Home › Class 12 › NCERT Solutions › Chapter 6: Application of Derivatives › Exercise 6.1
NCERT Solutions · Class 12 · Chapter 6: Application of Derivatives
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
Exercise 6.1: Rate of change of quantities. \(\tfrac{dy}{dx}\) is the rate of change of y with respect to x. When two quantities change with time, differentiate the relation between them with respect to t (chain rule: \(\tfrac{dA}{dt} = \tfrac{dA}{dr} \cdot \tfrac{dr}{dt}\)) and substitute the instant's values only after differentiating. Marginal cost / revenue is the derivative of total cost / revenue.
18 questions, 20 partsEvery answer checked by computer algebra Free, no sign-in
Show all solutions Try each question first, then open its solution. Our own step-by-step solutions, set out for step marks.
Exercise 6.1 questions and solutions
Exercise 6.1, Question 1
Rate of change of the area of a circle with respect to its radius r when:
(a) \(r = 3\) cm
Show solution
\[\begin{aligned}&A = \pi r^2 \\ \Rightarrow\ &\dfrac{dA}{dr} = 2\pi r\end{aligned}\] Answer: \(6\pi\) cm\(^2\) per cm
(b) \(r = 4\) cm
Show solution
\(2\pi(4)\). Answer: \(8\pi\) cm\(^2\) per cm
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 2
A cube's volume increases at 8 cm\(^3\)/s. How fast is its surface area increasing when an edge is 12 cm?
Show solution
\(V = x^3\): \[\begin{aligned}&3x^2\dfrac{dx}{dt} = 8 \\ \Rightarrow\ &\dfrac{dx}{dt} = \dfrac{8}{3x^2}\end{aligned}\] \(S = 6x^2\): \[\begin{aligned}\dfrac{dS}{dt} &= 12x\dfrac{dx}{dt} \\ &= \dfrac{32}{x} \\ &= \dfrac{32}{12}\end{aligned}\] Answer: \(\tfrac83\) cm\(^2\)/s
Practise this: Step 2, Board standard →
Exercise 6.1, Question 3
A circle's radius grows at 3 cm/s. Find the rate of increase of its area when \(r = 10\) cm.
Show solution
\[\begin{aligned}\dfrac{dA}{dt} &= 2\pi r\dfrac{dr}{dt} \\ &= 2\pi(10)(3)\end{aligned}\] Answer: \(60\pi\) cm\(^2\)/s
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 4
A cube's edge grows at 3 cm/s. How fast is the volume increasing when the edge is 10 cm?
Show solution
\[\begin{aligned}\dfrac{dV}{dt} &= 3x^2\dfrac{dx}{dt} \\ &= 3(100)(3)\end{aligned}\] Answer: 900 cm\(^3\)/s
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 5
Circular waves spread at 5 cm/s. How fast is the enclosed area increasing when the radius is 8 cm?
Show solution
\[\begin{aligned}\dfrac{dA}{dt} &= 2\pi r\dfrac{dr}{dt} \\ &= 2\pi(8)(5)\end{aligned}\] Answer: \(80\pi\) cm\(^2\)/s
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 6
A circle's radius grows at 0.7 cm/s. Find the rate of increase of the circumference.
Show solution
\[\begin{aligned}&C = 2\pi r \\ \Rightarrow\ &\dfrac{dC}{dt} = 2\pi(0.7)\end{aligned}\] Answer: \(1.4\pi\) cm/s, i.e. \(\tfrac{7\pi}{5}\) cm/s
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 7
A rectangle's length x decreases at 5 cm/min and its width y increases at 4 cm/min. When \(x = 8\) cm and \(y = 6\) cm, find the rate of change of:
(a) the perimeter
Show solution
\(P = 2(x + y)\): \(\dfrac{dP}{dt} = 2(-5 + 4)\). Answer: \(-2\) cm/min (decreasing at 2 cm/min)
(b) the area
Show solution
\(A = xy\): \[\begin{aligned}\dfrac{dA}{dt} &= x\dfrac{dy}{dt} + y\dfrac{dx}{dt} \\ &= 8(4) + 6(-5)\end{aligned}\] Answer: 2 cm\(^2\)/min (increasing)
Practise this: Step 2, Board standard →
Exercise 6.1, Question 8
A spherical balloon is inflated at 900 cm\(^3\)/s. Find the rate at which the radius increases when \(r = 15\) cm.
Show solution
\[\begin{aligned}&V = \tfrac43\pi r^3 \\ \Rightarrow\ &\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}\end{aligned}\] \(\dfrac{dr}{dt} = \dfrac{900}{4\pi(225)}\). Answer: \(\dfrac{1}{\pi}\) cm/s
Practise this: Step 2, Board standard →
Exercise 6.1, Question 9
A spherical balloon has variable radius. Find the rate of change of its volume with respect to the radius when \(r = 10\) cm.
Show solution
\(\dfrac{dV}{dr} = 4\pi r^2 = 4\pi(100)\). Answer: \(400\pi\) cm\(^3\) per cm
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 10
A 5 m ladder leans on a wall; its foot slides away at 2 cm/s. How fast is the top sliding down when the foot is 4 m from the wall?
Show solution
\(x^2 + y^2 = 25\) (x foot distance, y height): \(x\dfrac{dx}{dt} + y\dfrac{dy}{dt} = 0\). At \(x = 4\), \(y = 3\): \(\dfrac{dy}{dt} = -\dfrac{4 \cdot 2}{3}\) cm/s (units of the rate carry through). Answer: The height decreases at \(\tfrac83\) cm/s
Practise this: Step 2, Board standard →
Exercise 6.1, Question 11
A particle moves on \(6y = x^3 + 2\). Find the points where the y-coordinate changes 8 times as fast as the x-coordinate.
Show solution
\(6\dfrac{dy}{dt} = 3x^2\dfrac{dx}{dt}\); with \(\dfrac{dy}{dt} = 8\dfrac{dx}{dt}\): \(48 = 3x^2\), so \(x = \pm4\). \(x = 4\): \(y = \tfrac{66}{6} = 11\). \(x = -4\): \(y = \tfrac{-62}{6} = -\tfrac{31}{3}\). Answer: \((4, 11)\) and \(\left(-4, -\tfrac{31}{3}\right)\)
Practise this: Step 2, Board standard →
Exercise 6.1, Question 12
An air bubble's radius grows at \(\tfrac12\) cm/s. How fast is its volume increasing when \(r = 1\) cm?
Show solution
\[\begin{aligned}\dfrac{dV}{dt} &= 4\pi r^2\dfrac{dr}{dt} \\ &= 4\pi(1)\left(\tfrac12\right)\end{aligned}\] Answer: \(2\pi\) cm\(^3\)/s
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 13
A spherical balloon has diameter \(\tfrac32(2x + 1)\). Find the rate of change of its volume with respect to x.
Show solution
\(r = \tfrac34(2x + 1)\), so \[\begin{aligned}V &= \tfrac43\pi r^3 \\ &= \tfrac{9}{16}\pi(2x + 1)^3\end{aligned}\] \[\dfrac{dV}{dx} = \tfrac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot 2\] Answer: \(\dfrac{27\pi}{8}(2x + 1)^2\)
Practise this: Step 2, Board standard →
Exercise 6.1, Question 14
Sand pours at 12 cm\(^3\)/s into a cone whose height is always one-sixth of the base radius. How fast is the height increasing when it is 4 cm?
Show solution
\(r = 6h\), so \(V = \tfrac13\pi(6h)^2h = 12\pi h^3\). \[\begin{aligned}&\dfrac{dV}{dt} = 36\pi h^2\dfrac{dh}{dt} \\ \Rightarrow\ &12 = 36\pi(16)\dfrac{dh}{dt}\end{aligned}\] Answer: \(\dfrac{1}{48\pi}\) cm/s
Practise this: Step 3, Full marks on long answers →
Exercise 6.1, Question 15
\(C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000\). Find the marginal cost when 17 units are produced.
Show solution
\(C'(x) = 0.021x^2 - 0.006x + 15\). \[\begin{aligned}C'(17) &= 0.021(289) - 0.102 + 15 \\ &= 6.069 - 0.102 + 15\end{aligned}\] Answer: Rs 20.967
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 16
\(R(x) = 13x^2 + 26x + 15\). Find the marginal revenue when \(x = 7\).
Show solution
\(R'(x) = 26x + 26\); \(R'(7) = 182 + 26\). Answer: Rs 208
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 17
The rate of change of the area of a circle with respect to r at \(r = 6\) cm is: (A) \(10\pi\) (B) \(12\pi\) (C) \(8\pi\) (D) \(11\pi\)
Show solution
\(2\pi r = 12\pi\). Answer: (B)
Practise this: Step 1, Secure the basics →
Exercise 6.1, Question 18
\(R(x) = 3x^2 + 36x + 5\). The marginal revenue at \(x = 15\) is: (A) 116 (B) 96 (C) 90 (D) 126
Show solution
\(R'(x) = 6x + 36 = 90 + 36\). Answer: (D) 126
Practise this: Step 1, Secure the basics →
Done the NCERT exercises? The board paper asks more
Application of Derivatives has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.
Application of Derivatives in our sample papers: Sample paper 1 (questions 11, 23, 37) · Sample paper 2 (questions 11, 12, 23, 33) · Sample paper 3 (questions 10, 23, 37) · Sample paper 4 (questions 11, 12, 23, 29, 33) · Sample paper 5 (questions 10, 33, 37).
Also useful: free MCQs and case studies for Application of Derivatives · formulas for this chapter (Class 12 formula sheet, free PDF) · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan
← Chapter 5: Continuity and Differentiability All Chapter 6 exercises Exercise 6.2 → Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in . Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log .