144=5 mSo, sec θ=1312 and tan θ=125From eq. 1, we getdθdt=-338×1.5123×2×13122 ×512=-338×1.510×169=-0.3 rad/sec

Page No 13.20:

Question 13:

A particle moves along the curve y = x2 + 2x. At what point(s) on the curve are the x and y coordinates of the particle changing at the same rate?

Answer:

Here, y=x2+2xdydt=2x+2dxdt2x+2=1              dydt=dxdt2x=-1x=-12Substituting x=-12 in y=x2+2x, we gety=-34Hence, the coordinates of the point are -12,-34.

Page No 13.20:

Question 14:

If y = 7xx3 and x increases at the rate of 4 units per second, how fast is the slope of the curve changing when x = 2?

Answer:

Here,y=7x-x3dydx=7x-x3Let s be the slope. Then,s=7-3x2dsdt=-6xdxdtdsdt=-642      x=2 and dxdt=4 units/secdsdt=-48

Page No 13.20:

Question 15:

A particle moves along the curve y = x3. Find the points on the curve at which the y-coordinate changes three times more rapidly than the x-coordinate.

Answer:

According to the question,dydt=3dxdtNow, y=x3dydt=3x2dxdt3dxdt=3x2dxdtx2=1x=±1Substituting x=±1 in y=x3, we gety=±1So the points are 1, 1 and -1, -1.

Page No 13.20:

Question 16:

Find an angle θ
(i) which increases twice as fast as its cosine.
(ii) whose rate of increase twice is twice the rate of decrease of its cosine.

Answer:

(i) Let x=cosθDifferentiating both sides with respect to t, we getdxdt=dcosθdt      =-sinθdθdtBut it is given that dθdt=2dxdtdxdt=-sinθ2dxdtsinθ=-12θ=π+π6=7π6Hence, θ=7π6.

(ii) Let x=cosθDifferentiating both sides with respect to t, we getdxdt=dcosθdt      =-sinθdθdtBut it is given that dθdt=-2dxdtdxdt=-sinθ-2dxdtsinθ=12θ=π6Hence, θ=π6.

Page No 13.20:

Question 17:

The top of a ladder 6 metres long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance?
How far is the foot from the wall when it and the top are moving at the same rate?

Answer:

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Here,

x2+y2=362xdxdt=-2ydydt                      ...1When x=4, y=36-16=252×4×0.5=-2×25dydt       dxdt=0.5 m/secdydt=-15m/secFrom eq. (1), we get2xdxdt=-2ydydt                           dxdt=dydt     x=-ySubstituting x=-y in x2+y2=36, we get      x2+x2=36x2=18x=32 m

Page No 13.20:

Question 18:

A balloon in the form of a right circular cone surmounted by a hemisphere, having a diametre equal to the height of the cone, is being inflated. How fast is its volume changing with respect to its total height h, when h = 9 cm.

Answer:

Let r be the radius of the hemisphere, h be the height and V be the volume of the cone.



Then,H=h+r  H=3r              h=2r dHdt=3drdtWhen H=9 cm, r=3 cmVolume = 13πr2h+23πr3Substituting h=2rV=23πr3 +23πr3V=43πr3dVdt=4πr2drdtdVdt=r23dHdtdVdH=323dVdH=12π cm3/sec

Page No 13.20:

Question 19:

Water is running into an inverted cone at the rate of π cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5 m. How fast the water level is rising when the water stands 7.5 m below the base.

Answer:

Let r be the radius, h be the height and V be the volume of the cone at any time t.


Then,V=13πr2hdVdt=13πr2dhdt+23πrhdrdtNow, hr=105 or r=h2 anddhdt=2drdtdVdt=13πh22dhdt+23πh2h12dhdtdVdt=π3h24dhdt+h22dhdtdVdt=π3×3h2dh4dtdVdt=πh24dhdtπh24dhdt=πdhdt=4h2dhdt=42.52dhdt=0.64 m/min

Page No 13.20:

Question 20:

A man 2 metres high walks at a uniform speed of 6 km/h away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:


Let AB be the lamp post. Let at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.

Since triangles ABE and CDE are similar,ABCD=AECE

62=x+yyxy=62-1=2dydt=12dxdtdydt=126dydt=3 km/hr

Page No 13.20:

Question 21:

The surface area of a spherical bubble is increasing at the rate of 2 cm2/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

Answer:

Let r be the radius, S be the surface area and V be the volume of the sphere at any time t. Then,S=4πr2dSdt=8πrdrdtdrdt=18πrdSdtdrdt=28π×6drdt=124π cm/secNow, Volume of sphere=43πr3dVdt=4πr2drdtdVdt=4π6224πdVdt=6 cm3/sec

Page No 13.20:

Question 22:

The radius of a cylinder is increasing at the rate 2 cm/sec. and its altitude is decreasing at the rate of 3 cm/sec. Find the rate of change of volume when radius is 3 cm and altitude 5 cm.

Answer:

Let r be the radius, h be the height and V be the volume of the cylinder at any time t. Then, V=πr2hdVdt=2πrhdrdt+πr2dhdtdVdt=πr2hdrdt+rdhdtdVdt=π×32×5×2+3×-3dVdt=3π20-9dVdt=33π cm3/sec

Page No 13.20:

Question 23:

The volume of metal in a hollow sphere is constant. If the inner radius is increasing at the rate of 1 cm/sec, find the rate of increase of the outer radius when the radii are 4 cm and 8 cm respectively.

Answer:

Let r1 be the inner radius and r2 be the outer radius and V be the volume of the hollow sphere at any time t. Then,V=43πr13-r23dVdt=4πr12dr1dt-r22dr2dt r12dr1dt=r22dr2dt         dVdt=042×1=82dr2dtdr2dt=14 cm/sec

Page No 13.20:

Question 24:

Sand is being poured onto a conical pile at the constant rate of 50 cm3/ minute such that the height of the cone is always one half of the radius of its base. How fast is the height of the pile increasing when the sand is 5 cm deep.

Answer:

Let r be the radius, h be the height and be the volume of the conical pile at any time t. Then,V=13πr2hV=13π2h2h                            h=r2V=43πh3dVdt=4πh2dhdt50=4πh2dhdtdhdt=504π52dhdt=12π cm/min

Page No 13.20:

Question 25:

A kite is 120 m high and 130 m of string is out. If the kite is moving away horizontally at the rate of 52 m/sec, find the rate at which the string is being paid out.

Answer:

In the right triangle ABC,


Here,AB2+BC2=AC2x2+1202=y2    2xdxdt=2ydydtdydt=xydxdtdydt=50130×52               x=1302-1202=50dydt=20 m/sec

Page No 13.20:

Question 26:

A particle moves along the curve y = (2/3)x3 + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate.

Answer:

Here, y=23x3+1dydt=2x2dxdt2dxdt=2x2dxdt      dydt=2dxdtx=±1Substituting the value of x=1 and x=-1 in y=23x3+1, we gety=53 and y=13So, the points are 1,53 and -1,13.

Page No 13.20:

Question 27:

Find the point on the curve y2 = 8x for which the abscissa and ordinate change at the same rate.

Answer:

Here, y2=8x     ...1⇒2ydydt=8dxdt⇒2y=8            dydt=dxdty=4x=y28           From eq. 1x=168=2So, the point is 2,4.

Page No 13.20:

Question 28:

The volume of a cube is increasing at the rate of 9 cm3/sec. How fast is the surface area increasing when the length of an edge is 10 cm?

Answer:

Let x be the side and V be the volume of the cube at any time t. Then,V=x3dVdt=3x2dxdt⇒9=3102dxdt       x=10 cm and dVdt=cm3/secdxdt=0.03 cm/secLet S be the surface area of the cube at any time t. Then, S=6x2dSdt=12xdxdtdSdt=12×10×0.03            x=10 cm and dxdt= 0.03 cm/secdSdt=3.6 cm2/sec

Page No 13.20:

Question 29:

The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when radius is 5 cm.

Answer:

Let r be the radius and V be the volume of the sphere at any time t. Then, V=43πr3dVdt=4πr2drdtdrdt=14πr2dVdtdrdt=254π52              r=5 cm anddVdt=25 cm3/secdrdt=14πcm/secNow, let S be the surface area of the sphere at any time t. Then, Sr2dSdt=8πrdrdtdSdt=8π5×14π       r=5 cm anddrdt=14π cm/secdSdt=10 cm2/sec

Page No 13.20:

Question 30:

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (i) the perimeter (ii) the area of the rectangle.

Answer:

i Let P be the perimeter of the rectangle at any time t. Then,P=2x+ydPdt=2dxdt+dydtdPdt=2-5+4           dxdt=-5 cm/min and dydt=4 cm/mindPdt=-2 cm/minii Let A be the area of the rectangle at any time t. Then,A=xydAdt=xdydt+ydxdtdAdt=84+6-5     x=8 cm, y=6 cmdxdt=-5 cm/min and dydt=4 cm/mindAdt=32-30dAdt=2 cm2/min



Page No 13.21:

Question 31:

A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/sec. Find the rate at which its area is increasing when radius is 3.2 cm.

Answer:

Let r be the radius and A be the area of the circular disc at any time t. Then, Ar2dAdt=2πrdrdtdAdt=2π×3.2×0.05     r=3.2 cm and drdt=0.05 cm/secdAdt=0.32π cm2/sec



Page No 13.24:

Question 1:

If a particle moves in a straight line such that the distance travelled in time t is given by s = t3 − 6t2 + 9t + 8. Find the initial velocity of the particle.

Answer:

s=t3-6t2+9t+8dsdt=3t2-12t+9Initial velocity=Velocity at t=0dsdt=302-120+9dsdt=9 units/unit time

Page No 13.24:

Question 2:

The volume of a sphere is increasing at 3 cubic centimeter per second. Find the rate of increase of the radius, when the radius is 2 cms.

Answer:

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=4πr2drdtdrdt=14πr2dVdtdrdt=34π22       r=2 cm and dVdt=3 cm3/secdrdt=316π cm/sec

Page No 13.24:

Question 3:

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. How far is the area increasing when the side is 10 cms?

Answer:

 Let x be the side and A be the area of the equilateral triangle at any time t. Then, A=34x2dAdt=2×34xdxdtdAdt=32×2×10dAdt=103 cm2/sec

Page No 13.24:

Question 4:

The side of a square is increasing at the rate of 0.1 cm/sec. Find the rate of increase of its perimeter.

Answer:

Let x be the side and P be the perimeter of the square at any time t. Then,  P=4xdPdt=4dxdtdPdt=4×0.1             ∵ dxdt=0.1 cm/secdPdt=0.4 cm/sec

Page No 13.24:

Question 5:

The radius of a circle is increasing at the rate of 0.5 cm/sec. Find the rate of increase of its circumference.

Answer:

Let r be the radius and C be the circumference of the circle at any time t. Then, C=2πrdCdt=2πdrdtdCdt=2π×0.5                 ∵ drdt=0.5 cm/sec dCdt=π cm/sec

Page No 13.24:

Question 6:

The side of an equilateral triangle is increasing at the rate of 13 cm/sec. Find the rate of increase of its perimeter.

Answer:

Let x be the side and P be the perimeter of the equilateral triangle at any time t. Then, P=3xdPdt=3dxdtdPdt=3×13            dxdt=13cm/secdPdt=1 cm/sec

Page No 13.24:

Question 7:

Find the surface area of a sphere when its volume is changing at the same rate as its radius.

Answer:

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=4πr2drdtdVdt=4πr2dVdt     dVdt=drdt4πr2=1 Surface area of sphere =1 square unit

Page No 13.24:

Question 8:

If the rate of change of volume of a sphere is equal to the rate of change of its radius, find the radius of the sphere.

Answer:

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=4πr2drdt4πr2=1                  dVdt=drdtr2=14πr=14πr=12π units

Page No 13.24:

Question 9:

The amount of pollution content added in air in a city due to x diesel vehicles is given by P(x) = 0.005x3 + 0.02x2 + 30x. Find the marginal increase in pollution content when 3 diesel vehicles are added and write which value is indicated in the above questions.

Answer:

Since, marginal increase in the pollution content is the rate of change of total pollution with respect to the number of diesel vehicles, we have

Marginal increase in pollution = dPdx=0.015x2+0.04x+30

When x = 3, marginal increase in pollution = 0.0159+0.043+30=0.135+0.12+30=30.255

Hence, the required marginal increase in pollution is 30.255 units.

It indicates the pollution level due to x diesel vehicles.

Page No 13.24:

Question 10:

A ladder, 5 metre long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides down wards at the rate of 10 cm/sec, then find the rate at which the angle between the floor and ladder is decreasing when lower end of ladder is 2 metres from the wall.

Answer:

Length of the ladder=500cmLet the horizontal length covered between the wall and the ladder be x and vertical length covered between the wall and the ladder be y.And let the angle between the floor and ladder be θ.Then, sinθ=y500On differentiating with respect to t, we getcosθdθdt=1500dydt          ...(1)It is given that dydt=-10 cm/sec.        ...(2)Also,cosθ=x500When x=200 cm, cosθ=200500=25        ...(3)Substituting (2) and (3) in (1), we get25dθdt=1500-10dθdt=-120 radian/secondHence, the angle between the floor and the ladder is decreasing at the rate of 120 radian/second.

Page No 13.24:

Question 1:

If V=43πr3, at what rate in cubic units is V increasing when r = 10 and drdt=0.01?
(a) π
(b) 4π
(c) 40π
(d) 4π/3

Answer:

(b) 4π
Given: V=43πr3, r=10 and drdt=0.01dVdt=4πr2drdtdVdt=4π102×0.01dVdt=4π

Page No 13.24:

Question 2:

Side of an equilateral triangle expands at the rate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is
(a) 102 cm2/sec
(b) 103 cm2/sec
(c) 10 cm2/sec
(d) 5 cm2/sec

Answer:

(b) 103 cm2/sec

 Let x be the side and A be the area of the equilateral triangle at any time t. Then, A=34x2dAdt=2×34xdxdtdAdt=32×2×10dAdt=103 cm2/sec

Page No 13.24:

Question 3:

The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm is
(a) 8π cm2/sec
(b) 12π cm2/sec
(c) 160π cm2/sec
(d) 200 cm2/sec

Answer:

(c) 160π cm2/sec

Let r be the radius and S be the surface area of the sphere at any time t. Then,S=r2dSdt=8πrdrdtdSdt=8π2000.1dSdt=160π cm2/sec

Page No 13.24:

Question 4:

A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm3/sec. At what rate is the radius increasing when its circular base area is 1 m2?
(a) 1 mm/sec
(b) 0.001 cm/sec
(c) 2 mm/sec
(d) 0.002 cm/sec

Answer:

(d) 0.002 cm/sec

Let r be the radius, h be the height and V be the volume of the cone at any time t. Then,V=13πr2hV=23πr3                         h=2rdV dt=2×104drdt            πr2=1 m2 or 104 cm2 dr dt=12×104dV dtdr dt=402×104dr dt=0.002 cm/sec

Page No 13.24:

Question 5:

A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25 π m3/minute. The rate at which the surface of the oil is rising, is
(a) 1 m/minute
(b) 2 m/minute
(c) 5 m/minute
(d) 1.25 m/minute

Answer:

(a) 1 m/minute

Let r be the radius, h be the height and V be the volume of the cylindrical vessel at any time t. Then, V=πr2hdV dt=πr2dhdtdh dt=1πr2dV dtdh dt=0.25ππ0.52dh dt=0.250.25dh dt=1 m/min



Page No 13.25:

Question 6:

The distance moved by the particle in time t is given by x = t3− 12t2 + 6t + 8. At the instant when its acceleration is zero, the velocity is
(a) 42
(b) −42
(c) 48
(d) −48

Answer:

(b) −42

x=t3-12t2+6t+8dx dt=3t2-24t+6d2x dt2=6t-246t-24=0           acceleration is zerot=4So, Velocity at t=4dx dt=342-24×4+6dx dt=48-96+6dx dt=-42

Page No 13.25:

Question 7:

The altitude of a cone is 20 cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of
(a) 30 cm/sec
(b) 1603 cm/sec
(c) 10 cm/sec
(d) 160 cm/sec

Answer:

(b) 1603 cm/sec

Let r be the radius, h be the height and α be the semi-vertical angle of the cone.



Then,tan α=rhsec2αdαdt=drh dtdr dt=h×sec2αdαdtdr dt=20×sec230×2      h=20 cm, α=30° and dαdt=2° per seconddr dt=40×232dr dt=1603 cm/sec

Page No 13.25:

Question 8:

For what values of x is the rate of increase of x3− 5x2 + 5x + 8 is twice the rate of increase of x?
(a) -3, -13

(b) -3, 13

(c) 3, -13

(d) 3, 13

Answer:

(d) 3, 13

Let y=x3-5x2+5x+8dydt=3x2-10x+5dxdtAccording to the question,2dxdt=3x2-10x+5dxdt3x2-10x+5=23x2-10x+3=03x2-9x-x+3=03xx-3-1x-3=0x-3=0 or 3x-1=0x=3 or x=13

Page No 13.25:

Question 9:

The coordinates of the point on the ellipse 16x2 + 9y2 = 400 where the ordinate decreases at the same rate at which the abscissa increases, are
(a) (3, 16/3)
(b) (−3, 16/3)
(c) (3, −16/3)
(d) (3, −3)

Answer:

(a) (3, 16/3)

According to the question,dydt=-dxdt16x2+9y2=40032xdxdt+18ydydt=032xdxdt=-18ydydt32x=18yx=9y16         ...1Now,169y162+9y2=40081y216+9y2=40081y2+144y2=6400225y2=6400y2=6400225y=6400225y=163 or -163So,x=916×163    Using 1 orx=-916×163x=3  or -3So, the required point is 3,163.

Page No 13.25:

Question 10:

The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is
(a) 54π cm2/min
(b) 7π cm2/min
(c) 27 cm2/min
(d) none of these

Answer:

Let r be the radius, h be the height and S be the lateral surface area of the cone at any time t. 

Given: drdt=3 cm/min and dhdt=-4 cm/minHere,l2=h2+r2l=242+72l=625l=25S=πrlS2=πrl2S2=π2r2h2+r2S2=π2r4+π2h2r22SdSdt=4π2r3drdt+2π2r2hdhdt+2π2h2rdrdt2πrldSdt=2π2rh2r2hdrdt +rdhdt+hdrdt25dSdt=24π27224×3 -7×4+24×3                             Given: r=7, h=2425dSdt=24π494-28+7225dSdt=24π49+288-1124dSdt=24π225100dSdt=24π2.25dSdt=54π cm2/sec

Page No 13.25:

Question 11:

The radius of a sphere is increasing at the rate of 0.2 cm/sec. The rate at which the volume of the sphere increase when radius is 15 cm, is
(a) 12π cm3/sec
(b) 180π cm3/sec
(c) 225π cm3/sec
(d) 3π cm3/sec

Answer:

(b) 180π cm3/sec

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=4πr2drdtdVdt=152×0.2dVdt=180π cm3/sec

Page No 13.25:

Question 12:

The volume of a sphere is increasing at 3 cm3/sec. The rate at which the radius increases when radius is 2 cm, is
(a) 332πcm/sec

(b) 316πcm/sec

(c) 348πcm/sec

(d) 124πcm/sec

Answer:

(b) 316πcm/sec

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=4πr2drdtdrdt=1r2dVdtdrdt=322drdt=316π cm/sec

Page No 13.25:

Question 13:

The distance moved by a particle travelling in straight line in t seconds is given by s = 45t + 11t2t3. The time taken by the particle to come to rest is
(a) 9 sec
(b) 5/3 sec
(c) 3/5 sec
(d) 2 sec

Answer:

(a) 9 sec

s=45t+11t2-t3dsdt=45+22t-3t2According to the question,3t2-22t-45=03t2-27t+5t-45=03tt-9+5t-9=0t-93t +5=0t-9=0 or 3t +5=0As time can't be negative, t=9 sec

Page No 13.25:

Question 14:

The volume of a sphere is increasing at the rate of 4π cm3/sec. The rate of increase of the radius when the volume is 288 π cm3, is
(a) 1/4
(b) 1/12
(c) 1/36
(d) 1/9

Answer:

(c) 1/36

Let r be the radius and be the volume of the sphere at any time t. Then,V=43πr343πr3=288πr3=288×34r3=216r=6dVdt=4πr2drdtdVdt=4π62drdt 4π=144πdrdtdrdt=136

Page No 13.25:

Question 15:

If the rate of change of volume of a sphere is equal to the rate of change of its radius, then its radius is equal to

(a) 1 unit

(b) 2π units

(c) 12π unit

(d) 12π unit

Answer:

(d) 12π unit

Let r be the radius and V be the volume of the sphere at any time t. Then,V=43πr3dVdt=433πr2drdtdVdt=4πr2drdt4πr2=1    dVdt=drdtr2=14πr=14πr=12π unit

Page No 13.25:

Question 16:

If the rate of change of area of a circle is equal to the rate of change of its diameter, then its radius is equal to
(a) 2π unit

(b) 1π unit

(c) π2 units

(d) π units

Answer:

(b) 1π unit

Let r be the radius and Abe the area of the circle at any time t. Then,A = πr2 A=πD24                      r=D2dAdt=πD2dDdtdDdt=πD2dDdt          dAdt=dDdtD2=1πr=1π units

Page No 13.25:

Question 17:

Each side of an equilateral triangle is increasing at the rate of 8 cm/hr. The rate of increase of its area when side is 2 cm, is

(a) 83 cm2/hr

(b) 43 cm2/hr

(c) 38 cm2/hr

(d) none of these

Answer:

(a) 83 cm2/hr

Let x be the side and A be the area of the equilateral triangle at any time t. Then,A=34x2dAdt=32xdxdtdAdt=3228dAdt=83 cm2/hr

Page No 13.25:

Question 18:

If s = t3 − 4t2 + 5 describes the motion of a particle, then its velocity when the acceleration vanishes, is

(a) 169 unit/sec

(b) -323 unit/sec

(c) 43 unit/sec

(d) -163 unit/sec

Answer:

(d) -163 unit/sec

According to the question,s=t3-4t2+5dsdt=3t2-8td2sdt2=6t-86t-8=0        As velocity deminishes, thend2sdt2=0 t=43Now, dsdtt=43=3432-843dsdt=163-323dsdt=-163unit/sec

Page No 13.25:

Question 19:

The equation of motion of a particle is s = 2t2 + sin 2t, where s is in metres and t is in seconds. The velocity of the particle when its acceleration is 2 m/sec2, is
(a) π+3 m/sec

(b) π3+3 m/sec

(c) 2π3+3 m/sec

(d) π3+13 m/sec

Answer:

(b) π3+3 m/sec

According to the question,s=2t2+sin 2tdsdt=4t+2 cos 2td2sdt2=4-4 sin 2t4-4 sin 2t=24 sin 2t=2sin 2t=122t=π6Now, dsdt=4π12+2 cosπ6ds dt=π3+3  m/sec



Page No 13.26:

Question 20:

The radius of a circular plate is increasing at the rate of 0.01 cm/sec. The rate of increase of its area when the radius is 12 cm, is
(a) 144 π cm2/sec
(b) 2.4 π cm2/sec
(c) 0.24 π cm2/sec
(d) 0.024 π cm2/sec

Answer:

(c) 0.24 π cm2/sec

Let r be the radius and A be the area of the circular plate at any time t. Then,A=πr2dAdt=2πrdrdtdAdt=2π120.01dAdt=0.24π cm2/sec

Page No 13.26:

Question 21:

The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is π, the rate of increase of its area is
(a) π cm2/sec
(b) 2π cm2/sec
(c) π2 cm2/sec
(d) 2π2 cm2/sec2

Answer:

(c) π2 cm2/sec

 Let D be the diameter and be the area of the circle at any time t. Then,A=πr2 where r is the radius of the cicleAD24                    r=D2dAdt=2πD4dDdtdAdt=π2×2π ×1     dDdt=1 cm/secdAdt=π2 cm2/sec

Page No 13.26:

Question 22:

A man 2 metres tall walks away from a lamp post 5 metres height at the rate of 4.8 km/hr. The rate of increase of the length of his shadow is
(a) 1.6 km/hr
(b) 6.3 km/hr
(c) 5 km/hr
(d) 3.2 km/hr

Answer:

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.



Since triangles ABE and CDE are similar,ABCD=AECE

52=x+yyxy=52-1xy=32y=23xdydt=23dxdtdydt=23×4.8dy dt=3.2 km/hr

Page No 13.26:

Question 23:

A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp fixed at 15 ft height. The length of his shadow is increasing at the rate of
(a) 15 ft/sec
(b) 9 ft/sec
(c) 6 ft/sec
(d) none of these

Answer:

(c) 6 ft/sec

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y ft be the length of his shadow CE.


Since the triangles ABE and CDE are similar,ABCD=AECE

156=x+yyxy=156-1xy=32y=23xdydt=23dxdtdydt=23×9dydt=6 ft/sec 

Page No 13.26:

Question 24:

In a sphere the rate of change of volume is
(a) π times the rate of change of radius
(b) surface area times the rate of change of diameter
(c) surface area times the rate of change of radius
(d) none of these

Answer:

(c) surface area times the rate of change of radius

Let r be the radius and V be the volume of sphere at any time t. Then,V=43πr3dVdt=433πr2drdtdVdt=4πr2drdtThus, the rate of change of volume is surface area times the rate of change of the radius.

Page No 13.26:

Question 25:

In a sphere the rate of change of surface area is
(a) 8π times the rate of change of diameter
(b) 2π times the rate of change of diameter
(c) 2π times the rate of change of radius
(d) 8π times the rate of change of radius

Answer:

(d) 8π times the rate of change of radius

Let r be the radius and S be the surface area of the sphere at any time t. Then,S=4πr2dSdt=8πrdrdt The rate of change of surface area is 8π times the rate of change of the radius.

Page No 13.26:

Question 26:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
(a) 1 m/hr
(b) 0.1 m/hr
(c) 1.1 m/hr
(d) 0.5 m/hr

Answer:

(a) 1 m/hr

Let r, h and V be the radius, height and volume of the cylinder at any time t. Then, V=πr2hdVdt=πr2dhdt314=3.14 × 102dhdtdhdt=314314dhdt=1 m/hr



Page No 13.4:

Question 1:

Find the rate of change of the total surface area of a cylinder of radius r and height h, when the radius varies.

Answer:

Let T be the total surface area of a cylinder. Then,

T = 2πrr+h

Since the radius varies, we differentiate the total surface area w.r.t. radius r.

Now,

dTdr=ddr2πrr+hdTdr=ddr2πr2+ddr2πrhdTdr=4πr+2πhdTdr=2πr+h

Page No 13.4:

Question 2:

Find the rate of change of the volume of a sphere with respect to its diameter.

Answer:

Let V and r be the volume and diameter of the sphere, respectively. Then,
V = 43πradius3

 V=43πr23=16πr3

dVdr=12πr2

Page No 13.4:

Question 3:

Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2 cm.

Answer:

Let V be the volume of the sphere. Then,

V = 43πr3

dVdr=4πr2

Let S be the total surface area of sphere. Then,

S = 4πr2

dSdr=8πr
 dVdS=dVdrdSdrdVdS=4πr28πr=r2dVdSr=2 =22                        =1 cm

Page No 13.4:

Question 4:

Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3 cm.

Answer:

Let A be the area of the circular disc. Then,

A = πr2

dAdr=2πr
Let C be the circumference of the circular disc. Then,

C = 2πr

dCdr=2π

 dAdC=dAdrdCdrdAdC=2πr2π=rdAdCr=3=3 cm                      

Page No 13.4:

Question 5:

Find the rate of change of the volume of a cone with respect to the radius of its base.

Answer:

Let V be the volume of the cone. Then,

V = 13πr2h

dVdr=23πrh
 

Page No 13.4:

Question 6:

Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.

Answer:

Let A be area of the circle. Then,
A = πr2

dAdr=2πr
Hence, the rate of change of the area of the circle is 2πr.
When r = 5 cm,

dAdrr=5=2π5               =10π cm2/cm                                    

Page No 13.4:

Question 7:

Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2 cm?

Answer:

Let V be the volume of the spherical ball. Then,
V = 43πr3

dVdr=4πr2
Thus, the rate of change of the volume of the sphere is 4πr2.

When r=2 cm, dVdrr=2=4π22               =16π cm3/cm                                    

Page No 13.4:

Question 8:

The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.007x3 − 0.003x2 + 15x + 4000. Find the marginal cost when 17 units are produced.

Answer:

Since the marginal cost is the rate of change of total cost with respect to its output,

Marginal Cost (MC) = dCdxx=ddx0.007x3-0.003x2+15x+4000=0.021x2-0.006x+15
When x = 17,
Marginal Cost (MC) = =0.021172-0.006(17)+15=6.069-0.102+15=Rs 20.967

Page No 13.4:

Question 9:

The total revenue received from the sale of x units of a product is given by R (x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.

Answer:

Since the marginal revenue is the rate of change of total revenue with respect to its output,
Marginal Revenue (MR) = dRdxx=ddx13x2+26x+15=26x+26

When x = 7,
Marginal Revenue (MR)=26(7)+26=182+26=Rs 208

Page No 13.4:

Question 10:

The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (Marginal revenue). If the total revenue (in rupees) recieved from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5, find the marginal revenue, when x = 5, and write which value does the question indicate.

Answer:

Since, marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have

Marginal revenue (MR)=dRdx=6x+36When x=5, MR=65+36=66

Hence, the required marginal revenue is â‚¹66.

It indicates the extra money spent when number of employees increased from 5 to 6.



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