Question 1:
The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is \displaystyle 8 cm long?

Answer:

\displaystyle \text{Given: } A = x^2 \text{ and } \frac{dx}{dt} = 4\,\text{cm/min}.

\displaystyle \text{Let } x \text{ be the side of the square and } A \text{ be its area at any time } t.

\displaystyle A = x^2

\displaystyle \Rightarrow \frac{dA}{dt} = 2x \frac{dx}{dt}

\displaystyle \Rightarrow \frac{dA}{dt} = 2 \times 8 \times 4 \quad \left[\because x = 8\,\text{cm and } \frac{dx}{dt} = 4\,\text{cm/min}\right]

\displaystyle \Rightarrow \frac{dA}{dt} = 64\,\text{cm}^2/\text{min}.

Question 2:
An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when the edge is \displaystyle 10 cm long?

Answer:

\displaystyle \text{Let } x \text{ be the side and } V \text{ be the volume of the cube at any time } t.

\displaystyle V = x^3

\displaystyle \Rightarrow \frac{dV}{dt} = 3x^2 \frac{dx}{dt}

\displaystyle \Rightarrow \frac{dV}{dt} = 3 \times (10)^2 \times 3 \quad \left[\because x = 10\,\text{cm and } \frac{dx}{dt} = 3\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dV}{dt} = 900\,\text{cm}^3/\text{sec}.

Question 3:
The side of a square is increasing at the rate of 0.2 cm/sec. Find the rate of increase of the perimeter of the square.

Answer:

\displaystyle \text{Let } x \text{ be the side and } P \text{ be the perimeter of the square at any time } t.

\displaystyle P = 4x

\displaystyle \Rightarrow \frac{dP}{dt} = 4 \frac{dx}{dt}

\displaystyle \Rightarrow \frac{dP}{dt} = 4 \times 0.2 \quad \left[\because \frac{dx}{dt} = 0.2\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dP}{dt} = 0.8\,\text{cm/sec}.

Question 4:
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference? 

Answer:

\displaystyle \text{Let } r \text{ be the radius and } C \text{ be the circumference of the circle at any time } t.

\displaystyle C = 2\pi r

\displaystyle \Rightarrow \frac{dC}{dt} = 2\pi \frac{dr}{dt}

\displaystyle \Rightarrow \frac{dC}{dt} = 2\pi \times 0.7 \quad \left[\because \frac{dr}{dt} = 0.7\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dC}{dt} = 1.4\pi\,\text{cm/sec}.

Question 5:
The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area when the radius is \displaystyle 7 cm.

Answer:

\displaystyle \text{Let } r \text{ be the radius and } S \text{ be the surface area of the spherical ball at any time } t.

\displaystyle S = 4\pi r^2

\displaystyle \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt}

\displaystyle \Rightarrow \frac{dS}{dt} = 8\pi \times 7 \times 0.2 \quad \left[\because r = 7\,\text{cm and } \frac{dr}{dt} = 0.2\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dS}{dt} = 11.2\pi\,\text{cm}^2/\text{sec}.

Question 6:
A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is \displaystyle 15 cm. 

Answer:

\displaystyle \text{Let } r \text{ be the radius and } V \text{ be the volume of the spherical balloon at any time } t.

\displaystyle V = \frac{4}{3}\pi r^3

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi r^2}\,\frac{dV}{dt}

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{900}{4\pi (15)^2} \quad \left[\because r = 15\,\text{cm and } \frac{dV}{dt} = 900\,\text{cm}^3/\text{sec}\right]

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{900}{900\pi}

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{\pi}\,\text{cm/sec}.

Question 7:
The radius of an air bubble is increasing at the rate of 0.5 cm/sec. At what rate is the volume of the bubble increasing when the radius is \displaystyle 1 cm?

Answer:

\displaystyle \text{Let } r \text{ be the radius and } V \text{ be the volume of the air bubble at any time } t.

\displaystyle V = \frac{4}{3}\pi r^3

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi (1)^2 \times 0.5 \quad \left[\because r = 1\,\text{cm and } \frac{dr}{dt} = 0.5\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dV}{dt} = 2\pi\,\text{cm}^3/\text{sec}.

Question 8:
A man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:

\displaystyle \text{Let } AB \text{ be the lamp post. Suppose at any time } t, \text{ the man } CD \text{ is at a distance of } \\ x \text{ km from the lamp post and } y \text{ m be the length of his shadow } CE.

\displaystyle \text{Since } \triangle ABE \text{ and } \triangle CDE \text{ are similar},

\displaystyle \frac{AB}{CD} = \frac{AE}{CE}

\displaystyle \Rightarrow \frac{6}{2} = \frac{x + y}{y}

\displaystyle \Rightarrow 3y = x + y

\displaystyle \Rightarrow x = 2y

\displaystyle \Rightarrow \frac{dx}{dt} = 2\frac{dy}{dt}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{2}\frac{dx}{dt}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{2}(5) \quad \left[\because \frac{dx}{dt} = 5\,\text{km/hr}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{5}{2}\,\text{km/hr}.

Question 9:
A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is \displaystyle 10 cm, how fast is the enclosed area increasing?

Answer:

\displaystyle \text{Let } r \text{ be the radius and } A \text{ be the area of the circle at any time } t.

\displaystyle A = \pi r^2

\displaystyle \Rightarrow \frac{dA}{dt} = 2\pi r \frac{dr}{dt}

\displaystyle \Rightarrow \frac{dA}{dt} = 2\pi \times 4 \times 10 \quad \left[\because r = 4\,\text{cm and } \frac{dr}{dt} = 10\,\text{cm/sec}\right]

\displaystyle \Rightarrow \frac{dA}{dt} = 80\pi\,\text{cm}^2/\text{sec}.

Question 10:
A man 160 cm tall walks away from a source of light situated at the top of a pole 6 m high at the rate of 1.1 m/sec. How fast is the length of his shadow increasing when he is \displaystyle 1 m away from the pole?

Answer:

\displaystyle \text{Let } AB \text{ be the lamp post. Suppose at any time } t, \text{ the man } \\ CD \text{ is at a distance of } x \text{ m from the lamp post and } y \text{ m is the length of his shadow } CE.

\displaystyle \text{Since } \triangle ABE \text{ and } \triangle CDE \text{ are similar},

\displaystyle \frac{AB}{CD} = \frac{AE}{CE}

\displaystyle \Rightarrow \frac{6}{1.6} = \frac{x + y}{y}

\displaystyle \Rightarrow \frac{x}{y} = \frac{6}{1.6} - 1

\displaystyle \Rightarrow \frac{x}{y} = \frac{4.4}{1.6}

\displaystyle \Rightarrow y = \frac{16}{44}x

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{16}{44}\frac{dx}{dt}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{16}{44} \times 1.1 \quad \left[\because \frac{dx}{dt} = 1.1\,\text{m/sec}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = 0.4\,\text{m/sec}.

Question 11:
A man 180 cm tall walks at a rate of 2 m/sec away from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is \displaystyle 3 m away from the base of the light?

Answer:

\displaystyle \text{Let } AB \text{ be the lamp post. Suppose at any time } t, \text{ the man } \\ CD \text{ is at a distance } x \text{ m from the lamp post and } y \text{ m is the length of his shadow } CE.

\displaystyle \text{Since } \triangle ABE \text{ and } \triangle CDE \text{ are similar},

\displaystyle \frac{AB}{CD} = \frac{AE}{CE}

\displaystyle \Rightarrow \frac{9}{1.8} = \frac{x + y}{y}

\displaystyle \Rightarrow \frac{x}{y} = \frac{9}{1.8} - 1

\displaystyle \Rightarrow \frac{x}{y} = \frac{7.2}{1.8}

\displaystyle \Rightarrow x = 4y

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{4}\frac{dx}{dt}

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{4} \times 2 \quad \left[\because \frac{dx}{dt} = 2\,\text{m/sec}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = 0.5\,\text{m/sec}.

Question 12:
A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. How fast is the angle \displaystyle \theta between the ladder and the ground changing when the foot of the ladder is \displaystyle 12 m away from the wall?

Answer:

\displaystyle \text{Let the bottom of the ladder be at a distance of } x \text{ m from the wall and its top be at a height of } \\ y \text{ m from the ground.}

\displaystyle \text{Then } \tan\theta = \frac{y}{x} \text{ and } x^2 + y^2 = 13^2.

\displaystyle \Rightarrow x^2 \left(1 + \tan^2\theta\right) = 169

\displaystyle \Rightarrow x^2 \sec^2\theta = 169

\displaystyle \Rightarrow \sec^2\theta = \frac{169}{x^2}

\displaystyle \Rightarrow 2\sec^2\theta \tan\theta \frac{d\theta}{dt} = 169\left(-\frac{2}{x^3}\right)\frac{dx}{dt}

\displaystyle \Rightarrow \frac{d\theta}{dt} = \frac{-338}{x^3 \, 2\sec^2\theta \tan\theta}\,\frac{dx}{dt} \quad \text{(1)}

\displaystyle \text{When } x = 12,\; y = \sqrt{169 - 144} = 5\,\text{m}.

\displaystyle \text{So, } \sec\theta = \frac{13}{12} \text{ and } \tan\theta = \frac{5}{12}.

\displaystyle \text{From (1),}

\displaystyle \frac{d\theta}{dt} = \frac{-338 \times 1.5}{(12)^3 \times 2 \times \left(\frac{13}{12}\right)^2 \times \frac{5}{12}}

\displaystyle \Rightarrow \frac{d\theta}{dt} = -0.3\,\text{rad/sec}.

Question 13:
A particle moves along the curve \displaystyle y = x^2 + 2x. At what point(s) on the curve are the \displaystyle x and \displaystyle y coordinates changing at the same rate?

Answer:

\displaystyle \text{Here, } y = x^2 + 2x.

\displaystyle \Rightarrow \frac{dy}{dt} = (2x + 2)\frac{dx}{dt}.

\displaystyle \Rightarrow 2x + 2 = 1 \quad \left[\because \frac{dy}{dt} = \frac{dx}{dt}\right]

\displaystyle \Rightarrow 2x = -1

\displaystyle \Rightarrow x = -\frac{1}{2}.

\displaystyle \text{Substituting } x = -\frac{1}{2} \text{ in } y = x^2 + 2x, \text{ we get}

\displaystyle y = \frac{1}{4} - 1 = -\frac{3}{4}.

\displaystyle \text{Hence, the coordinates of the point are } \left(-\frac{1}{2}, -\frac{3}{4}\right).

Question 14:
If \displaystyle y = 7x - x^3 and \displaystyle x increases at the rate of 4 units per second, how fast is the slope of the curve changing when \displaystyle x = 2?

Answer:

\displaystyle \text{Here, } y = 7x - x^3.

\displaystyle \Rightarrow \frac{dy}{dx} = 7 - 3x^2.

\displaystyle \text{Let } s \text{ be the slope. Then, } s = 7 - 3x^2.

\displaystyle \Rightarrow \frac{ds}{dt} = -6x \frac{dx}{dt}.

\displaystyle \Rightarrow \frac{ds}{dt} = -6 \times 2 \times 4 \quad \left[\because x = 2 \text{ and } \frac{dx}{dt} = 4\,\text{units/sec}\right]

\displaystyle \Rightarrow \frac{ds}{dt} = -48.

Question 15:
A particle moves along the curve \displaystyle y = x^3. Find the points on the curve at which the \displaystyle y-coordinate changes three times more rapidly than the \displaystyle x-coordinate.

Answer:

\displaystyle \text{According to the question, } \frac{dy}{dt} = 3\frac{dx}{dt}.

\displaystyle \text{Now, } y = x^3.

\displaystyle \Rightarrow \frac{dy}{dt} = 3x^2 \frac{dx}{dt}.

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

\displaystyle \Rightarrow x^2 = 1.

\displaystyle \Rightarrow x = \pm 1.

\displaystyle \text{Substituting } x = \pm 1 \text{ in } y = x^3, \text{ we get}

\displaystyle y = \pm 1.

\displaystyle \text{So the points are } (1,1) \text{ and } (-1,-1).

Question 16:
Find an angle \displaystyle \theta
(i) which increases twice as fast as its cosine
(ii) whose rate of increase is twice the rate of decrease of its cosine

Answer:

(i)

\displaystyle \text{Let } x = \cos\theta.

\displaystyle \text{Differentiating both sides with respect to } t, \text{ we get}

\displaystyle \frac{dx}{dt} = \frac{d(\cos\theta)}{dt} = -\sin\theta \frac{d\theta}{dt}.

\displaystyle \text{But it is given that } \frac{d\theta}{dt} = 2\frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dx}{dt} = -\sin\theta \left(2\frac{dx}{dt}\right).

\displaystyle \Rightarrow 1 = -2\sin\theta \quad \left[\because \frac{dx}{dt} \neq 0\right]

\displaystyle \Rightarrow \sin\theta = -\frac{1}{2}.

\displaystyle \Rightarrow \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.

\displaystyle \text{Hence, } \theta = \frac{7\pi}{6}.

(ii)

\displaystyle \text{Let } x = \cos\theta.

\displaystyle \text{Differentiating both sides with respect to } t, \text{ we get}

\displaystyle \frac{dx}{dt} = \frac{d(\cos\theta)}{dt} = -\sin\theta \frac{d\theta}{dt}.

\displaystyle \text{But it is given that } \frac{d\theta}{dt} = -2\frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dx}{dt} = -\sin\theta \left(-2\frac{dx}{dt}\right).

\displaystyle \Rightarrow 1 = 2\sin\theta \quad \left[\because \frac{dx}{dt} \neq 0\right]

\displaystyle \Rightarrow \sin\theta = \frac{1}{2}.

\displaystyle \Rightarrow \theta = \frac{\pi}{6}.

\displaystyle \text{Hence, } \theta = \frac{\pi}{6}.

Question 17:
The top of a ladder 6 metres long rests against a vertical wall. When the foot of the ladder is 4 metres from the wall, it is sliding away at the rate of 0.5 m/sec.
(a) How fast is the top sliding downwards at this instant?
(b) How far is the foot from the wall when the top and foot move at the same rate?

Answer:

\displaystyle \text{Let the bottom of the ladder be at a distance of } x \text{ m from the wall and its top be at a height of } \\ y \text{ m from the ground.}

\displaystyle \text{Here, } x^2 + y^2 = 36.

\displaystyle \Rightarrow 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad \text{or} \quad 2x\frac{dx}{dt} = -2y\frac{dy}{dt}. \quad (1)

\displaystyle \text{When } x = 4,\; y = \sqrt{36 - 16} = 2\sqrt{5}.

\displaystyle \Rightarrow 2 \times 4 \times 0.5 = -2 \times 2\sqrt{5}\,\frac{dy}{dt} \quad \left[\because \frac{dx}{dt} = 0.5\,\text{m/sec}\right]

\displaystyle \Rightarrow \frac{dy}{dt} = -\frac{1}{\sqrt{5}}\,\text{m/sec}.

\displaystyle \text{From (1),}

\displaystyle 2x\frac{dx}{dt} = -2y\frac{dy}{dt} \Rightarrow x\frac{dx}{dt} = -y\frac{dy}{dt}.

\displaystyle \Rightarrow x = -y \quad \left[\because \frac{dx}{dt} = \frac{dy}{dt}\right].

\displaystyle \text{Substituting } x = -y \text{ in } x^2 + y^2 = 36, \text{ we get}

\displaystyle x^2 + x^2 = 36

\displaystyle \Rightarrow x^2 = 18

\displaystyle \Rightarrow x = 3\sqrt{2}\,\text{m}.

Question 18:
A balloon is in the form of a right circular cone surmounted by a hemisphere. The diameter equals the height of the cone. How fast is its volume changing with respect to its total height \displaystyle h when \displaystyle h = 9 cm?

Answer:

\displaystyle \text{Let } r \text{ be the radius of the hemisphere, } h \text{ be the height of the cone and } V \text{ be the total volume.}

\displaystyle \text{Then, } H = h + r.

\displaystyle \Rightarrow H = 3r \quad \left[\because h = 2r\right].

\displaystyle \Rightarrow \frac{dH}{dt} = 3\frac{dr}{dt}.

\displaystyle \text{When } H = 9\,\text{cm}, \; r = 3\,\text{cm}.

\displaystyle \text{Volume } V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3.

\displaystyle \text{Substituting } h = 2r, \text{ we get}

\displaystyle V = \frac{1}{3}\pi r^2 (2r) + \frac{2}{3}\pi r^3.

\displaystyle \Rightarrow V = \frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3.

\displaystyle \Rightarrow V = \frac{4}{3}\pi r^3.

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{4\pi r^2}{3}\frac{dH}{dt}.

\displaystyle \Rightarrow \frac{dV}{dH} = \frac{4\pi (3)^2}{3}.

\displaystyle \Rightarrow \frac{dV}{dH} = 12\pi\,\text{cm}^3/\text{cm}.

Question 19:
Water is running into an inverted cone at the rate of \displaystyle \pi cubic metres per minute. The cone has height 10 m and base radius 5 m. How fast is the water level rising when the water stands 7.5 m below the base?

Answer:

\displaystyle \text{Let } r \text{ be the radius, } h \text{ be the height and } V \text{ be the volume of the cone at any time } t.

\displaystyle \text{Then, } V = \frac{1}{3}\pi r^2 h.

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{1}{3}\pi r^2 \frac{dh}{dt} + \frac{2}{3}\pi r h \frac{dr}{dt}.

\displaystyle \text{Now, } \frac{h}{r} = \frac{10}{5} \Rightarrow r = \frac{h}{2} \text{ and } \frac{dh}{dt} = 2\frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 \frac{dh}{dt} + \frac{2}{3}\pi \left(\frac{h}{2}\right) h \left(\frac{1}{2}\frac{dh}{dt}\right).

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{\pi}{3}\left(\frac{h^2}{4}\frac{dh}{dt} + \frac{h^2}{2}\frac{dh}{dt}\right).

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{\pi}{3} \times \frac{3h^2}{4}\frac{dh}{dt}.

\displaystyle \Rightarrow \frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}.

\displaystyle \Rightarrow \frac{\pi h^2}{4}\frac{dh}{dt} = \pi.

\displaystyle \Rightarrow \frac{dh}{dt} = \frac{4}{h^2}.

\displaystyle \Rightarrow \frac{dh}{dt} = \frac{4}{(2.5)^2}.

\displaystyle \Rightarrow \frac{dh}{dt} = 0.64\,\text{m/min}.

Question 20:
A man 2 metres high walks at a uniform speed of 6 km/h away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

Answer:

\displaystyle \text{Let } AB \text{ be the lamp post. Let at any time } t, \text{ the man } CD \text{ be at a distance of } \\ x \text{ km from the lamp post and } y \text{ be the length of his shadow } CE.

\displaystyle \text{Since } \triangle ABE \text{ and } \triangle CDE \text{ are similar},

\displaystyle \frac{AB}{CD} = \frac{AE}{CE}.

\displaystyle \Rightarrow \frac{6}{2} = \frac{x + y}{y}.

\displaystyle \Rightarrow \frac{x}{y} = \frac{6}{2} - 1 = 2.

\displaystyle \Rightarrow x = 2y.

\displaystyle \Rightarrow \frac{dx}{dt} = 2\frac{dy}{dt}.

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{2}\frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{1}{2}(6) \quad \left[\because \frac{dx}{dt} = 6\,\text{km/hr}\right].

\displaystyle \Rightarrow \frac{dy}{dt} = 3\,\text{km/hr}.

Question 21:
The surface area of a spherical bubble is increasing at the rate of 2 cm²/sec. When the radius is \displaystyle 6 cm, at what rate is the volume increasing? (CBSE 2005)

Answer:

\displaystyle \text{Let } r \text{ be the radius, } S \text{ be the surface area and } V \text{ be the volume of the sphere at any time } t.

\displaystyle S = 4\pi r^2.

\displaystyle \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{8\pi r}\frac{dS}{dt}.

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{2}{8\pi \times 6}.

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{24\pi}\,\text{cm/sec}.

\displaystyle \text{Now, volume of the sphere } V = \frac{4}{3}\pi r^3.

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi (6)^2 \times \frac{1}{24\pi}.

\displaystyle \Rightarrow \frac{dV}{dt} = 6\,\text{cm}^3/\text{sec}.

Question 22:
The radius of a cylinder increases at 2 cm/sec while its height decreases at 3 cm/sec. Find the rate of change of volume when radius is \displaystyle 3 cm and height is \displaystyle 5 cm. (CBSE 2017)

Answer:

\displaystyle \text{Let } r \text{ be the radius, } h \text{ be the height and } V \text{ be the volume of the cylinder at any time } t.

\displaystyle V = \pi r^2 h.

\displaystyle \Rightarrow \frac{dV}{dt} = 2\pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}.

\displaystyle \Rightarrow \frac{dV}{dt} = \pi r \left(2h\frac{dr}{dt} + r\frac{dh}{dt}\right).

\displaystyle \Rightarrow \frac{dV}{dt} = \pi \times 3 \left(2 \times 5 \times 2 + 3 \times (-3)\right) \quad \left[\because r = 3,\; h = 5,\; \frac{dr}{dt} = 2,\; \frac{dh}{dt} = -3\right].

\displaystyle \Rightarrow \frac{dV}{dt} = 3\pi (20 - 9).

\displaystyle \Rightarrow \frac{dV}{dt} = 33\pi\,\text{cm}^3/\text{sec}.

Question 23:
The volume of metal in a hollow sphere is constant. If the inner radius increases at 1 cm/sec, find the rate of increase of the outer radius when the radii are \displaystyle 4 cm and \displaystyle 8 cm.

Answer:

\displaystyle \text{Let } r_1 \text{ be the inner radius and } r_2 \text{ be the outer radius and } V \text{ be the volume of the hollow sphere at any time } t.

\displaystyle V = \frac{4}{3}\pi \left(r_2^3 - r_1^3\right).

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi \left(r_2^2 \frac{dr_2}{dt} - r_1^2 \frac{dr_1}{dt}\right).

\displaystyle \Rightarrow r_2^2 \frac{dr_2}{dt} = r_1^2 \frac{dr_1}{dt} \quad \left[\because \frac{dV}{dt} = 0\right].

\displaystyle \Rightarrow (8)^2 \frac{dr_2}{dt} = (4)^2 \times 1.

\displaystyle \Rightarrow \frac{dr_2}{dt} = \frac{16}{64}.

\displaystyle \Rightarrow \frac{dr_2}{dt} = \frac{1}{4}\,\text{cm/sec}.

Question 24:
Sand is poured onto a conical pile at the rate of 50 cm³/min. The height of the cone is always half the radius of its base. How fast is the height increasing when the sand is 5 cm deep?

Answer:

\displaystyle \text{Let } r \text{ be the radius, } h \text{ be the height and } V \text{ be the volume of the conical pile at any time } t.

\displaystyle V = \frac{1}{3}\pi r^2 h.

\displaystyle \Rightarrow V = \frac{1}{3}\pi (2h)^2 h \quad \left[\because r = 2h\right].

\displaystyle \Rightarrow V = \frac{4}{3}\pi h^3.

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi h^2 \frac{dh}{dt}.

\displaystyle \Rightarrow 50 = 4\pi h^2 \frac{dh}{dt}.

\displaystyle \Rightarrow \frac{dh}{dt} = \frac{50}{4\pi (5)^2}.

\displaystyle \Rightarrow \frac{dh}{dt} = \frac{1}{2\pi}\,\text{cm/min}.

Question 25:
A kite is 120 m high and 130 m of string is out. If the kite moves horizontally at 52 m/sec, find the rate at which the string is being paid out.

Answer:

\displaystyle \text{In the right triangle } ABC,

\displaystyle \text{Here, } AB^2 + BC^2 = AC^2.

\displaystyle \Rightarrow x^2 + (120)^2 = y^2.

\displaystyle \Rightarrow 2x\frac{dx}{dt} = 2y\frac{dy}{dt}.

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{x}{y}\frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dy}{dt} = \frac{50}{130} \times 52 \quad \left[\because x = \sqrt{(130)^2 - (120)^2} = 50\right].

\displaystyle \Rightarrow \frac{dy}{dt} = 20\,\text{m/sec}.

Question 26:
A particle moves along the curve \displaystyle y = \frac{2}{3}x^3 + 1. Find the points where the \displaystyle y-coordinate changes twice as fast as the \displaystyle x-coordinate.

Answer:

\displaystyle \text{Here, } y = \frac{2}{3}x^3 + 1.

\displaystyle \Rightarrow \frac{dy}{dt} = 2x^2 \frac{dx}{dt}.

\displaystyle \Rightarrow 2\frac{dx}{dt} = 2x^2 \frac{dx}{dt} \quad \left[\because \frac{dy}{dt} = 2\frac{dx}{dt}\right].

\displaystyle \Rightarrow x^2 = 1.

\displaystyle \Rightarrow x = \pm 1.

\displaystyle \text{Substituting } x = 1 \text{ and } x = -1 \text{ in } y = \frac{2}{3}x^3 + 1, \text{ we get}

\displaystyle y = \frac{5}{3} \text{ and } y = \frac{1}{3}.

\displaystyle \text{So, the points are } \left(1,\frac{5}{3}\right) \text{ and } \left(-1,\frac{1}{3}\right).

Question 27:
Find the point on the curve \displaystyle y^2 = 8x where the abscissa and ordinate change at the same rate. (CBSE 2002C)

Answer:

\displaystyle \text{Here, } y^2 = 8x. \quad (1)

\displaystyle \Rightarrow 2y\frac{dy}{dt} = 8\frac{dx}{dt}.

\displaystyle \Rightarrow 2y = 8 \quad \left[\because \frac{dy}{dt} = \frac{dx}{dt}\right].

\displaystyle \Rightarrow y = 4.

\displaystyle \Rightarrow x = \frac{y^2}{8} \quad \text{[from (1)]}.

\displaystyle \Rightarrow x = \frac{16}{8} = 2.

\displaystyle \text{So, the point is } (2,4).

Question 28:
The volume of a cube increases at the rate of 9 cm³/sec. How fast is the surface area increasing when the edge length is \displaystyle 10 cm?

Answer:

\displaystyle \text{Let } x \text{ be the side and } V \text{ be the volume of the cube at any time } t.

\displaystyle V = x^3.

\displaystyle \Rightarrow \frac{dV}{dt} = 3x^2 \frac{dx}{dt}.

\displaystyle \Rightarrow 9 = 3(10)^2 \frac{dx}{dt} \quad \left[\because x = 10\,\text{cm and } \frac{dV}{dt} = 9\,\text{cm}^3/\text{sec}\right].

\displaystyle \Rightarrow \frac{dx}{dt} = 0.03\,\text{cm/sec}.

\displaystyle \text{Let } S \text{ be the surface area of the cube at any time } t.

\displaystyle S = 6x^2.

\displaystyle \Rightarrow \frac{dS}{dt} = 12x \frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dS}{dt} = 12 \times 10 \times 0.03 \quad \left[\because x = 10\,\text{cm and } \frac{dx}{dt} = 0.03\,\text{cm/sec}\right].

\displaystyle \Rightarrow \frac{dS}{dt} = 3.6\,\text{cm}^2/\text{sec}.

Question 29:
The volume of a spherical balloon increases at the rate of 25 cm³/sec. Find the rate of change of surface area when the radius is \displaystyle 5 cm. (CBSE 2004, 2017)

Answer:

\displaystyle \text{Let } r \text{ be the radius and } V \text{ be the volume of the sphere at any time } t.

\displaystyle V = \frac{4}{3}\pi r^3.

\displaystyle \Rightarrow \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}.

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{25}{4\pi (5)^2} \quad \left[\because r = 5\,\text{cm and } \frac{dV}{dt} = 25\,\text{cm}^3/\text{sec}\right].

\displaystyle \Rightarrow \frac{dr}{dt} = \frac{1}{4\pi}\,\text{cm/sec}.

\displaystyle \text{Now, let } S \text{ be the surface area of the sphere at any time } t.

\displaystyle S = 4\pi r^2.

\displaystyle \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dS}{dt} = 8\pi \times 5 \times \frac{1}{4\pi} \quad \left[\because r = 5\,\text{cm and } \frac{dr}{dt} = \frac{1}{4\pi}\,\text{cm/sec}\right].

\displaystyle \Rightarrow \frac{dS}{dt} = 10\,\text{cm}^2/\text{sec}.

Question 30:
The length \displaystyle x of a rectangle decreases at 5 cm/min and the width \displaystyle y increases at 4 cm/min. When \displaystyle x = 8 cm and \displaystyle y = 6 cm, find the rate of change of
(i) perimeter
(ii) area. (CBSE 2009)

Answer:
(i)

\displaystyle \text{Let at any instant of time } t, \text{ the length of the rectangle be } x, \text{ the breadth be } y, \text{ the perimeter be } P \text{ and the area be } A.

\displaystyle P = 2(x + y). \quad (1)

\displaystyle \text{We have } \frac{dx}{dt} = 5\,\text{cm/min and } \frac{dy}{dt} = 4\,\text{cm/min}.

\displaystyle \text{Differentiating (1) with respect to } t, \text{ we get}

\displaystyle \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right).

\displaystyle \Rightarrow \frac{dP}{dt} = 2(5 + 4).

\displaystyle \Rightarrow \frac{dP}{dt} = 18\,\text{cm/min}.

\displaystyle \text{Therefore, the perimeter of the rectangle is increasing at a rate of } 18\,\text{cm/min}.

(ii)

\displaystyle A = xy. \quad (i)

\displaystyle \text{Differentiating (i) with respect to } t, \text{ we get}

\displaystyle \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}.

\displaystyle \Rightarrow \frac{dA}{dt} = (8\,\text{cm})(4\,\text{cm/min}) + (6\,\text{cm})(-5\,\text{cm/min}).

\displaystyle \Rightarrow \frac{dA}{dt} = 32 - 30.

\displaystyle \Rightarrow \frac{dA}{dt} = 2\,\text{cm}^2/\text{min}.

\displaystyle \text{Therefore, the area of the rectangle is increasing at a rate of } 2\,\text{cm}^2/\text{min}.

Question 31:
A circular disc is heated so that its radius increases at 0.05 cm/sec. Find the rate at which the area is increasing when the radius is \displaystyle 3.2 cm.

Answer:

\displaystyle \text{Let } r \text{ be the radius and } A \text{ be the area of the circular disc at any time } t.

\displaystyle A = \pi r^2.

\displaystyle \Rightarrow \frac{dA}{dt} = 2\pi r \frac{dr}{dt}.

\displaystyle \Rightarrow \frac{dA}{dt} = 2\pi \times 3.2 \times 0.05 \quad \left[\because r = 3.2\,\text{cm and } \frac{dr}{dt} = 0.05\,\text{cm/sec}\right].

\displaystyle \Rightarrow \frac{dA}{dt} = 0.32\pi\,\text{cm}^2/\text{sec}.

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