NCERT Class 12 Mathematics — Application of Derivatives, Exercise 6.1. All 18 questions solved.

Every question here is one idea in two disguises. A derivative $\dfrac{dy}{dx}$ measures how fast $y$ changes as $x$ changes; when both depend on time, the chain rule links their rates:

$$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

That is all a “related rates” question is. The formulas you need are the ones from mensuration — circle, cube, sphere, cone — differentiated once.

Key insight. Read the units in the question and they tell you which derivative is wanted. “cm/s” or “cm³/s” means a rate with respect to time, so the chain rule is involved. “cm²/cm” or “cm³/cm” — as in questions 1, 9 and 13 — means a rate with respect to the variable itself, and no time appears at all.

Questions 1 and 3 look identical and are not: question 1 asks for $\dfrac{dA}{dr}$ and question 3 for $\dfrac{dA}{dt}$. Getting this distinction right is worth more marks in this exercise than any algebra.

Question 1

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

Solution. The area is $A = \pi r^2$, so

$$\frac{dA}{dr} = 2\pi r$$

No time is involved — the answer is a rate per centimetre of radius, which is why the units come out as cm²/cm.

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. With edge $x$, we have $V = x^3$ and $S = 6x^2$. Differentiating the volume with respect to time:

$$\frac{dV}{dt} = 3x^2\frac{dx}{dt} = 8 \quad\Longrightarrow\quad \frac{dx}{dt} = \frac{8}{3x^2}$$

Now feed that into the surface area:

$$\frac{dS}{dt} = 12x\frac{dx}{dt} = 12x \cdot \frac{8}{3x^2} = \frac{32}{x}$$

At $x = 12$ this is $\tfrac{32}{12}$.

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.

Solution. $A = \pi r^2$, so by the chain rule

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(10)(3)$$

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?

Solution. $V = x^3$, so

$$\frac{dV}{dt} = 3x^2\frac{dx}{dt} = 3(100)(3)$$

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?

Solution. The wave’s radius grows at $\tfrac{dr}{dt} = 5$ cm/s, so

$$\frac{dA}{dt} = 2\pi r\frac{dr}{dt} = 2\pi(8)(5)$$

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. $C = 2\pi r$, and the derivative is a constant multiple, so the answer does not depend on $r$:

$$\frac{dC}{dt} = 2\pi\frac{dr}{dt} = 2\pi(0.7)$$

Question 7

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 (a) the perimeter, and (b) the area of the rectangle.

Solution. “Decreasing” means the derivative is negative: $\tfrac{dx}{dt} = -5$ and $\tfrac{dy}{dt} = 4$.

(a) $P = 2(x + y)$, so

$$\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2$$

The perimeter is decreasing at $2$ cm/minute.

(b) $A = xy$ needs the product rule:

$$\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 8(4) + 6(-5) = 32 – 30$$

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.

Solution. $V = \tfrac43\pi r^3$, so

$$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 900$$

At $r = 15$, $4\pi(225) = 900\pi$, so

$$\frac{dr}{dt} = \frac{900}{900\pi}$$

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 latter is $10$ cm.

Solution. “With the radius”, not with time — so differentiate with respect to $r$ and stop:

$$\frac{dV}{dr} = 4\pi r^2 = 4\pi(100)$$

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 $2$ cm/s. How fast is its height on the wall decreasing when the foot of the ladder is $4$ m away from the wall?

Solution. Let $x$ be the distance of the foot from the wall and $y$ the height on the wall. The ladder’s length is fixed, so Pythagoras gives a relation that holds at every instant:

$$x^2 + y^2 = 25$$

Differentiating with respect to time — this is the step that turns a static picture into a rate:

$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$$

When $x = 4$, $y = \sqrt{25 – 16} = 3$, and $\tfrac{dx}{dt} = 2$:

$$\frac{dy}{dt} = -\frac{4}{3}(2) = -\frac83$$

The negative sign confirms that the height is decreasing.

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.

Solution. Differentiate the curve with respect to time:

$$6\frac{dy}{dt} = 3x^2\frac{dx}{dt}$$

The condition is $\tfrac{dy}{dt} = 8\tfrac{dx}{dt}$. Substituting:

$$48\frac{dx}{dt} = 3x^2\frac{dx}{dt} \quad\Longrightarrow\quad x^2 = 16 \quad\Longrightarrow\quad x = \pm4$$

Both signs give a point, and both must be reported. At $x = 4$: $6y = 66$, so $y = 11$. At $x = -4$: $6y = -62$, so $y = -\tfrac{31}{3}$.

Question 12

The radius of an air bubble is increasing at the rate of $\tfrac12$ cm/s. At what rate is the volume of the bubble increasing when the radius is $1$ cm?

Solution. A bubble is a sphere, so

$$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 4\pi(1)\left(\frac12\right)$$

Question 13

A balloon, which always remains spherical, has a variable diameter $\tfrac32(2x+1)$. Find the rate of change of its volume with respect to $x$.

Solution. The question gives the diameter, so halve it first — this is where most of the marks are lost:

$$r = \frac34(2x+1)$$

$$V = \frac43\pi r^3 = \frac43\pi \cdot \frac{27}{64}(2x+1)^3 = \frac{9\pi}{16}(2x+1)^3$$

$$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x+1)^2 \cdot 2 = \frac{27\pi}{8}(2x+1)^2$$

Question 14

Sand is pouring from a pipe at the rate of $12$ cm³/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?

Solution. The constraint $h = \tfrac{r}{6}$ lets you eliminate $r$ before differentiating, which is much easier than differentiating both:

$$r = 6h, \qquad V = \frac13\pi r^2 h = \frac13\pi(36h^2)h = 12\pi h^3$$

$$\frac{dV}{dt} = 36\pi h^2\frac{dh}{dt} = 12$$

At $h = 4$:

$$\frac{dh}{dt} = \frac{12}{36\pi(16)} = \frac{1}{48\pi}$$

Question 15

The total cost $\mathrm{C}(x)$ in Rupees associated with the production of $x$ units of an item is given by $\mathrm{C}(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$. Find the marginal cost when $17$ units are produced.

Solution. Marginal cost is the derivative of total cost — the cost of producing one more unit, at the margin:

$$\mathrm{MC} = \frac{d\mathrm{C}}{dx} = 0.021x^2 – 0.006x + 15$$

At $x = 17$:

$$0.021(289) – 0.006(17) + 15 = 6.069 – 0.102 + 15$$

Question 16

The total revenue in Rupees received from the sale of $x$ units of a product is given by $\mathrm{R}(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.

Solution. Marginal revenue is $\tfrac{d\mathrm{R}}{dx} = 26x + 26$. At $x = 7$ this is $182 + 26$.

Choose the correct answer for questions 17 and 18.

Question 17

The rate of change of the area of a circle with respect to its radius $r$ at $r = 6$ cm is

Solution. $\tfrac{dA}{dr} = 2\pi r = 2\pi(6)$.

Question 18

The total revenue in Rupees received from the sale of $x$ units of a product is given by $\mathrm{R}(x) = 3x^2 + 36x + 5$. The marginal revenue, when $x = 15$, is

Solution. $\tfrac{d\mathrm{R}}{dx} = 6x + 36$, which at $x = 15$ gives $90 + 36$.

Option (C) is what you get by stopping at $6x$ and forgetting the constant term’s contribution.

Common mistakes

• Differentiating with respect to the wrong variable. Questions 1, 9 and 13 want $\tfrac{d}{dr}$ or $\tfrac{d}{dx}$; questions 3, 5 and 12 want $\tfrac{d}{dt}$. The units in the question settle it.
• Using the diameter as the radius. Question 13 gives the diameter. Halving it changes the answer by a factor of $8$.
• Dropping the sign on a decreasing quantity. In question 7 the length is decreasing, so $\tfrac{dx}{dt} = -5$, not $5$. The area would come out as $62$ instead of $2$.
• Substituting the numerical value before differentiating. In question 10 you must differentiate $x^2 + y^2 = 25$ first and only then put $x = 4$. Putting $x = 4$ first makes $x$ a constant and destroys the relation.
• Forgetting that both roots of $x^2 = 16$ count. Question 11 has two points; reporting only $(4, 11)$ loses half the marks.
• Confusing marginal cost with average cost. Marginal cost is $\tfrac{d\mathrm{C}}{dx}$, not $\tfrac{\mathrm{C}(x)}{x}$.

Practise next

• Exercise 6.2 — the sign of the derivative, used to decide where a function increases or decreases.
• Exercise 6.3 — maxima and minima, where the derivative is set to zero rather than evaluated.