CBSE Class 12 Mathematics HOTs Application of Derivatives Set 02

Very Short Answer Type Questions

Question. The surface area of a cube increases at the rate of \( 72\text{ cm}^2/\text{sec} \). Find the rate of change of its volume when the edge of the cube measures \( 3\text{ cm} \).
Answer: Let \( a \) be the edge, \( S \) be the surface area and \( V \) be volume of cube.
Given, \( \frac{dS}{dt} = 72\text{ cm}^2/\text{s} \) and \( a = 3\text{ cm} \)
Surface area of cube \( = 6a^2 \)
\( \Rightarrow \frac{dS}{dt} = 12a \cdot \frac{da}{dt} \)
\( \Rightarrow \frac{da}{dt} = \frac{72}{12a} = \frac{6}{a} \)
Now, volume of cube, \( V = a^3 \)
On differentiating w.r.t. \( t \), we get
\( \frac{dV}{dt} = 3a^2 \cdot \frac{da}{dt} = 3a^2 \cdot \frac{6}{a} \Rightarrow \frac{dV}{dt} = 18a \)
\( \Rightarrow \left(\frac{dV}{dt}\right)_{a=3} = 18 \times 3 = 54\text{ cm}^3/\text{s} \)

Question. Find the points on the curve \( 6y = x^3 + 2 \) at which ordinate is changing \( 8 \) times as fast as abscissa.
Answer: Given, \( 6y = x^3 + 2 \)
\( \Rightarrow 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \)
Also, given \( \frac{dy}{dt} = 8 \frac{dx}{dt} \)
\( \therefore 6\left(8 \frac{dx}{dt}\right) = 3x^2 \frac{dx}{dt} \)
\( \Rightarrow 48 = 3x^2 \Rightarrow x^2 = 16 \)
\( \Rightarrow x = \pm 4 \)
When \( x = 4 \), then \( 6y = (4)^3 + 2 \)
\( \Rightarrow 6y = 66 \)
\( \Rightarrow y = 11 \)
and when \( x = -4 \), then \( 6y = (-4)^3 + 2 \)
\( \Rightarrow 6y = -62 \)
\( \Rightarrow y = -\frac{31}{3} \)
\( \therefore \) Required points are \( (4, 11) \) and \( \left(-4, -\frac{31}{3}\right) \).

Question. A particle moves along the curve \( 3y = ax^3 + 1 \) such that at a point with \( x \)-coordinate \( 1 \), \( y \)-coordinate is changing twice as fast as \( x \)-coordinate. Find the value of \( a \).
Answer: Given, \( 3y = ax^3 + 1 \) ...(i)
and \( \frac{dy}{dt} = 2 \frac{dx}{dt} \) ...(ii)
Now, from Eq. (i), \( 3 \frac{dy}{dt} = a\left(3x^2\right) \frac{dx}{dt} \)
\( \Rightarrow 3\left(2 \frac{dx}{dt}\right) = a\left(3x^2\right) \frac{dx}{dt} \) [using Eq. (ii)]
\( \Rightarrow 2 = ax^2 \)
When \( x = 1 \), then \( a(1)^2 = 2 \Rightarrow a = 2 \).

Question. The radius of a circle increases at a rate of \( 0.2\text{ cm/s} \). Calculate the rate of the increase of the area, when the radius is \( 5\text{ cm} \).
Answer: Let \( x \) be the radius and \( y \) be the area of circle.
Then, \( y = \pi x^2 \) ...(i)
Also, rate of change of radius with respect to time is \( 0.2\text{ cm/s} \).
\( \dots \frac{dx}{dt} = 0.2 \) ...(ii)
Now, on differentiating both sides of Eq. (i) w.r.t. \( t \), we get
\( \frac{dy}{dt} = \pi \cdot 2x \cdot \frac{dx}{dt} \) [by chain rule]
\( = 2\pi x(0.2) = 0.4\pi x \) [using Eq. (ii)]
When radius \( x = 5\text{ cm} \), then
\( \left(\frac{dy}{dt}\right) = 0.4 \pi \times 5 = 2\pi\text{ cm}^2/\text{s} \)
Hence, the required rate of increase of the area is \( 2\pi\text{ cm}^2/\text{s} \).

Question. The sides of an equilateral triangle are increasing at the rate of \( 2\text{ cm/s} \). Find the rate at which the area increases, when the side is \( 10\text{ cm} \).
Answer: Let the side of triangle be \( a \).
Given, \( \frac{da}{dt} = 2\text{ cm/s} \)
Now, area of equilateral triangle having side \( a \) is given by
\( A = \frac{\sqrt{3}a^2}{4} \)
On differentiating both sides w.r.t. \( t \), we get
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} (2a) \frac{da}{dt} \)
On putting \( \frac{da}{dt} = 2\text{ cm/s} \) and \( a = 10\text{ cm} \), we get
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2 \times 10 \times 2 = 10\sqrt{3}\text{ cm}^2/\text{s} \).

Question. For the curve \( y = 5x - 2x^3 \), if \( x \) increases at the rate of \( 2\text{ units/s} \), then find the rate of change of the slope of curve when \( x = 3 \).
Answer: Given curve is \( y = 5x - 2x^3 \) and \( \frac{dx}{dt} = 2\text{ units/s} \) ...(i)
Now, slope of the curve \( \frac{dy}{dx} = 5 - 6x^2 = M \) (say)
Rate of change of the slope,
\( \frac{dM}{dt} = -6 \frac{d}{dt}(x^2) \Rightarrow \frac{dM}{dt} = -12x \frac{dx}{dt} \)
When \( x = 3 \), then \( \frac{dM}{dt} = -12 \times 3 \times 2 = -72\text{ units/s} \)
Thus, the slope is decreasing at the rate of \( 72\text{ units/s} \).

Question. 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?
Answer: Let \( r \) be the radius and \( C \) be the circumference of circle.
Given, \( \frac{dr}{dt} = 0.7\text{ cm/s} \)
Now, circumference of a circle, \( C = 2\pi r \)
On differentiating both sides w.r.t. \( t \), we get
\( \frac{dC}{dt} = 2\pi \frac{dr}{dt} \)
\( \Rightarrow \frac{dC}{dt} = 2\pi \times 0.7 = 1.4\pi\text{ cm/s} \)
Hence, the rate of increase of circumference is \( 1.4\pi\text{ cm/s} \).

Short Answer Type Questions

Question. Sand is pouring from a pipe at the rate of \( 15\text{ cm}^3/\text{min} \). The falling sand forms a cone on the ground such that the height of the cone is always one third of the radius of the base. How fast is the height of the sand cone increasing at the instant when the height is \( 4\text{ cm} \)?
Answer: Let \( r \) be the radius, \( h \) be the height and \( V \) be the volume of the sand cone.
Given, \( \frac{dV}{dt} = 15\text{ cm}^3/\text{min} \) and \( h = \frac{1}{3}r \)
\( \Rightarrow r = 3h \) and \( h = 4\text{ cm} \)
Volume of sand cone, \( V = \frac{1}{3}\pi r^2 h \)
\( \Rightarrow V = \frac{1}{3}\pi (3h)^2 h = \frac{1}{3}\pi \times 9h^2 \times h = 3\pi h^3 \)
On differentiating both sides w.r.t. \( t \), we get
\( \frac{dV}{dt} = 3\pi \times 3h^2 \frac{dh}{dt} = 9\pi h^2 \frac{dh}{dt} \)
\( \Rightarrow 15 = 9\pi(4)^2 \frac{dh}{dt} \)
\( \Rightarrow \frac{dh}{dt} = \frac{15}{9\pi \times 16} = \frac{5}{48\pi}\text{ cm/min} \)
Hence, the height of the sand cone is increasing at the rate of \( \frac{5}{48\pi}\text{ cm/min} \), when the height is \( 4\text{ cm} \).

Question. A man is walking at the rate of \( 6.5\text{ km/hr} \) towards the foot of the tower \( 120\text{ m} \) high. At what rate is he approaching the top of the tower when he is \( 50\text{ m} \) away from the tower?
Answer: Let \( AB \) be the tower and \( C \) is the man. Height of the man will be negligible as compared to the tower.
If man is walking towards tower then \( x \) is decreasing
\( \Rightarrow \frac{dx}{dt} = -6.5\text{ km/hr} \).
We have to find the rate of approaching the top of tower, so we have to find \( \frac{dy}{dt} \).
Applying Pythagoras in \( \Delta ABC \),
\( y^2 = (120)^2 + x^2 \Rightarrow y = \sqrt{(120)^2 + x^2} \)
On differentiating w.r.t. \( t \), we get
\( \frac{dy}{dt} = \frac{1}{2\sqrt{(120)^2 + x^2}} \frac{d}{dt}\left((120)^2 + x^2\right) \)
\( = \frac{1}{2\sqrt{(120)^2 + x^2}} \cdot 2x \cdot \frac{dx}{dt} = \frac{x}{\sqrt{(120)^2 + x^2}} \frac{dx}{dt} \)
On putting \( x = 50\text{ m} \) and \( \frac{dx}{dt} = -6.5\text{ km/hr} \), we get
\( \frac{dy}{dt} = \frac{50}{\sqrt{(120)^2 + (50)^2}} (-6.5) = \frac{50}{130} \times (-6.5)\text{ km/hr} \)
\( = -2.5\text{ km/hr} \).

Question. The volume of a sphere is increasing at the rate of \( 3\text{ cm}^3/\text{s} \). Find the rate of increase of its surface area, when the radius is \( 2\text{ cm} \).
Answer: Let \( r \) be the radius of sphere and \( V \) be its volume.
Then, \( V = \frac{4}{3}\pi r^3 \)
Given, \( \frac{dV}{dt} = 3\text{ cm}^3/\text{s} \)
\( \therefore \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 3 \)
\( \Rightarrow \frac{4}{3}\pi \left(3r^2\right) \frac{dr}{dt} = 3 \)
\( \Rightarrow \left(4\pi r^2\right) \frac{dr}{dt} = 3 \)
\( \Rightarrow \frac{dr}{dt} = \frac{3}{4\pi r^2} \) ...(i)
Now, let \( S \) be the surface area of sphere.
Then, \( S = 4\pi r^2 \)
\( \Rightarrow \frac{dS}{dt} = 4\pi (2r) \frac{dr}{dt} \)
\( = 8\pi r \left(\frac{3}{4\pi r^2}\right) \) [using Eq. (i)]
\( \Rightarrow \frac{dS}{dt} = \frac{6}{r} \)
When \( r = 2 \), then \( \frac{dS}{dt} = \frac{6}{2} = 3\text{ cm}^2/\text{s} \).

Question. A balloon which always remains spherical has a variable radius, find the rate at which its volume is increasing with respect to its radius, when the radius is \( 7\text{ cm} \).
Answer: Here, we have to find rate of change of volume with respect to radius, so volume is dependent quantity and radius is independent quantity.
Let \( r \) be the radius and \( V \) be the volume of spherical balloon.
Then, \( V = \frac{4}{3}\pi r^3 \).
On differentiating both sides w.r.t. \( r \), we get
\( \frac{dV}{dr} = \frac{4}{3}\pi (3r^2) = 4\pi r^2 \) ...(i)
Now, we have to find rate of change of volume, when radius is \( 7\text{ cm} \). So, on putting \( r = 7 \) in Eq. (i), we get
\( \left(\frac{dV}{dr}\right)_{r=7} = 4\pi(7)^2 = 4\pi(49) = 196\pi\text{ cm}^3/\text{cm} \)
Hence, the volume is increasing with respect to its radius at the rate of \( 196\pi\text{ cm}^3/\text{cm} \), when the radius is \( 7\text{ cm} \).

Question. The length \( x \) of a rectangle is decreasing at the rate of \( 6\text{ cm/min} \) and the width \( y \) is increasing at the rate of \( 5\text{ cm/min} \). Find the rate of change of the perimeter and the area of the rectangle when \( x = 10\text{ cm} \) and \( y = 8\text{ cm} \).
Answer: Given that \( \frac{dx}{dt} = -6\text{ cm/min} \) and \( \frac{dy}{dt} = 5\text{ cm/min} \)
Let \( P \) and \( A \) be the perimeter and area of rectangle.
Then, \( P = 2(x + y) \)
On differentiating w.r.t. \( t \), we get
\( \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-6 + 5) = -2\text{ cm/min} \)
And \( A = xy \)
On differentiating w.r.t. \( t \), we get
\( \frac{dA}{dt} = \frac{dx}{dt}y + x\frac{dy}{dt} \)
\( = [-6 \times 8 + 10 \times 5] \)
\( = -48 + 50 = 2\text{ cm}^2/\text{min} \)
Hence, the perimeter of a rectangle is decreasing at the rate of \( 2\text{ cm/min} \) and area of rectangle is increasing at the rate of \( 2\text{ cm}^2/\text{min} \).

Question. A balloon which always remains spherical has a variable diameter \( \frac{3}{2}(2x + 1) \). Then, find the rate of change of its volume with respect to \( x \).
Answer: Given, diameter of the balloon \( = \frac{3}{2}(2x + 1) \)
\( \therefore \) Radius of the balloon \( = \frac{\text{Diameter}}{2} = \frac{1}{2} \left[ \frac{3}{2}(2x + 1) \right] = \frac{3}{4}(2x + 1) \)
For the balloon, the volume \( V \) is given by
\( V = \frac{4}{3}\pi(\text{radius})^3 = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 1) \right]^3 = \frac{9\pi}{16}(2x + 1)^3 \)
For the rate of change of volume, differentiate w.r.t. \( x \), we get
\( \frac{dV}{dx} = \frac{9\pi}{16} \times 3(2x + 1)^2 \times 2 = \frac{27\pi}{8}(2x + 1)^2 \)
Thus, the rate of change of volume is \( \frac{27\pi}{8}(2x + 1)^2 \).

Question. A spherical ball of salt is dissolving in water in such a manner that the rate of decreasing of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.
Answer: Let the radius of spherical ball of the salt be \( r \).
\( \therefore \) Volume of the ball, \( V = \frac{4}{3}\pi r^3 \) and surface area, \( S = 4\pi r^2 \)
\( \therefore -\frac{dV}{dt} = -\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = -\frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} = -4\pi r^2 \frac{dr}{dt} \)
[here, we take negative sign because salt is dissolving]
According to the given condition,
\( -\frac{dV}{dt} \propto S \Rightarrow -\frac{dV}{dt} = kS \),
where \( k \) is proportionality constant.
\( \Rightarrow -4\pi r^2 \frac{dr}{dt} = k \cdot 4\pi r^2 \Rightarrow \frac{dr}{dt} = -k \)
Hence proved.

Question. A man \( 2\text{ m} \) tall, walks at a uniform speed of \( 6\text{ km/h} \) away from a lamp post \( 6\text{ m} \) high. Find the rate at which the length of his shadow increases.
Answer: Let \( AB \) be the lamp post and a man \( CD \) be at a distance \( x \) from the lamp post and let \( CE = y \) be his shadow.
Given that \( \frac{dx}{dt} = 6\text{ km/h} \), \( AB = 6\text{ m} \) and \( CD = 2\text{ m} \).
Here, \( \Delta ABE \sim \Delta CDE \)
\( \therefore \frac{AB}{CD} = \frac{AE}{CE} \) [by property of similar triangles]
\( \Rightarrow \frac{6}{2} = \frac{x+y}{y} \Rightarrow 3 = \frac{x+y}{y} \)
\( \Rightarrow 3y = x + y \Rightarrow 3y - y = x \Rightarrow 2y = x \)
On differentiating both sides w.r.t. \( t \), we get
\( 2\frac{dy}{dt} = \frac{dx}{dt} \)
\( \Rightarrow 2\frac{dy}{dt} = 6 \) [since \( \frac{dx}{dt} = 6 \)]
\( \Rightarrow \frac{dy}{dt} = \frac{6}{2} = 3\text{ km/h} \)
Hence, the shadow increases at the rate of \( 3\text{ km/h} \).

Question. Show that the function \( f(x) = 4x^3 - 18x^2 + 27x - 7 \) is always increasing on \( R \).
Answer: We have, \( f(x) = 4x^3 - 18x^2 + 27x - 7 \)
On differentiating both sides w.r.t. \( x \), we get
\( f'(x) = 12x^2 - 36x + 27 \)
\( \Rightarrow f'(x) = 3(4x^2 - 12x + 9) \)
\( \Rightarrow f'(x) = 3(2x - 3)^2 \)
\( \Rightarrow \) For any \( x \in R \), \( (2x - 3)^2 \geq 0 \) [since, a perfect square number cannot be negative.]
\( \therefore 3(2x - 3)^2 \geq 0 \), \( \forall x \in R \)
\( \dots f'(x) \geq 0 \), \( \forall x \in R \)
\( \therefore \) Given function \( f(x) \) is an increasing function on \( R \).

Question. Find the interval in which the function \( f(x) = -2x^3 - 9x^2 - 12x + 1 \) is strictly increasing or strictly decreasing.
Answer: Given, \( f(x) = -2x^3 - 9x^2 - 12x + 1 \)
On differentiating both sides w.r.t. \( x \), we get
\( f'(x) = -6x^2 - 18x - 12 \)
\( \Rightarrow f'(x) = -6(x^2 + 3x + 2) \)
\( \Rightarrow f'(x) = -6(x^2 + 2x + x + 2) \)
\( = -6[x(x + 2) + 1(x + 2)] \)
\( = -6(x + 2)(x + 1) \)
Now, put \( f'(x) = 0 \)
\( \Rightarrow -6(x + 2)(x + 1) = 0 \Rightarrow x = -2, -1 \)
The points \( x = -2 \) and \( x = -1 \) divide the real line into three disjoint intervals \( (-\infty, -2) \), \( (-2, -1) \) and \( (-1, \infty) \).
The nature of function in these intervals are given below.
- In \( (-\infty, -2) \), sign of \( f'(x) = (-)(-)(-) = (-)<0 \), so \( f(x) \) is strictly decreasing.
- In \( (-2, -1) \), sign of \( f'(x) = (-)(+)(-) = (+)>0 \), so \( f(x) \) is strictly increasing.
- In \( (-1, \infty) \), sign of \( f'(x) = (-)(+)(+) = (-)<0 \), so \( f(x) \) is strictly decreasing.
Hence, \( f(x) \) is strictly increasing in the interval \( (-2, -1) \) and \( f(x) \) is strictly decreasing in the interval \( (-\infty, -2) \cup (-1, \infty) \).

Long Answer Type Questions

Question. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool, t seconds after the pool has been plugged off to drain and \( L = 200(10-t)^2 \). How fast is the water running out at the end of 5 s and what is the average rate at which the water flows out during the first 5 s?
Answer: Given that \( L \) represents the number of litres of water in the pool, \( t \) seconds after the pool has been plugged off to drain:
\( L = 200(10-t)^2 \)
The rate at which the water is running out is given by \( -\frac{dL}{dt} \):
\( \frac{dL}{dt} = 200 \cdot 2(10-t) \cdot (-1) = -400(10-t) \)
[Here, we take the negative sign because the water in the pool is decreasing.]
Therefore, the rate of water running out is \( R(t) = 400(10-t) \text{ L/s} \).
(i) At the end of \( 5\text{ s} \):
Rate at which the water is running out \( = 400(10-5) = 2000 \text{ L/s} \).
(ii) Average rate during the first \( 5\text{ s} \):
Initial rate (at \( t = 0 \)) \( = 400(10-0) = 4000 \text{ L/s} \).
Final rate (at \( t = 5 \)) \( = 2000 \text{ L/s} \).
Average rate \( = \frac{\text{Initial rate} + \text{Final rate}}{2} = \frac{4000 + 2000}{2} = 3000 \text{ L/s} \).
Hence, the water flows out during the first 5 s with an average rate of 3000 L/s.

Question. A kite is moving horizontally at a height of 151.5 m. If the speed of kite is 10 m/s, then how fast is the string being let out, when the kite is 250 m away from the boy who is flying the kite, if the height of boy is 1.5 m?
Answer: Let \( CD \) be the height of the kite and \( AB \) be the height of the boy.
Then, height of the kite \( CD = 151.5 \text{ m} \).
Let \( DB = x \text{ m} = EA \).
Since the kite is 250 m away from the boy, \( AC = y = 250 \text{ m} \).
Speed of the kite, \( v = \frac{dx}{dt} = 10 \text{ m/s} \).
From the geometry of the figure:
\( EC = 151.5 - 1.5 = 150 \text{ m} \).
In right-angled triangle \( \Delta CEA \), by Pythagoras theorem:
\( AE^2 + EC^2 = AC^2 \)
\( \Rightarrow x^2 + (150)^2 = y^2 \) ... (i)
Substituting \( y = 250 \) in Eq. (i):
\( x^2 + (150)^2 = (250)^2 \)
\( \Rightarrow x^2 = (250)^2 - (150)^2 = (250 + 150)(250 - 150) = 400 \times 100 = 40000 \)
\( \Rightarrow x = 200 \text{ m} \).
On differentiating both sides of Eq. (i) w.r.t. \( t \), we get:
\( 2x\frac{dx}{dt} = 2y\frac{dy}{dt} \Rightarrow y\frac{dy}{dt} = x\frac{dx}{dt} \)
\( \Rightarrow \frac{dy}{dt} = \frac{x}{y}\frac{dx}{dt} \)
Substituting the known values:
\( \frac{dy}{dt} = \frac{200}{250} \times 10 = 8 \text{ m/s} \).
Hence, the required rate at which the string is being let out is 8 m/s.

Question. Two men A and B start with velocity v at the same time from the junction of two roads inclined at \( 45^\circ \) to each other. If they travel by different roads, then find the rate at which they are being separated.
Answer: Let two men start from the junction point \( C \) with velocity \( v \) each at the same time.
Given that \( \angle BCA = 45^\circ \).
Since A and B move with the same velocity \( v \), they cover the same distance in the same time.
Therefore, \( \Delta ABC \) is an isosceles triangle with \( AC = BC \).
Draw \( CD \perp AB \).
Let at any instant \( t \), the distance between them be \( AB = y \).
Let \( AC = BC = x \).
In \( \Delta ACD \) and \( \Delta BCD \), we have:
\( \angle CAD = \angle CBD \) [since \( BC = AC \Rightarrow \angle A = \angle B \)]
\( \angle CDA = \angle CDB = 90^\circ \)
\( \dots \angle ACD = \angle BCD = \frac{1}{2} \angle ACB = \frac{45^\circ}{2} = 22.5^\circ \).
In \( \Delta ACD \):
\( \sin 22.5^\circ = \frac{AD}{AC} = \frac{y/2}{x} \)
\( \Rightarrow y = 2x \sin 22.5^\circ \).
On differentiating both sides w.r.t. \( t \), we get:
\( \frac{dy}{dt} = 2 \sin 22.5^\circ \frac{dx}{dt} \) [Since \( \frac{dx}{dt} = v \)]
\( \Rightarrow \frac{dy}{dt} = 2v \sin 22.5^\circ \).
Since \( \sin 22.5^\circ = \frac{\sqrt{2-\sqrt{2}}}{2} \), we have:
\( \frac{dy}{dt} = 2v \cdot \frac{\sqrt{2-\sqrt{2}}}{2} = v\sqrt{2-\sqrt{2}} \).
Hence, the rate at which they are being separated is \( v\sqrt{2-\sqrt{2}} \).

Question. Water is running into a conical vessel, 15 cm deep and 5 cm in radius at the rate of \( 0.1 \text{ cm}^3\text{/s} \), when the water is 6 cm deep. Find at what rate is (i) the water level rising? (ii) the water surface area increasing? (iii) the wetted surface of the vessel increasing?
Answer: Let \( r \) be the radius, \( h \) be the height, and \( V \) be the volume of water in the cone at any time \( t \).
Given, height of conical vessel \( H = 15 \text{ cm} \), radius \( R = 5 \text{ cm} \).
Rate of flow of water \( \frac{dV}{dt} = 0.1 \text{ cm}^3\text{/s} \).
By similar triangles, we have:
\( \frac{r}{h} = \frac{R}{H} = \frac{5}{15} = \frac{1}{3} \Rightarrow r = \frac{h}{3} \) ... (i)
Volume of water \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{\pi h^3}{27} \) ... (ii)
Differentiating Eq. (ii) w.r.t. \( t \):
\( \frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} \Rightarrow \frac{dV}{dt} = \frac{\pi h^2}{9} \frac{dh}{dt} \)
When \( h = 6 \text{ cm} \):
\( 0.1 = \frac{\pi (6)^2}{9} \frac{dh}{dt} \Rightarrow 0.1 = 4\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{40\pi} \text{ cm/s} \).

(i) Therefore, the rate at which the water level is rising is \( \frac{1}{40\pi} \text{ cm/s} \).

(ii) Water surface area \( A = \pi r^2 = \pi \left(\frac{h}{3}\right)^2 = \frac{\pi h^2}{9} \).
Differentiating w.r.t. \( t \):
\( \frac{dA}{dt} = \frac{2\pi h}{9} \frac{dh}{dt} \)
At \( h = 6 \text{ cm} \) and \( \frac{dh}{dt} = \frac{1}{40\pi} \text{ cm/s} \):
\( \frac{dA}{dt} = \frac{2\pi(6)}{9} \times \frac{1}{40\pi} = \frac{12\pi}{360\pi} = \frac{1}{30} \text{ cm}^2\text{/s} \).
Therefore, the water surface area is increasing at the rate of \( \frac{1}{30} \text{ cm}^2\text{/s} \).

(iii) Wetted surface area of the vessel \( S = \pi r l = \pi r \sqrt{r^2 + h^2} \).
Substituting \( r = \frac{h}{3} \):
\( S = \pi \left(\frac{h}{3}\right) \sqrt{\left(\frac{h}{3}\right)^2 + h^2} = \frac{\pi h}{3} \sqrt{\frac{10h^2}{9}} = \frac{\pi \sqrt{10} h^2}{9} \).
Differentiating w.r.t. \( t \):
\( \frac{dS}{dt} = \frac{2\pi \sqrt{10} h}{9} \frac{dh}{dt} \)
At \( h = 6 \text{ cm} \) and \( \frac{dh}{dt} = \frac{1}{40\pi} \text{ cm/s} \):
\( \frac{dS}{dt} = \frac{2\pi \sqrt{10} (6)}{9} \times \frac{1}{40\pi} = \frac{\sqrt{10}}{30} \text{ cm}^2\text{/s} \).
Therefore, the wetted surface of the vessel is increasing at the rate of \( \frac{\sqrt{10}}{30} \text{ cm}^2\text{/s} \).

Question. A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is \( \tan^{-1}(0.5) \). Water is poured into it at a constant rate of 5 \( \text{m}^3\text{/h} \). Find the rate at which the level of the water is rising at the instant, when the depth of water in the tank is 4 m.
Answer: Let \( r \) be the radius, \( h \) be the height, and \( V \) be the volume of water in the conical tank at any time \( t \).
Given, the semi-vertical angle \( \alpha = \tan^{-1}(0.5) \).
Thus, \( \tan \alpha = \frac{r}{h} \Rightarrow 0.5 = \frac{r}{h} \Rightarrow r = \frac{h}{2} \).
The volume of water is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12} \).
Differentiating w.r.t. \( t \):
\( \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt} \).
Given \( \frac{dV}{dt} = 5 \text{ m}^3\text{/h} \) and \( h = 4 \text{ m} \):
\( 5 = \frac{\pi (4)^2}{4} \frac{dh}{dt} \Rightarrow 5 = 4\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{5}{4\pi} \text{ m/h} \).
Using \( \pi \approx \frac{22}{7} \):
\( \frac{dh}{dt} = \frac{5}{4 \times \frac{22}{7}} = \frac{35}{88} \text{ m/h} \).

Question. Show that for \( a \ge 1 \), \( f(x) = \sqrt{3}\sin x - \cos x - 2ax + b \) is decreasing in \( \mathbb{R} \).
Answer: Given, \( f(x) = \sqrt{3}\sin x - \cos x - 2ax + b \) for \( a \ge 1 \).
On differentiating w.r.t. \( x \), we get:
\( f'(x) = \sqrt{3}\cos x + \sin x - 2a \).
Multiplying and dividing the first two terms by 2:
\( f'(x) = 2\left(\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x\right) - 2a \)
\( \Rightarrow f'(x) = 2\left(\cos\frac{\pi}{6}\cos x + \sin\frac{\pi}{6}\sin x\right) - 2a \)
\( \Rightarrow f'(x) = 2\cos\left(x - \frac{\pi}{6}\right) - 2a \).
We know that the maximum value of \( \cos\left(x - \frac{\pi}{6}\right) \) is 1.
Thus, \( 2\cos\left(x - \frac{\pi}{6}\right) \le 2 \).
Since \( a \ge 1 \), we have \( 2a \ge 2 \Rightarrow -2a \le -2 \).
Therefore, \( 2\cos\left(x - \frac{\pi}{6}\right) - 2a \le 2 - 2 \Rightarrow f'(x) \le 0 \).
Hence, \( f(x) \) is a decreasing function in \( \mathbb{R} \).

Question. Find the intervals in which the function \( f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12 \) is (a) strictly increasing, (b) strictly decreasing.
Answer: We have, \( f(x) = \frac{x^4}{4} - x^3 - 5x^2 + 24x + 12 \).
On differentiating w.r.t. \( x \), we get:
\( f'(x) = x^3 - 3x^2 - 10x + 24 \)
\( \Rightarrow f'(x) = (x-2)(x^2 - x - 12) \)
\( \Rightarrow f'(x) = (x-2)(x-4)(x+3) \).
For critical points, put \( f'(x) = 0 \):
\( (x-2)(x-4)(x+3) = 0 \Rightarrow x = -3, 2, 4 \).
These points divide the real line into four disjoint intervals: \( (-\infty, -3), (-3, 2), (2, 4), (4, \infty) \).
Let's check the sign of \( f'(x) \) in these intervals:
- In \( (-\infty, -3) \): \( f'(x) < 0 \) (strictly decreasing)
- In \( (-3, 2) \): \( f'(x) > 0 \) (strictly increasing)
- In \( (2, 4) \): \( f'(x) < 0 \) (strictly decreasing)
- In \( (4, \infty) \): \( f'(x) > 0 \) (strictly increasing)
Thus:
(a) \( f(x) \) is strictly increasing in \( (-3, 2) \cup (4, \infty) \).
(b) \( f(x) \) is strictly decreasing in \( (-\infty, -3) \cup (2, 4) \).