12th NCERT Application of Derivatives Exercise 6.1 Questions 18
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Question (1)

Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm and (b) r =4 cm

Solution

The rate of change of area with respect to radius = dA/dr
(a) Area of circle A = πr2
Differentiate with respect to r
$\frac{{dA}}{{dr}} = 2\pi r$ ${\left( {\frac{{dA}}{{dr}}} \right)_{r = 3}} = 2\pi \left( 3 \right) = 6\pi$ (b) when r = 4 cm.
${\left( {\frac{{dA}}{{dr}}} \right)_{r = 4}} = 2\pi \left( 4 \right) = 8\pi$

Question (2)

The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Solution

volume of a cube is increasing at the rate of 8 cm3/s that is per unit time
Thus dV/dt =8 cm3/s Let sides of the cube be l
Volume of cube V = l3
Differentiate with respect to 't'
$\frac{{dV}}{{dt}} = 3{l^2}\frac{{dl}}{{dt}}$ $8 = 3{\left( {12} \right)^2}\frac{{dl}}{{dt}}$ $\frac{{dl}}{{dt}} = \frac{8}{{3 \times 12 \times 12}} = \frac{1}{{54}}cm/\sec$ Area of cube A = 6l2
Differentiate with respect to 't'
$\frac{{dA}}{{dt}} = 6\left( {2l} \right)\frac{{dl}}{{dt}}$ $\frac{{dA}}{{dt}} = \require{cancel} \cancel{6}\left( {2 \times \cancel{12}^4} \right) \times \frac{1}{{\cancel{54}_\cancel{9}3}}$ $\frac{{dA}}{{dt}} = \frac{8}{3}\frac{{c{m^2}}}{{\sec }}$

Question (3)

The radius of a circle is increasing uniformly at the rate of 3cm/s. Find the rate at which the area of the circle is increasing when the radius is 10cm

Solution

The radius of a circle is increasing uniformly at the rate of 3cm/s = dr/dt
Area of circle A = πr2
Differentiate with respect to 't'
$\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}$ $\frac{{dA}}{{dt}} = 2\pi \left( {10} \right)3$ $\frac{{dA}}{{dt}} = 60\pi \frac{{c{m^2}}}{{\sec }}$

Question (4)

An edge of a variable cube is increasing at the rate of 3cm/s. how fast is the volume of the cube increasing when the edge is 10cm long?

Solution

An edge of a variable cube is increasing at the rate of 3cm/s = dl/dt
given l = 10
Volume of cube V= l3
Differentiate with respect to 't' $\frac{{dV}}{{dt}} = 3{l^2}\frac{{dl}}{{dt}}$ $\frac{{dV}}{{dt}} = 3{\left( {10} \right)^2}3$ $\frac{{dV}}{{dt}} = 900\frac{{c{m^3}}}{{\sec }}$

Question (5)

A stone is dropped into a quiet lake and waves move in circle at the speed of 4cm/s. At the instant when the radius of the circular wave is 8cm, how fast is the enclosed area increasing?

Solution

waves move in circle at the speed of 4cm/s = dr/dt
To find dA/dt
Area of circle A = πr2
Differentiate with respect to 't' $\frac{{dA}}{{dt}} = 2\pi r\frac{{dr}}{{dt}}$ $\frac{{dA}}{{dt}} = 2\pi \left( 8 \right) \times 5$ $\frac{{dA}}{{dt}} = 80\pi \frac{{c{m^2}}}{{\sec }}$

Question (6)

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

Solution

radius of a circle is increasing at the rate of 0.7 cm/s = dr/dt
To find dC/dt, C = circumference
The circumference of circle C = 2πr
Differentiate with respect to 't' $\frac{{dC}}{{dt}} = 2\pi \frac{{dr}}{{dt}}$ $\frac{{dC}}{{dt}} = 2\pi \left( {0.7} \right)$ $\frac{{dC}}{{dt}} = 2 \times \frac{{22}}{\cancel{7}} \times \left( {\frac{\cancel{7}}{{10}}} \right)$ $\frac{{dC}}{{dt}} = 4.4\frac{{cm}}{{\sec }}$

Question (7)

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

Solution

length x of a rectangle is decreasing at the rate of 5cm/minute
width y is increasing at the rate of 4cm/minute
(a) Perimiter of rectangle P = 2(l+b)
∴ P= 2(x+y)
Differentiate with respect to 't' $\frac{{dP}}{{dt}} = 2\left( {\frac{{dx}}{{dt}} + \frac{{dy}}{{dt}}} \right)$ $\frac{{dP}}{{dt}} = - 2\frac{{cm}}{{\min }}$ (b) Area of rectangle A = lb
A = xy
Differentiate with respect to 't' $\frac{{dA}}{{dt}} = x\frac{{dy}}{{dt}} + y\frac{{dx}}{{dt}}$ $\frac{{dA}}{{dt}} = 8\left( 4 \right) + 6\left( { - 5} \right)$ $\frac{{dA}}{{dt}} = 32 - 30$ $\frac{{dA}}{{dt}} = 2\frac{{c{m^2}}}{{\min }}$

Question (8)

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

Solution

balloon inflated by pumping in 900 cubic centimeters of gas per second = dV/dt
rate at which the radius of the balloon increases = dr/dt at r = 15 cm
Volume of sphere V = (4/3) πr3
Differentiate with respect to 't' $\frac{{dV}}{{dt}} = \frac{4}{3}\pi \left( {3{r^2}} \right)\frac{{dr}}{{dt}}$ $\frac{{dV}}{{dt}} = 4\pi {r^2}\frac{{dr}}{{dt}}$ $900 = 4\pi {\left( {15} \right)^2}\frac{{dr}}{{dt}}$ $\frac{{dr}}{{dt}} = \frac{{\cancel{900}^\cancel{225}}}{{\cancel{4}\pi \times \cancel{225}}}$ $\frac{{dr}}{{dt}} = \frac{1}{\pi }\frac{{cm}}{{\sec }}$

Question (9)

A balloon, which always remains spherical has a variable radius. Find the rate of which its volume is increasing with the radius when the later is 10cm

Solution

To find dV/dr
Volume of sphere = (4/3) π r3
Differentiate with respect to 'r' $\frac{{dV}}{{dr}} = \frac{4}{3}\pi \times \left( {3{r^2}} \right)$ $\frac{{dV}}{{dr}} = 4\pi \times \left( {{r^2}} \right)$ $\frac{{dV}}{{dr}} = 4\pi \times \left( {{{10}^2}} \right)$ $\frac{{dV}}{{dr}} = 400\pi \frac{{c{m^3}}}{{\sec }}$

Question (10)

A ladder 5m long is leaning against a wall. the bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from the wall?

Solution Let AC is ladder of 5 m leans against wall AB
Let AB=y and BC = x
Ladder is pulled along the ground at rate 2 cm/sec, dx/dt = 2cm/sec, x = 4 m = 400cm
To find the height on the wall decreases at rate dy/dt
In triangle ABC, ∠B = 90
By Pythagoras' theorem
AC2 = AB2 + BC2
∴ x2 + y2 = 25
when x = 4
16 + y2 = 25
y2 = 9
y = 3m. = 300 cm.
Now x2 + y2 = 25
Differentiate with respect to 't' $2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 0$ $x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}} = 0$ $400\left( 2 \right) + 300\frac{{dy}}{{dt}} = 0$ $\frac{{dy}}{{dt}} = - \frac{{400 \times 2}}{{300}} = - \frac{8}{3}$ So it decreases at rate 8/3 cm/sec

Question (11)

A particle moves along the curve 6y=x3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Solution

curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. dy/dt = 8(dx/dt)
6y=x3+2.
Differentiate with respect to 't' $6\frac{{dy}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}$ $\text{As} \quad \frac{{dy}}{{dt}} = 8\frac{{dx}}{{dt}}$ $6\left( {8\frac{{dx}}{{dt}}} \right) = 3{x^2}\frac{{dx}}{{dt}}$ 48 = 3x2
x2 = 16
x = ±4
When x = 4
6y = 43 + 2
6y = 66
y = 11 ∴ (x, y) = (4, 11)
If x = -4
6y = (-4)3 + 2
6y = -62
y = -62/6 = -31/3
(x, y) = (-4, -31/3)
So the points on the curve are (4, 11) and (-4, -31/3).

Question (12)

The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1cm?

Solution

given dr/dt = ½ cm/s
To find dV/dt at r = 1
Volume of bubble = (4/3) π r3
Differentiate with respect to 't' $\frac{{dV}}{{dt}} = \frac{4}{3}\pi \left( {3{r^2}} \right)\frac{{dr}}{{dt}}$ $\frac{{dV}}{{dt}} = 4\pi \left( {{r^2}} \right)\frac{{dr}}{{dt}}$ $\frac{{dV}}{{dt}} = 4\pi \left( {{1^2}} \right) \times \frac{1}{2}$ $\frac{{dV}}{{dt}} = 2\pi \frac{{c{m^3}}}{{\sec }}$

Question (13)

A balloon, which always remains spherical, has a variable diameter (3/2)(2x+1). Find the rate of change of its volume with respect to x

Solution

rate of change of its volume with respect to x , To find dV/dx
$d = \frac{3}{2}\left( {2x + 1} \right)$ To find dv/dr $r = \frac{d}{2} = \frac{3}{4}\left( {2x + 1} \right)$ Volume of sphere V $V = \frac{4}{3}\pi {r^3}$ $V = \frac{4}{3}\pi {\left( {\frac{3}{4}\left( {2x + 1} \right)} \right)^3}$ $V = \frac{\cancel{4}}{\cancel{3}}\pi \frac{{\cancel{27}^9}}{{\cancel{64}_{16}}}{\left( {2x + 1} \right)^3}$ $V = \frac{9}{{16}}\pi {\left( {2x + 1} \right)^3}$ Differentiate with respect to 'x' $\frac{{dV}}{{dx}} = \frac{9}{{16}}\pi \times 3{\left( {2x + 1} \right)^2}\frac{d}{{dx}}\left( {2x + 1} \right)$ $\frac{{dV}}{{dx}} = \frac{{27\pi }}{{16}}{\left( {2x + 1} \right)^2} \times 2$ $\frac{{dV}}{{dx}} = \frac{{27\pi }}{8}{\left( {2x + 1} \right)^2}$

Question (14)

Sand is pouring from a pipe at the rate of 12cm3s. the falling sand forms a cone on the ground in such a way that the height of the cone is always on-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4cm?

Solution

pipe at the rate of 12cm3s = dV/dt
height h = (1/6) r
To find how fast is the height of the sand cone increasing when the height is 4cm = dh/dt at h =4
Volume of cone V $V = \frac{1}{3}\pi {r^2}h$ as r = 6h $V = \frac{1}{3}\pi {\left( {6h} \right)^2}h$ $V = \frac{1}{3}\pi \times 36{h^2} \times h$ V = 12πh3
Differentiate with respect to 't' $\frac{{dV}}{{dt}} = 12\pi \times 3{h^2}\frac{{dh}}{{dt}}$ $12 = 12\pi \times 3{\left( 4 \right)^2}\frac{{dh}}{{dt}}$ $\frac{{dh}}{{dt}} = \frac{{\cancel{12}}}{{\cancel{12}\pi \times 3 \times 16}}$ $\frac{{dh}}{{dt}} = \frac{1}{{48\pi }}\frac{{cm}}{{\sec }}$

Question (15)

The total cost C(x) in Rupees associate 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 unit are produced

Solution

Marginal cost is rate of change of total cost per unit
$\text{Marginal cost} = \frac{d}{{dx}}C$ C(x) = 0.007x3 - 0.003x2 +15x+4000
Differentiate with respect to 'x' $\frac{d}{{dx}}C\left( x \right) = 0.007\left( {3{x^2}} \right) - 0.003\left( {2x} \right) + 15 + 0$ $\frac{d}{{dx}}C\left( x \right) = 0.021{x^2} - 0.006x + 15$ ${\left[ {\frac{d}{{dx}}C\left( x \right)} \right]_{x = 17}} = 0.021{\left( {17} \right)^2} - 0.006\left( {17} \right) + 15$ ${\left[ {\frac{d}{{dx}}C\left( x \right)} \right]_{x = 17}} = 0.021\left( {289} \right) - 0.103 + 15$ ${\left[ {\frac{d}{{dx}}C\left( x \right)} \right]_{x = 17}} = 6.069 - 0.102 + 15$ ${\left[ {\frac{d}{{dx}}C\left( x \right)} \right]_{x = 17}} = 20.967$ Marginal cost when 17 units are produced = Rs 20.97

Question (16)

The total revenue in Rupees 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

Solution

Marginal revenue is rate of change of total revenue per unit produced. $\text{Marginal Revenue} = \frac{d}{{dx}}R\left( x \right)$ R(x) = 13x2 + 26x + 15
Differentiate with respect to 'x' $\frac{d}{{dx}}R\left( x \right) = 26x + 26$ ${\left[ {\frac{d}{{dx}}R\left( x \right)} \right]_{x = 7}} = 26\left( 7 \right) + 26$ ${\left[ {\frac{d}{{dx}}R\left( x \right)} \right]_{x = 7}} = 182 + 26$ ${\left[ {\frac{d}{{dx}}R\left( x \right)} \right]_{x = 7}} = 208$ marginal revenue = Rs. 208

### Choose the correct answer in the Exercises 17 and 18

Question (17)

The rate of change of the area of a circle with respect to its radius r at r=6cm is
(A) 10π     (B) 12π
(C) 8 π     (D) 11π

Solution

To find dA/dr when r = 6
Area of circle A = πr2
Differentiate with respect to 'r' $\frac{{dA}}{{dr}} = 2\pi r$ ${\left( {\frac{{dA}}{{dr}}} \right)_{r = 6}} = 2\pi \left( {6} \right)$ ${\left( {\frac{{dA}}{{dr}}} \right)_{r = 6}} = 12\pi$ So B is the correct option.

Question (18)

The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 3x2 + 36x+5. The marginal revenue, when x=15 is
(A) 116     (B) 96
(C) 90     (D) 126

Solution

The marginal revenue is the derivative of total revenue.
$R(x) = 3{x^2} + 36x + 5$ $Diff.w.r.t.x$ $R'(x) = 6x + 36$ The marginal revenue at x = 15,
$R'(15) = 6(15) + 36$ $= 90 + 36 = 126$ So D is the correct option.