NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
Class 12 Maths Chapter 6, Application of Derivatives (AOD), board exams ke liye ek highly scoring topic hai. Exercise 6.1 strictly Rate of Change of Quantities par focus karti hai. Juggling board prep and competitive exams can feel overwhelming, but clearing these fundamental concepts abhi se will keep your foundation strong. CBSE 2026-27 session ke liye yeh chapter directly calculus weightage ko boost karta hai. Chaliye, step-by-step har ek question ko decode karte hain!
What You Will Learn in This Exercise
- Rate of Change: How one quantity changes with respect to another (e.g., area changing with radius).
- Time Derivatives: Calculating the rate of change of variables with respect to time ($dt$).
- Marginal Cost & Revenue: Practical applications of derivatives in economics and business.
- Chain Rule Application: Using differentiation to link varying geometric dimensions (like volume and surface area).
- Real-Life Physics Context: Understanding velocity and dimensional growth in spheres, circles, and cones.
Key Formulas Used in This Exercise
| Shape / Concept | Formula | Rate of Change Equation |
|---|---|---|
| Circle (Area) | $A = \pi r^2$ | $$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$ |
| Circle (Circumference) | $C = 2\pi r$ | $$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$$ |
| Cube (Volume) | $V = x^3$ | $$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$$ |
| Cube (Surface Area) | $S = 6x^2$ | $$\frac{dS}{dt} = 12x \frac{dx}{dt}$$ |
| Sphere (Volume) | $V = \frac{4}{3}\pi r^3$ | $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ |
| Cone (Volume) | $V = \frac{1}{3}\pi r^2 h$ | Depends on relation of $r$ & $h$ |
| Marginal Cost (MC) | $C(x)$ | $$MC = \frac{dC}{dx}$$ |
| Marginal Revenue (MR) | $R(x)$ | $$MR = \frac{dR}{dx}$$ |
Exercise 6.1 Solutions
Question 1: Find the rate of change of the area of a circle with respect to its radius $r$ when: (a) $r=3 \text{ cm}$ (b) $r=4 \text{ cm}$
Given:
Radius of the circle = $r$
Area of the circle = $A$
Formula Used:
$$A = \pi r^2$$
Step 1: Differentiate the area with respect to the radius $r$.
$$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$$
Step 2: Calculate the rate of change for the given values of $r$.
(a) When $r=3 \text{ cm}$:
$$\frac{dA}{dr} = 2\pi (3) = 6\pi \text{ cm}^2/\text{cm}$$
(b) When $r=4 \text{ cm}$:
$$\frac{dA}{dr} = 2\pi (4) = 8\pi \text{ cm}^2/\text{cm}$$
Final Answer:
(a) $6\pi \text{ cm}^2/\text{cm}$ and (b) $8\pi \text{ cm}^2/\text{cm}$
Question 2: The volume of a cube is increasing at the rate of $8 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is 12 cm?
Given:
Rate of change of volume = $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$
Edge length = $x = 12 \text{ cm}$
Formula Used:
Volume of cube $V = x^3$
Surface Area $S = 6x^2$
Step 1: Find the rate of change of the edge $\frac{dx}{dt}$.
$$V = x^3$$
Differentiating w.r.t $t$:
$$\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$$
$$8 = 3x^2 \cdot \frac{dx}{dt}$$
$$\frac{dx}{dt} = \frac{8}{3x^2}$$
Step 2: Find the rate of change of the surface area $\frac{dS}{dt}$.
$$S = 6x^2$$
$$\frac{dS}{dt} = 12x \cdot \frac{dx}{dt}$$
Substitute $\frac{dx}{dt}$:
$$\frac{dS}{dt} = 12x \left(\frac{8}{3x^2}\right) = \frac{32}{x}$$
Step 3: Evaluate at $x = 12 \text{ cm}$.
$$\frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{s}$$
Final Answer:
The surface area is increasing at $8/3 \text{ cm}^2/\text{s}$.
Question 3: The radius of a circle is increasing uniformly at the rate of $3 \text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Given:
$\frac{dr}{dt} = 3 \text{ cm/s}$
$r = 10 \text{ cm}$
Formula Used:
$$A = \pi r^2$$
Step 1: Differentiate area w.r.t time $t$.
$$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$$
Step 2: Substitute the known values.
$$\frac{dA}{dt} = 2\pi (10) \cdot (3)$$
$$\frac{dA}{dt} = 60\pi$$
Final Answer:
The area is increasing at $60\pi \text{ cm}^2/\text{s}$.
Question 4: An edge of a variable cube is increasing at the rate of $3 \text{ cm/s}$. How fast is the volume of the cube increasing when the edge is 10 cm long?
Given:
$\frac{dx}{dt} = 3 \text{ cm/s}$
$x = 10 \text{ cm}$
Formula Used:
$$V = x^3$$
Step 1: Differentiate volume w.r.t time $t$.
$$\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$$
Step 2: Substitute the known values.
$$\frac{dV}{dt} = 3(10)^2 \cdot (3)$$
$$\frac{dV}{dt} = 3(100) \cdot 3 = 900$$
Final Answer:
The volume is increasing at $900 \text{ cm}^3/\text{s}$.
Question 5: A stone is dropped into a quiet lake and waves move in circles at the speed of $5 \text{ cm/s}$. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Given:
Speed of wave (rate of change of radius) $\frac{dr}{dt} = 5 \text{ cm/s}$
Radius $r = 8 \text{ cm}$
Formula Used:
$$A = \pi r^2$$
Step 1: Differentiate area w.r.t time $t$.
$$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$$
Step 2: Substitute the values.
$$\frac{dA}{dt} = 2\pi (8) \cdot (5)$$
$$\frac{dA}{dt} = 80\pi$$
Final Answer:
The enclosed area is increasing at $80\pi \text{ cm}^2/\text{s}$.
Question 6: The radius of a circle is increasing at the rate of $0.7 \text{ cm/s}$. What is the rate of increase of its circumference?
Given:
$\frac{dr}{dt} = 0.7 \text{ cm/s}$
Formula Used:
$$C = 2\pi r$$
Step 1: Differentiate circumference w.r.t time $t$.
$$\frac{dC}{dt} = 2\pi \cdot \frac{dr}{dt}$$
Step 2: Substitute the value of $\frac{dr}{dt}$.
$$\frac{dC}{dt} = 2\pi (0.7) = 1.4\pi$$
Final Answer:
The rate of increase of circumference is $1.4\pi \text{ cm/s}$.
Question 7: The length $x$ of a rectangle is decreasing at the rate of $5 \text{ cm/min}$ and the width $y$ is increasing at the rate of $4 \text{ cm/min}$. When $x=8 \text{ cm}$ and $y=6 \text{ cm}$, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Given:
$\frac{dx}{dt} = -5 \text{ cm/min}$ (decreasing)
$\frac{dy}{dt} = 4 \text{ cm/min}$ (increasing)
$x = 8 \text{ cm}, y = 6 \text{ cm}$
Formula Used:
Perimeter $P = 2(x+y)$
Area $A = xy$
Step 1: Rate of change of perimeter.
$$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right)$$
$$\frac{dP}{dt} = 2(-5 + 4) = 2(-1) = -2 \text{ cm/min}$$
Step 2: Rate of change of area.
$$\frac{dA}{dt} = x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} \text{ (Product Rule)}$$
$$\frac{dA}{dt} = (8)(4) + (6)(-5)$$
$$\frac{dA}{dt} = 32 - 30 = 2 \text{ cm}^2/\text{min}$$
Final Answer:
(a) The perimeter is decreasing at $2 \text{ cm/min}$.
(b) The area is increasing at $2 \text{ cm}^2/\text{min}$.
Question 8: A balloon, which always remains spherical on inflation, is being inflated by pumping in $900 \text{ cubic centimeters of gas per second}$. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Given:
$\frac{dV}{dt} = 900 \text{ cm}^3/\text{s}$
$r = 15 \text{ cm}$
Formula Used:
Volume of a sphere $V = \frac{4}{3}\pi r^3$
Step 1: Differentiate volume w.r.t time $t$.
$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$
Step 2: Isolate $\frac{dr}{dt}$ and substitute values.
$$900 = 4\pi (15)^2 \frac{dr}{dt}$$
$$900 = 4\pi (225) \frac{dr}{dt}$$
$$900 = 900\pi \frac{dr}{dt}$$
$$\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi}$$
Final Answer:
The radius is increasing at $1/\pi \text{ cm/s}$.
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.
Given:
$r = 10 \text{ cm}$
Formula Used:
$$V = \frac{4}{3}\pi r^3$$
Step 1: Differentiate volume w.r.t radius $r$ (not time!).
$$\frac{dV}{dr} = \frac{4}{3}\pi (3r^2) = 4\pi r^2$$
Step 2: Substitute $r = 10$.
$$\frac{dV}{dr} = 4\pi (10)^2 = 400\pi$$
Final Answer:
The volume is increasing at $400\pi \text{ cm}^3/\text{cm}$.
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 $2 \text{ cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Given:
Length of ladder $L = 5 \text{ m} = 500 \text{ cm}$
$\frac{dx}{dt} = 2 \text{ cm/s}$
Distance of foot from wall $x = 4 \text{ m} = 400 \text{ cm}$
Formula Used:
Pythagoras Theorem: $x^2 + y^2 = L^2$
Step 1: Find initial height $y$.
$$(400)^2 + y^2 = (500)^2$$
$$y^2 = 250000 - 160000 = 90000$$
$$y = 300 \text{ cm}$$
Step 2: Differentiate Pythagorean equation w.r.t time $t$.
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$
$$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$
Step 3: Substitute the values to find $\frac{dy}{dt}$.
$$(400)(2) + (300)\frac{dy}{dt} = 0$$
$$800 + 300\frac{dy}{dt} = 0$$
$$\frac{dy}{dt} = \frac{-800}{300} = \frac{-8}{3} \text{ cm/s}$$
Final Answer:
The height on the wall is decreasing at the rate of $8/3 \text{ cm/s}$.
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.
Given:
Equation of curve: $6y = x^3 + 2$
Condition: $\frac{dy}{dt} = 8 \frac{dx}{dt}$
Step 1: Differentiate the curve equation w.r.t time $t$.
$$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$$
Step 2: Substitute the given condition $\frac{dy}{dt} = 8 \frac{dx}{dt}$.
$$6 \left( 8 \frac{dx}{dt} \right) = 3x^2 \frac{dx}{dt}$$
$$48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$$
Dividing by $\frac{dx}{dt}$ (assuming it's non-zero):
$$3x^2 = 48$$
$$x^2 = 16 \implies x = \pm 4$$
Step 3: Find corresponding $y$ values.
When $x = 4$:
$$6y = (4)^3 + 2 = 64 + 2 = 66 \implies y = 11$$
When $x = -4$:
$$6y = (-4)^3 + 2 = -64 + 2 = -62 \implies y = -31/3$$
Final Answer:
The required points are $(4, 11)$ and $(-4, -31/3)$.
Question 12: The radius of an air bubble is increasing at the rate of $1/2 \text{ cm/s}$. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Given:
$\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}$
$r = 1 \text{ cm}$
Formula Used:
Volume of a sphere $V = \frac{4}{3}\pi r^3$
Step 1: Differentiate volume w.r.t time $t$.
$$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$
Step 2: Substitute values.
$$\frac{dV}{dt} = 4\pi (1)^2 \left(\frac{1}{2}\right) = 2\pi$$
Final Answer:
The volume is increasing at $2\pi \text{ cm}^3/\text{s}$.
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 $x$.
Given:
Diameter $D = \frac{3}{2}(2x+1)$
Radius $r = \frac{D}{2} = \frac{3}{4}(2x+1)$
Formula Used:
$$V = \frac{4}{3}\pi r^3$$
Step 1: Express volume $V$ in terms of $x$.
$$V = \frac{4}{3}\pi \left[ \frac{3}{4}(2x+1) \right]^3$$
$$V = \frac{4}{3}\pi \left(\frac{27}{64} \right) (2x+1)^3$$
$$V = \frac{9\pi}{16} (2x+1)^3$$
Step 2: Differentiate $V$ w.r.t $x$.
$$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot \frac{d}{dx}(2x + 1)$$
$$\frac{dV}{dx} = \frac{27\pi}{16} (2x+1)^2 \cdot 2$$
$$\frac{dV}{dx} = \frac{27\pi}{8} (2x+1)^2$$
Final Answer:
The rate of change of volume is $\frac{27\pi}{8}(2x+1)^2$.
Question 14: Sand is pouring from a pipe at the rate of $12 \text{ cm}^3/\text{s}$. 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?
Given:
$\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$
$h = \frac{r}{6} \implies r = 6h$
$h = 4 \text{ cm}$
Formula Used:
Volume of a cone $V = \frac{1}{3}\pi r^2 h$
Step 1: Express volume entirely in terms of $h$.
$$V = \frac{1}{3}\pi (6h)^2 h$$
$$V = \frac{1}{3}\pi (36h^2) h = 12\pi h^3$$
Step 2: Differentiate w.r.t time $t$.
$$\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$$
Step 3: Substitute the known values to find $\frac{dh}{dt}$.
$$12 = 36\pi (4)^2 \frac{dh}{dt}$$
$$12 = 36\pi (16) \frac{dh}{dt}$$
$$12 = 576\pi \frac{dh}{dt}$$
$$\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}$$
Final Answer:
The height is increasing at the rate of $1 / (48\pi) \text{ cm/s}$.
Question 15: The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$. Find the marginal cost when 17 units are produced.
Given:
$C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000$
$x = 17$
Formula Used:
Marginal Cost $MC = \frac{dC}{dx}$
Step 1: Differentiate $C(x)$ w.r.t $x$.
$$MC = \frac{d}{dx}(0.007x^3 - 0.003x^2 + 15x + 4000)$$
$$MC = 0.021x^2 - 0.006x + 15$$
Step 2: Substitute $x = 17$
$$MC = 0.021(17)^2 - 0.006(17) + 15$$
$$MC = 0.021(289) - 0.102 + 15$$
$$MC = 6.069 - 0.102 + 15 = 20.967$$
Final Answer:
The Marginal Cost is Rs. 20.967.
Question 16: The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x=7$.
Given:
$R(x) = 13x^2 + 26x + 15$
$x = 7$
Formula Used:
Marginal Revenue $MR = \frac{dR}{dx}$
Step 1: Differentiate $R(x)$ w.r.t $x$.
$$MR = \frac{d}{dx}(13x^2 + 26x + 15)$$
$$MR = 26x + 26$$
Step 2: Substitute $x = 7$.
$$MR = 26(7) + 26 = 182 + 26 = 208$$
Final Answer:
The Marginal Revenue is Rs. 208.
Question 17: The rate of change of the area of a circle with respect to its radius $r$ at $r=6 \text{ cm}$ is:
(A) $10\pi$
(B) $12\pi$
(C) $8\pi$
(D) $11\pi$
Given: $A = \pi r^2$ and $r = 6 \text{ cm}$
Step 1: Differentiate area w.r.t $r$.
$$\frac{dA}{dr} = 2\pi r$$
Step 2: Substitute $r = 6$
$$\frac{dA}{dr} = 2\pi(6) = 12\pi$$
Final Answer: (B) $12\pi$
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
Given: $R(x) = 3x^2 + 36x + 5$ and $x = 15$
Step 1: Differentiate $R(x)$ w.r.t $x$.
$$MR = \frac{d}{dx}(3x^2 + 36x + 5) = 6x + 36$$
Step 2: Substitute $x = 15$.
$$MR = 6(15) + 36 = 90 + 36 = 126$$
Final Answer: (D) 126
Common Doubts & FAQs
Differentiation kab radius ($r$) ke respect mein karni hai, aur kab time ($t$) ke?
Question 10 (Ladder problem) mein $\frac{dx}{dt}$ positive kyun hai aur $\frac{dy}{dt}$ negative kyun aati hai?
Marginal Cost wale questions itne ajeeb kyun lagte hain?
Chain rule lagana bhool gaye toh exam mein marks katenge?
Q14 mein $r$ ko hi substitute kyun kiya, $h$ ko kyun nahi?
Why is the derivative of Marginal Revenue calculated w.r.t units ($x$)?
What is the best way to practice this exercise?
Conclusion: AOD Chapter 6 initially thoda overwhelming lag sakta hai, especially agar aap baki entrance exams ko bhi parallel balance kar rahe ho. But Exercise 6.1 aapke basic derivatives application ki neev pakki karti hai. Chahe Agra ke weather kitna bhi extreme ho, apne study room mein shaant dimaag se in calculations ko step-by-step master karte jao. Aap really lucky ho ki abhi aapke paas aise structured resources available hain! Practice regularly, especially wo ladder aur cone wale conceptual sums. Revise all 3D shape formulas, aur confidence ke sath in NCERT questions ko khud ek baar dobara solve zaroor karo. Best of luck for your boards!