NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

NCERT solutions for exercise 6.1 Class 12 Maths chapter 6 surrounds by questions related to the topic rate of change quantities. Students can prepare exercise 6.1 Class 12 Maths using the NCERT solutions for Class 12 Maths chapter 6 exercise 6.1. The Class 12th Maths chapter 6 exercise 6.1 are curated by Mathematics faculties according to the CBSE pattern. All these NCERT solutions given in Class 12 Maths chapter 6 exercise 6.1 give a more deep knowledge on the concept of rate of change of quantities that is discussed in the NCERT book. There are 6 exercises in Class 12 NCERT chapter applications of derivatives. Other than Exercise 6.1 Class 12 Maths, the following exercise, are also present in the NCERT syllabus of this chapter.

Application of Derivatives Exercise 6.2
Application of Derivatives Exercise 6.3
Application of Derivatives Exercise 6.4
Application of Derivatives Exercise 6.5
Application of Derivatives Miscellaneous Exercise

Application of Derivatives Class 12 Chapter 6 Exercise 6.1

Question:1 b) Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm

Answer:

Area of the circle (A) = $\pi r^{2}$
Rate of change of the area of a circle with respect to its radius r = $\frac{dA}{dr}$ = $\frac{d(\pi r^{2})}{dr}$ = $2 \pi r$
So, when r = 4, Rate of change of the area of a circle = $2 \pi (4)$ = $8 \pi$
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is $8 \pi$

Question:2 . The volume of a cube is increasing at the rate of $8 cm^3 /s$ . How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

The volume of the cube(V) = $x^{3}$ where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of $8 cm^3 /s$

we can write $\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ ( By chain rule)

$\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}$

$\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8$ $\Rightarrow 3x^{2}.\frac{dx}{dt} = 8$

$\frac{dx}{dt} = \frac{8}{3x^{2}}$ - (i)
Now, we know that the surface area of the cube(A) is $6x^{2}$

$\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt}$ - (ii)

from equation (i) we know that $\frac{dx}{dt} = \frac{8}{3x^{2}}$

put this value in equation (i)
We get,
$\frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}$
It is given in the question that the value of edge length(x) = 12cm
So,
$\frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} cm^2/s$

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

Answer:

Radius of a circle is increasing uniformly at the rate $\left ( \frac{dr}{dt} \right )$ = 3 cm/s
Area of circle(A) = $\pi r^{2}$
$\frac{dA}{dt} =\frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dA}{dt} =\frac{d \pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 3 = 6\pi r$
It is given that the value of r = 10 cm
So,
$\frac{dA}{dt} = 6\pi \times 10 = 60\pi \ cm^{2}/s$
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is $60\pi \ cm^{2}/s$

Question:4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

It is given that the rate at which edge of cube increase $\left ( \frac{dx}{dt} \right )$ = 3 cm/s
The volume of cube = $x^{3}$
$\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ (By chain rule)
$\frac{dV}{dt} = \frac{dx^{3}}{dx}.\frac{dx}{dt} = 3x^{2}.\frac{dx}{dt} = 3x^{2}\times 3 = 9x^{2} cm^{3}/s$
It is given that the value of x is 10 cm
So,
$\frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900 \ cm^{3}/s$
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is $900 \ cm^{3}/s$

Question:5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Answer:

Given = $\frac{dr}{dt} = 5 \ cm/s$

To find = $\frac{dA}{dt}$ at r = 8 cm

Area of the circle (A) = $\pi r^{2}$
$\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dA}{dt} = \frac{d\pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ cm^{2}/s$
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is $80\pi \ cm^{2}/s$

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?

Answer:

Given = $\frac{dr}{dt} = 0.7 \ cm/s$
To find = $\frac{dC}{dt}$ , where C is circumference
Solution :-

we know that the circumference of the circle (C) = $2\pi r$
$\frac{dC}{dt} = \frac{dC}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dC}{dt} = \frac{d2\pi r}{dr}.\frac{dr}{dt} = 2\pi \times 0.7 = 1.4\pi \ cm/s$
Hence, the rate of increase of its circumference is $1.4\pi \ cm/s$

Question:7(a) . 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 = 8cm and y = 6cm, find the rate of change of

the perimeter of rectangle

Answer:

Given = Length x of a rectangle is decreasing at the rate $(\frac{dx}{dt})$ = -5 cm/minute (-ve sign indicates decrease in rate)
the width y is increasing at the rate $(\frac{dy}{dt})$ = 4 cm/minute
To find = $\frac{dP}{dt}$ and at x = 8 cm and y = 6 cm , where P is perimeter
Solution:-

Perimeter of rectangle(P) = 2(x+y)
$\frac{dP}{dt} = \frac{d(2(x+y))}{dt} = 2\left ( \frac{dx}{dt} + \frac{dy}{dt} \right ) = 2(-5+4) = -2 \ cm/minute$
Hence, Perimeter decreases at the rate of $2 \ cm/minute$

Question:7(b) 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 = 8cm and y = 6cm, find the rates of change of the area of the rectangle.

Answer:

Given same as previous question
Solution:-
Area of rectangle = xy
$\frac{dA}{dt} = \frac{d(xy)}{dt} = \left ( x\frac{dy}{dt} + y\frac{dx}{dt} \right ) = \left ( 8\times 4 + 6 \times (-5) \right ) = (32 -30) = 2 \ cm^{2}/minute$
Hence, the rate of change of area is $2 \ cm^{2}/minute$

Question:8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer:

Given = $\frac{dV}{dt} = 900 \ cm^{3}/s$
To find = $\frac{dr}{dt}$ at r = 15 cm
Solution:-

Volume of sphere(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})} {dr}.\frac{dr}{dt} = \frac{4}{3}\pi\times 3r^{2} \times \frac{dr}{dt}$

$\frac{dV}{dt}= 4 \pi r^{2} \times \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times(15)^{2}} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/s$
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is $\frac{1}{\pi} \ cm/s$

Question:9 . A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answer:

We need to find the value of $\frac{dV}{dr}$ at r =10 cm
The volume of the sphere (V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dr} = \frac{d(\frac{4}{3}\pi r^{3})}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi \ cm^{3}/s$
Hence, the rate at which its volume is increasing with the radius when the later is 10 cm is $400\pi \ cm^{3}/s$

Question:10. A ladder 5 m 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 4 m away from the wall?

Answer:

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given that $\frac{dx}{dt} = 2 \ cm/s$
We need to find the rate at which the height of the ladder decreases $(\frac{dh}{dt})$
length of ladder(L) = 5m and x = 4m (given)
By Pythagoras theorem, we can say that
$h^{2}+x^{2} = L^{2}$
$h^{2} = L^{2} - x^{2}$
$h$ $= \sqrt{L^{2} - x^{2}}$
Differentiate on both sides w.r.t. t
$\frac{dh}{dt} = \frac{d(\sqrt{L^{2} -x^{2}})}{dx}.\frac{dx}{dt} = \frac{1}{2}\frac{-2x}{\sqrt{5^{2}-x^{2}}}.\frac{dx}{dt} = \frac{-x}{\sqrt{25-x^{2}}}\frac{dx}{dt}$
at x = 4

$\frac{dh}{dt} = \frac{-4}{\sqrt{25-16}}\times 2 = \frac{-4}{3} \times 2 =\frac{-8}{3} \ cm/s$
Hence, the rate at which the height of ladder decreases is $\frac{8}{3} \ cm/s$

Question:11. A particle moves along the curve $6y = x^3 + 2$ Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

We need to find the point at which $\frac{dy}{dt} = 8\frac{dx}{dt}$
Given the equation of curve = $6y = x^3 + 2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0$
$= 3x^{2}.\frac{dx}{dt}$
$\frac{dy}{dt} = 8\frac{dx}{dt}$ (required condition)
$6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}$
$3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}$ $\Rightarrow x^{2} = \frac{48}{3} = 16$
$x = \pm 4$
when x = 4 , $y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11$
and
when x = -4 , $y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3}$
So , the coordinates are
$(4,11) \ and \ (-4,\frac{-31}{3})$

Question:12 The radius of an air bubble is increasing at the rate of 1 /2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that $\frac{dr}{dt} = \frac{1}{2} \ cm/s$
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ cm^{3}/s$
Hence, the rate of change in volume is $2\pi \ cm^{3}/s$

Question:13 A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}( 2x +1)$ Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = $\frac{4}{3}\pi r^{3}$
Diameter = $\frac{3}{2}(2x+1)$
So, radius(r) = $\frac{3}{4}(2x+1)$
$\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d(\frac{4}{3}\pi (\frac{3}{4}(2x+1))^{3})}{dx} = \frac{4}{3}\pi\times 3\times\frac{27}{64}(2x+1)^{2}\times 2$
$= \frac{27}{8}\pi (2x+1)^{2}$

Question:14 Sand is pouring from a pipe at the rate of 12 cm ³ /s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Given = $\frac{dV}{dt} = 12 \ cm^{3}/s$ and $h = \frac{1}{6}r$
To find = $\frac{dh}{dt}$ at h = 4 cm
Solution:-

Volume of cone(V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt} = \frac{d(\frac{1}{3}\pi (6h)^{2}h)}{dh}.\frac{dh}{dt} = \frac{1}{3}\pi\times36\times3h^{2}.\frac{dh}{dt} = 36\pi \times(4)^{2}.\frac{dh}{dt}$
$\frac{dV}{dt} = 576\pi.\frac{dh}{dt}$

Question:15 The total cost C(x) in Rupees associated with the production of x units of an
item is given by $C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = $\frac{dC}{dx}$
$C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$
$\frac{dC}{dx} = \frac{d(.007x^{3} - .003x^{2}+15x+400)}{dx} = 3\times .007x^{2} - 2\times.003x+15$
$= .021x^{2} - .006x + 15$
Now, at x = 17
MC $= .021(17)^{2} - .006(17) + 15$
$= 6.069 - .102 + 15$
$= 20.967$
Hence, marginal cost when 17 units are produced is 20.967

Question:16 The total revenue in Rupees received from the sale of x units of a product is
given by $R ( x) = 13 x^2 + 26 x + 15$

Find the marginal revenue when x = 7

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 13 x^2 + 26 x + 15$
$\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)$
at x = 7
$\frac{dR}{dx} = 26(7+1) = 26\times8 = 208$
Hence, marginal revenue when x = 7 is 208

Question:17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π

Answer:

Area of circle(A) = $\pi r^{2}$
$\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r$
Now, at r = 6cm
$\frac{dA}{dr}= 2\pi \times 6 = 12\pi cm^{2}/s$
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is $12\pi cm^{2}/s$
Hence, the correct answer is B

Question:18 The total revenue in Rupees received from the sale of x units of a product is given by $R(x) = 3x^2 + 36x + 5$ . The marginal revenue, when x = 15 is
(A) 116 (B) 96 (C) 90 (D) 126

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 3 x^2 + 36 x + 5$
$\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)$
at x = 15
$\frac{dR}{dx} = 6(15+6) = 6\times21 = 126$
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is D

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

Solving the Exercise 6.1 Class 12 Maths helps in the preparation of exams like JEE main
The concepts explained in Class 12th Maths chapter 6 exercise 6.1 are useful for higher studies in the field of Mathematics, Science and Engineering
The application of derivatives holds a good weightage for various board exams also.

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NCERT Exemplar Solutions Class 12 Maths Chapter 6
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NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

Application of Derivatives Class 12 Chapter 6 Exercise 6.1

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