OP Malhotra Class 12 Maths Solutions Chapter 11 Applications of Derivatives Exercise 11 (C)

Question 1. If \( y^2 = Ax \), find the rate at which y is changing with respect to x, when \( x = 4 \).
Answer:
Given the equation: \( y^2 = 4x \) ...(1)
We need to find the rate of change of y with respect to x, which is \( \frac{dy}{dx} \).
First, differentiate both sides of equation (1) with respect to x:
\( 2y \frac{dy}{dx} = 4 \)

\( \implies \frac{dy}{dx} = \frac{4}{2y} \)

\( \implies \frac{dy}{dx} = \frac{2}{y} \)
Now, we need to find the value of y when \( x = 4 \). Substitute \( x = 4 \) into the original equation (1):
\( y^2 = 4(4) \)
\( y^2 = 16 \)
\( y = \pm 4 \)
Substitute \( y = \pm 4 \) into the expression for \( \frac{dy}{dx} \):
When \( y = 4 \), \( \frac{dy}{dx} = \frac{2}{4} = \frac{1}{2} \)
When \( y = -4 \), \( \frac{dy}{dx} = \frac{2}{-4} = -\frac{1}{2} \)
So, the rate at which y is changing with respect to x when \( x = 4 \) is \( \pm \frac{1}{2} \). The derivative helps us understand how a function changes at any point.
In simple words: We start with the given equation and find how y changes as x changes. We then use the given value of x to find the matching y value and substitute it into our rate equation.

🎯 Exam Tip: Remember to consider both positive and negative values for y when solving \( y^2 = \text{constant} \), as both can be valid. Also, clearly show all differentiation steps.

Question 2. If the radius of a circle increases from 3 to 3.01 cm, find the increase in the area.
Answer:
Let r be the radius of the circle.
The area of the circle is given by \( A = \pi r^2 \).
To find how the area changes with respect to the radius, we differentiate A with respect to r:
\( \frac{dA}{dr} = 2\pi r \)
We are given that the initial radius \( r = 3 \) cm.
The radius increases to \( r + \delta r = 3.01 \) cm.
So, the change in radius is \( \delta r = 3.01 - 3 = 0.01 \) cm.
The approximate increase in area, \( \delta A \), can be found using the formula: \( \delta A \approx \frac{dA}{dr} \times \delta r \)
Substitute the values:
\( \delta A = (2\pi r) \times \delta r \)
\( \delta A = (2\pi \times 3) \times 0.01 \)
\( \delta A = 6\pi \times 0.01 \)
\( \delta A = 0.06\pi \text{ cm}^2 \)
Thus, the increase in the area of the circle is \( 0.06\pi \text{ cm}^2 \). This method uses differentiation to approximate the change, which is very useful for small changes.
In simple words: We figure out how the circle's area changes when its radius changes. Given a small increase in radius, we multiply this by how much the area changes for each unit of radius to find the total increase in area.

🎯 Exam Tip: For small changes, the differential \( dA = \frac{dA}{dr} dr \) provides a good approximation for the actual change in area \( \delta A \).

Question 3. The height of a right circular cylinder is 5 cm and remains constant. The cylinder is closed at both ends. The radius r cm of the base is decreasing at a constant rate of 0.04 cm s¯¹. Find the rate at which the total surface is decreasing when the radius is 2 cm, giving your answer correct to 3 significant figures. (Take \( \pi \) to be 3.142).
Answer:
Given height of cylinder, \( h = 5 \) cm (constant).
Rate of decrease of radius, \( \frac{dr}{dt} = -0.04 \) cm/s (negative because it's decreasing).
We need to find the rate of change of total surface area, \( \frac{dS}{dt} \), when \( r = 2 \) cm.
The total surface area (S) of a closed right circular cylinder is:
\( S = 2\pi r^2 + 2\pi rh \)
Since \( h = 5 \) cm, substitute this into the formula for S:
\( S = 2\pi r^2 + 2\pi r(5) \)
\( S = 2\pi r^2 + 10\pi r \)
Now, differentiate S with respect to time (t) to find \( \frac{dS}{dt} \):
\( \frac{dS}{dt} = \frac{d}{dt}(2\pi r^2 + 10\pi r) \)
\( \frac{dS}{dt} = 4\pi r \frac{dr}{dt} + 10\pi \frac{dr}{dt} \)
\( \frac{dS}{dt} = (4\pi r + 10\pi) \frac{dr}{dt} \)
Substitute the given values: \( r = 2 \) cm and \( \frac{dr}{dt} = -0.04 \) cm/s.
\( \frac{dS}{dt} = (4\pi(2) + 10\pi)(-0.04) \)
\( \frac{dS}{dt} = (8\pi + 10\pi)(-0.04) \)
\( \frac{dS}{dt} = (18\pi)(-0.04) \)
\( \frac{dS}{dt} = -0.72\pi \)
Using \( \pi = 3.142 \):
\( \frac{dS}{dt} = -0.72 \times 3.142 \)
\( \frac{dS}{dt} \approx -2.26224 \)
Rounding to 3 significant figures, \( \frac{dS}{dt} = -2.26 \text{ cm}^2/\text{s} \). The negative sign indicates that the total surface area is decreasing. This shows how quickly the total outer covering of the cylinder shrinks as its radius gets smaller.
Thus, the total surface area is decreasing at the rate of \( 2.26 \text{ cm}^2/\text{s} \).
In simple words: We calculate how fast the cylinder's total outside area is getting smaller. We know its height stays the same and its radius is shrinking at a certain speed. By using a formula for the surface area and its rate of change, we find the exact speed at which the area is decreasing.

🎯 Exam Tip: Remember to use a negative sign for rates of decrease. Also, pay attention to the value of \( \pi \) specified in the question (e.g., 3.142 or \( \frac{22}{7} \)).

Question 4. A balloon contains a certain mass of gas whose volume is V cm³ and whose pressure is p newton per square centimetre, and it is known that p and v obey the law \( pV = 1000 \). If the volume increases at the rate of 50 cm³ per minute, find the rate of change of p when \( v = 40 \).
Answer:
Given the law for pressure and volume: \( pV = 1000 \)
We can express p in terms of V: \( p = \frac{1000}{V} \) ...(1)
We are given the rate at which volume is increasing: \( \frac{dV}{dt} = 50 \text{ cm}^3/\text{min} \).
We need to find \( \frac{dp}{dt} \) when \( V = 40 \text{ cm}^3 \).
Differentiate equation (1) with respect to time (t):
\( \frac{dp}{dt} = \frac{d}{dt} \left(\frac{1000}{V}\right) \)
\( \frac{dp}{dt} = -1000V^{-2} \frac{dV}{dt} \)
\( \frac{dp}{dt} = -\frac{1000}{V^2} \frac{dV}{dt} \)
Substitute the given values: \( V = 40 \text{ cm}^3 \) and \( \frac{dV}{dt} = 50 \text{ cm}^3/\text{min} \).
\( \frac{dp}{dt} = -\frac{1000}{(40)^2} \times 50 \)
\( \frac{dp}{dt} = -\frac{1000}{1600} \times 50 \)
\( \frac{dp}{dt} = -\frac{10}{16} \times 50 \)
\( \frac{dp}{dt} = -\frac{5}{8} \times 50 \)
\( \frac{dp}{dt} = -\frac{250}{8} \)
\( \frac{dp}{dt} = -\frac{125}{4} \)
\( \frac{dp}{dt} = -31\frac{1}{4} \text{ N cm}^{-2}/\text{min} \)
The negative sign indicates that the pressure is decreasing. So, the rate of change of p is a decrease of \( 31\frac{1}{4} \text{ N cm}^{-2}/\text{min} \). This illustrates Boyle's Law where pressure and volume are inversely related.
In simple words: A balloon's pressure and volume are linked by a rule. If the balloon's volume grows, its pressure must shrink. We use a math method called differentiation to find out exactly how fast the pressure is going down when we know how fast the volume is expanding.

🎯 Exam Tip: Remember that if \( pV = \text{constant} \), then as V increases, p must decrease, and vice-versa. The derivative will naturally show this inverse relationship with a negative sign.

Question 5. The stone dropped into still water produces a series of continually enlarging concentric circles; it is required to find the rate per second at which the area of one of them is enlarging when its diameter is 12 cm supposing the wave to be then receding from the centre at the rate of 3 cm s¯¹.
Answer:
Let r be the radius of one of the concentric circles.
The area of the circle is given by \( A = \pi r^2 \).
We are given that the wave is receding from the center at the rate of 3 cm/s, which means the radius is increasing at this rate: \( \frac{dr}{dt} = 3 \text{ cm/s} \).
We need to find the rate at which the area is enlarging, \( \frac{dA}{dt} \), when the diameter is 12 cm.
If the diameter is 12 cm, then the radius \( r = \frac{12}{2} = 6 \) cm.
Differentiate the area formula with respect to time (t):
\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Substitute the values: \( r = 6 \) cm and \( \frac{dr}{dt} = 3 \text{ cm/s} \).
\( \frac{dA}{dt} = 2\pi (6) (3) \)
\( \frac{dA}{dt} = 36\pi \text{ cm}^2/\text{s} \)
Thus, the rate at which the area of one circle is enlarging is \( 36\pi \text{ cm}^2/\text{s} \). This calculation shows how the spread of a ripple increases its covered surface over time.
In simple words: When a stone falls in water, it makes circles that grow bigger. We want to know how fast the area of one circle is increasing when we know its size and how fast its edge is moving outwards.

🎯 Exam Tip: Always convert diameter to radius before using it in formulas involving circles, and ensure units are consistent throughout the problem.

Question 6. The volume of a cube is increasing at a constant rate. Prove that the surface increase varies inversely as the length of the edge.
Answer:
Let x be the length of the edge of the cube.
The volume of the cube is \( V = x^3 \).
The surface area of the cube is \( S = 6x^2 \).
We are given that the volume is increasing at a constant rate, so \( \frac{dV}{dt} = k \), where k is a constant.
Differentiate V with respect to time (t):
\( \frac{dV}{dt} = \frac{d}{dt}(x^3) \)
\( \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \)
Since \( \frac{dV}{dt} = k \):
\( k = 3x^2 \frac{dx}{dt} \)
Now, we can find \( \frac{dx}{dt} \):
\( \frac{dx}{dt} = \frac{k}{3x^2} \) ...(2)
Next, differentiate the surface area S with respect to time (t):
\( \frac{dS}{dt} = \frac{d}{dt}(6x^2) \)
\( \frac{dS}{dt} = 12x \frac{dx}{dt} \)
Substitute the expression for \( \frac{dx}{dt} \) from (2) into this equation:
\( \frac{dS}{dt} = 12x \left(\frac{k}{3x^2}\right) \)
\( \frac{dS}{dt} = \frac{12xk}{3x^2} \)
\( \frac{dS}{dt} = \frac{4k}{x} \)
Since \( 4k \) is a constant, this means \( \frac{dS}{dt} \propto \frac{1}{x} \). This proves that the rate at which the surface area is increasing varies inversely as the length of the edge. This relationship shows that as a cube gets larger, its surface area grows slower relative to its edge length.
In simple words: We are told a cube's volume grows at a steady speed. We need to show that its outside area grows slower as the cube gets bigger. We do this by using formulas for volume and surface area and seeing how their rates of change are connected.

🎯 Exam Tip: When proving inverse variation, your final expression should clearly show \( \frac{dS}{dt} \) as a constant divided by the variable (x in this case).

Question 7. In the function \( y = x^3 + 10 \), what is the value of x, when y increases 27 times as fast as x.
Answer:
Given the function: \( y = x^3 + 10 \)
We are told that y increases 27 times as fast as x, which means \( \frac{dy}{dt} = 27 \frac{dx}{dt} \) ...(i)
Now, differentiate the given function \( y = x^3 + 10 \) with respect to time (t):
\( \frac{dy}{dt} = \frac{d}{dt}(x^3 + 10) \)
\( \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \) ...(ii)
Now, we equate the expressions for \( \frac{dy}{dt} \) from (i) and (ii):
\( 27 \frac{dx}{dt} = 3x^2 \frac{dx}{dt} \)
Since \( \frac{dx}{dt} \) cannot be zero (because y is changing, implying x is also changing), we can divide both sides by \( \frac{dx}{dt} \):
\( 27 = 3x^2 \)
Divide by 3:
\( 9 = x^2 \)
Take the square root of both sides:
\( x = \pm 3 \)
So, the value of x is \( 3 \) or \( -3 \). This problem shows how rates of change can be used to find specific points on a curve.
In simple words: We have an equation where y depends on x. We are told that y grows 27 times faster than x. By using special math to find out how quickly y and x are changing, we can find the exact value of x that makes this statement true.

🎯 Exam Tip: When comparing rates of change like \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), always differentiate the original function with respect to time (t), not just x.

Question 8. A man 1.6 m high walks at the rate of 50 metres per minute away from a lamp which is 4 m above the ground. How fast is the man's shadow lengthening?
Answer:
Let AB be the lamp post with height \( AB = 4 \) m.
Let CD be the man's position with height \( CD = 1.6 \) m.
Let x be the distance of the man from the lamp post (BD).
Let y be the length of the man's shadow (DE).
The man is walking away from the lamp at a rate of 50 m/min, so \( \frac{dx}{dt} = 50 \text{ m/min} \).
We need to find how fast the shadow is lengthening, which is \( \frac{dy}{dt} \).


From the similar triangles \( \triangle CDE \) and \( \triangle ABE \):
\( \frac{CD}{AB} = \frac{DE}{BE} \)
\( \frac{CD}{AB} = \frac{DE}{BD + DE} \)
Substitute the known heights and variables:
\( \frac{1.6}{4} = \frac{y}{x+y} \)
Simplify the fraction:
\( \frac{16}{40} = \frac{y}{x+y} \)
\( \frac{2}{5} = \frac{y}{x+y} \)
Cross-multiply:
\( 2(x+y) = 5y \)
\( 2x + 2y = 5y \)
\( 2x = 3y \)
Now, differentiate this equation with respect to time (t):
\( \frac{d}{dt}(2x) = \frac{d}{dt}(3y) \)
\( 2 \frac{dx}{dt} = 3 \frac{dy}{dt} \)
Substitute the given rate \( \frac{dx}{dt} = 50 \text{ m/min} \):
\( 2(50) = 3 \frac{dy}{dt} \)
\( 100 = 3 \frac{dy}{dt} \)
\( \frac{dy}{dt} = \frac{100}{3} \text{ m/min} \)
\( \frac{dy}{dt} = 33\frac{1}{3} \text{ m/min} \)
Thus, the man's shadow is lengthening at the rate of \( 33\frac{1}{3} \text{ m/min} \). Understanding similar triangles is key to solving problems like this.
In simple words: A man walks away from a light pole, and his shadow grows. We use the idea of similar shapes (triangles) to set up a math problem. Then, knowing how fast the man is walking, we can figure out how quickly his shadow is getting longer.

🎯 Exam Tip: Always draw a clear diagram for related rates problems involving shadows, and correctly label all variables and their rates of change. Identify similar triangles to form the primary relationship.

Question 9. If water is poured into an inverted hollow cone whose semi-vertical angle is 30°, so that its depth (measured along the axis) increases at the rate of 1 cm s¯¹, find the rate at which the volume of water is increasing when the depth is 24 cm.
Answer:
Let r be the radius of the water surface and h be the depth (height) of the water in the cone.
The semi-vertical angle is \( 30^\circ \). In the right-angled triangle formed by the radius, height, and slant height of the water, we have:
\( \tan 30^\circ = \frac{r}{h} \)
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we get:
\( \frac{1}{\sqrt{3}} = \frac{r}{h} \)
\( r = \frac{h}{\sqrt{3}} \)
The volume of water (V) in the cone is given by:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute \( r = \frac{h}{\sqrt{3}} \) into the volume formula:
\( V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h \)
\( V = \frac{1}{3}\pi \left(\frac{h^2}{3}\right) h \)
\( V = \frac{\pi}{9}h^3 \)
We are given that the depth is increasing at \( \frac{dh}{dt} = 1 \text{ cm/s} \).
We need to find \( \frac{dV}{dt} \) when \( h = 24 \) cm.
Differentiate V with respect to time (t):
\( \frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{9}h^3\right) \)
\( \frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \frac{dh}{dt} \)
\( \frac{dV}{dt} = \frac{\pi}{3}h^2 \frac{dh}{dt} \)
Substitute the given values: \( h = 24 \) cm and \( \frac{dh}{dt} = 1 \text{ cm/s} \).
\( \frac{dV}{dt} = \frac{\pi}{3}(24)^2 (1) \)
\( \frac{dV}{dt} = \frac{\pi}{3}(576) \)
\( \frac{dV}{dt} = 192\pi \text{ cm}^3/\text{s} \)
Thus, the volume of water is increasing at the rate of \( 192\pi \text{ cm}^3/\text{s} \). This calculation helps in understanding how fast a container fills up.
In simple words: Water is flowing into a cone, making the water level rise. We use the cone's shape and how fast the water depth is increasing to figure out how quickly the total amount of water (volume) in the cone is growing at a specific depth.

🎯 Exam Tip: For cone problems, always use the relationship between radius and height (often involving \( \tan \) or \( \cot \) of the semi-vertical angle) to express the volume solely in terms of either r or h, before differentiating.

Question 10. The area of a circle is increasing at the uniform rate of 5 cm² per minute. Calculate the rate, in centimetre per minute, at which the radius is increasing when the circumference of the circle is 40 cm.
Answer:
Let r be the radius of the circle.
The area of the circle is \( A = \pi r^2 \).
The circumference of the circle is \( C = 2\pi r \).
We are given the rate of increase of the area: \( \frac{dA}{dt} = 5 \text{ cm}^2/\text{min} \).
We need to find the rate of increase of the radius, \( \frac{dr}{dt} \), when the circumference is 40 cm.
First, let's find the radius when the circumference is 40 cm:
\( 2\pi r = 40 \)
\( r = \frac{40}{2\pi} = \frac{20}{\pi} \text{ cm} \)
Now, differentiate the area formula with respect to time (t):
\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Substitute the given value for \( \frac{dA}{dt} = 5 \):
\( 5 = 2\pi r \frac{dr}{dt} \)
Now, substitute the value of r we found when the circumference is 40 cm:
\( 5 = 2\pi \left(\frac{20}{\pi}\right) \frac{dr}{dt} \)
\( 5 = (2 \times 20) \frac{dr}{dt} \)
\( 5 = 40 \frac{dr}{dt} \)
Solve for \( \frac{dr}{dt} \):
\( \frac{dr}{dt} = \frac{5}{40} \)
\( \frac{dr}{dt} = \frac{1}{8} \text{ cm/min} \)
Thus, the rate at which the radius is increasing is \( \frac{1}{8} \text{ cm/min} \). This illustrates the relationship between a circle's size and how its dimensions change over time.
In simple words: A circle's area is growing at a steady speed. We want to find out how fast its radius is growing when the circle's edge (circumference) is a specific length. We use formulas for area and circumference to link these rates together.

🎯 Exam Tip: When a question gives a condition (like a specific circumference), use it to find the relevant dimension (radius in this case) before substituting into the differentiated rate equation.

Question 11. A kite is 112 metres above the ground and has 130 metres string out. If the kite is travelling horizontally at 8 m/s directly away from the boy who is flying it, at what rate is the string out?
Answer:
Let h be the height of the kite above the ground, \( h = 112 \) m (constant).
Let x be the horizontal distance of the kite from the boy.
Let s be the length of the string out.
We are given that the kite is travelling horizontally away from the boy at \( \frac{dx}{dt} = 8 \text{ m/s} \).
We need to find the rate at which the string is being let out, \( \frac{ds}{dt} \), when the string length \( s = 130 \) m.


From the right-angled triangle formed by the boy, the point on the ground directly below the kite, and the kite itself:
\( x^2 + h^2 = s^2 \)
Since \( h = 112 \) (constant):
\( x^2 + (112)^2 = s^2 \) ...(i)
First, find x when \( s = 130 \) m:
\( x^2 + (112)^2 = (130)^2 \)
\( x^2 + 12544 = 16900 \)
\( x^2 = 16900 - 12544 \)
\( x^2 = 4356 \)
\( x = \sqrt{4356} \)
\( x = 66 \) m
Now, differentiate equation (i) with respect to time (t):
\( \frac{d}{dt}(x^2 + (112)^2) = \frac{d}{dt}(s^2) \)
\( 2x \frac{dx}{dt} + 0 = 2s \frac{ds}{dt} \)
\( 2x \frac{dx}{dt} = 2s \frac{ds}{dt} \)
\( x \frac{dx}{dt} = s \frac{ds}{dt} \)
Substitute the known values: \( x = 66 \) m, \( \frac{dx}{dt} = 8 \text{ m/s} \), and \( s = 130 \) m.
\( 66 \times 8 = 130 \times \frac{ds}{dt} \)
\( 528 = 130 \frac{ds}{dt} \)
\( \frac{ds}{dt} = \frac{528}{130} \)
\( \frac{ds}{dt} = \frac{264}{65} \)
\( \frac{ds}{dt} \approx 4.0615 \text{ m/s} \)
Rounding to two decimal places, the required rate at which the string is out is \( 4.06 \text{ m/s} \). This problem is a good example of related rates involving the Pythagorean theorem.
In simple words: A boy is flying a kite, which stays at a certain height but moves sideways. We know how fast the kite is moving horizontally. We want to find out how quickly the string needs to be let out (or pulled in) at a certain moment, using the lengths of the kite's height, horizontal distance, and string.

🎯 Exam Tip: When solving related rates problems, first identify all constants and variables, then find an equation relating these variables, and finally differentiate with respect to time.

Question 12. 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 m/sec. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
Answer:
Let x be the distance of the foot of the ladder from the wall.
Let y be the height of the ladder on the wall.
The length of the ladder is constant at 5 m.
We are given that the bottom of the ladder is pulled away from the wall at \( \frac{dx}{dt} = 2 \text{ m/s} \).
We need to find \( \frac{dy}{dt} \) when \( x = 4 \) m.


Using the Pythagorean theorem:
\( x^2 + y^2 = 5^2 \)
\( x^2 + y^2 = 25 \) ...(i)
First, find y when \( x = 4 \) m:
\( (4)^2 + y^2 = 25 \)
\( 16 + y^2 = 25 \)
\( y^2 = 25 - 16 \)
\( y^2 = 9 \)
\( y = 3 \) m (since height must be positive)
Now, differentiate equation (i) with respect to time (t):
\( \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25) \)
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)
Divide by 2:
\( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \)
Substitute the known values: \( x = 4 \) m, \( y = 3 \) m, and \( \frac{dx}{dt} = 2 \text{ m/s} \).
\( (4)(2) + (3) \frac{dy}{dt} = 0 \)
\( 8 + 3 \frac{dy}{dt} = 0 \)
\( 3 \frac{dy}{dt} = -8 \)
\( \frac{dy}{dt} = -\frac{8}{3} \text{ m/s} \)
The negative sign indicates that the height is decreasing. So, the height of the ladder on the wall is decreasing at the rate of \( \frac{8}{3} \text{ m/s} \). This problem showcases how the rate of change of one side of a right triangle impacts another.
In simple words: A ladder leans against a wall. As we pull the bottom of the ladder away from the wall, the top of the ladder slides down. We want to find out how fast the top is sliding down when the bottom is at a certain distance from the wall, knowing how fast the bottom is being pulled.

🎯 Exam Tip: In ladder problems, the length of the ladder is always constant, which means its derivative with respect to time is zero. Make sure to find the missing side length using Pythagoras theorem before substituting values into the differentiated equation.

Question 13. Water is dripping out from a conical funnel, at the uniform rate of 2 cm³/s through a tiny hole at the vertex of the bottom. When the slant height of the water is 4 cm, find the rate of decrease of the slant height of the water given that the vertical angle of the funnel is 60°.
Answer:
Let r be the radius, h be the height, and l be the slant height of the water in the conical funnel.
The vertical angle of the funnel is \( 60^\circ \), so the semi-vertical angle is \( \frac{60^\circ}{2} = 30^\circ \).
From the right-angled triangle formed by r, h, and l:
\( \sin 30^\circ = \frac{r}{l} \implies r = l \sin 30^\circ = l \left(\frac{1}{2}\right) = \frac{l}{2} \)
\( \cos 30^\circ = \frac{h}{l} \implies h = l \cos 30^\circ = l \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}l}{2} \)
The volume of water (V) in the cone is given by:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute the expressions for r and h in terms of l:
\( V = \frac{1}{3}\pi \left(\frac{l}{2}\right)^2 \left(\frac{\sqrt{3}l}{2}\right) \)
\( V = \frac{1}{3}\pi \left(\frac{l^2}{4}\right) \left(\frac{\sqrt{3}l}{2}\right) \)
\( V = \frac{\pi \sqrt{3}}{24}l^3 \)
We are given that water is dripping out at a uniform rate, so \( \frac{dV}{dt} = -2 \text{ cm}^3/\text{s} \) (negative because volume is decreasing).
We need to find \( \frac{dl}{dt} \) when \( l = 4 \) cm.
Differentiate V with respect to time (t):
\( \frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi \sqrt{3}}{24}l^3\right) \)
\( \frac{dV}{dt} = \frac{\pi \sqrt{3}}{24} \cdot 3l^2 \frac{dl}{dt} \)
\( \frac{dV}{dt} = \frac{\pi \sqrt{3}}{8}l^2 \frac{dl}{dt} \)
Substitute the known values: \( \frac{dV}{dt} = -2 \) and \( l = 4 \).
\( -2 = \frac{\pi \sqrt{3}}{8}(4)^2 \frac{dl}{dt} \)
\( -2 = \frac{\pi \sqrt{3}}{8}(16) \frac{dl}{dt} \)
\( -2 = 2\pi \sqrt{3} \frac{dl}{dt} \)
Solve for \( \frac{dl}{dt} \):
\( \frac{dl}{dt} = -\frac{2}{2\pi \sqrt{3}} \)
\( \frac{dl}{dt} = -\frac{1}{\pi \sqrt{3}} \text{ cm/s} \)
The negative sign indicates that the slant height is decreasing. Thus, the rate of decrease of the slant height of the water is \( \frac{1}{\pi \sqrt{3}} \text{ cm/s} \). It's helpful to express all dimensions in terms of a single variable, like slant height, before differentiating.
In simple words: Water is leaking from a cone, so its volume and the slant height of the water level are decreasing. We know how fast the water is leaving. We use the cone's angle and how its volume relates to its slant height to find out how quickly the slant height is getting smaller when it's at a specific length.

🎯 Exam Tip: For problems involving cones with fixed angles, always relate the radius and height to the slant height using trigonometric ratios of the semi-vertical angle. This simplifies the volume formula to one variable.

Question 14.
(i) The side of an equilateral triangle is a cm long and is increasing at the rate of k cm/hr. Find the rate at which the area increases, when the side is 10 cm.
(ii) The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which the area increases, when the side is 10 cm.
Answer:
(i) Let the side of the equilateral triangle be a cm.
The area of an equilateral triangle (A) is given by: \( A = \frac{\sqrt{3}}{4}a^2 \)
We are given the rate of increase of the side: \( \frac{da}{dt} = k \text{ cm/hr} \).
We need to find the rate at which the area increases, \( \frac{dA}{dt} \).
Differentiate A with respect to time (t):
\( \frac{dA}{dt} = \frac{d}{dt}\left(\frac{\sqrt{3}}{4}a^2\right) \)
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \frac{da}{dt} \)
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}a \frac{da}{dt} \)
Substitute \( \frac{da}{dt} = k \):
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}ak \text{ cm}^2/\text{hr} \)
Now, we need to find this rate when \( a = 10 \) cm:
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}(10)k \)
\( \frac{dA}{dt} = 5\sqrt{3}k \text{ cm}^2/\text{hr} \)
Thus, the area of the equilateral triangle is increasing at the rate of \( 5\sqrt{3}k \text{ cm}^2/\text{hr} \).

(ii) Let the side of the equilateral triangle be a cm.
The area of an equilateral triangle (A) is given by: \( A = \frac{\sqrt{3}}{4}a^2 \)
We are given the rate of increase of the side: \( \frac{da}{dt} = 2 \text{ cm/sec} \).
We need to find the rate at which the area increases, \( \frac{dA}{dt} \), when \( a = 10 \) cm.
Differentiate A with respect to time (t):
\( \frac{dA}{dt} = \frac{d}{dt}\left(\frac{\sqrt{3}}{4}a^2\right) \)
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \frac{da}{dt} \)
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}a \frac{da}{dt} \)
Substitute the given values: \( a = 10 \) cm and \( \frac{da}{dt} = 2 \text{ cm/sec} \).
\( \frac{dA}{dt} = \frac{\sqrt{3}}{2}(10)(2) \)
\( \frac{dA}{dt} = \sqrt{3}(10) \)
\( \frac{dA}{dt} = 10\sqrt{3} \text{ cm}^2/\text{sec} \)
Thus, the area of the equilateral triangle is increasing at the rate of \( 10\sqrt{3} \text{ cm}^2/\text{sec} \). Both parts illustrate how the area of a shape changes with its dimensions.
In simple words: We know the formula for the area of an equilateral triangle. If its sides are growing at a certain speed, we use that information to calculate how fast the triangle's area is growing when the side reaches a particular length.

🎯 Exam Tip: Remember the formula for the area of an equilateral triangle, \( A = \frac{\sqrt{3}}{4}a^2 \). Differentiate it correctly using the chain rule with respect to time.

Question 15. The diameter and altitude of a right circular cylinder are found at a certain instant to be 10 cm and 20 cm respectively. If the diameter is increasing at the rate of 2 cm per sec, what change in the altitude will keep the volume constant ?
Answer:
Let x be the diameter and h be the altitude (height) of the right circular cylinder.
The radius is \( r = \frac{x}{2} \).
The volume of the cylinder (V) is given by: \( V = \pi r^2 h \)
Substitute \( r = \frac{x}{2} \):
\( V = \pi \left(\frac{x}{2}\right)^2 h \)
\( V = \pi \frac{x^2}{4} h \)
\( V = \frac{\pi}{4}x^2 h \)
We are given that the volume is constant, so \( \frac{dV}{dt} = 0 \).
Differentiate V with respect to time (t):
\( \frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{4}x^2 h\right) \)
Using the product rule \( (uv)' = u'v + uv' \):
\( \frac{dV}{dt} = \frac{\pi}{4} \left( \frac{d}{dt}(x^2)h + x^2 \frac{d}{dt}(h) \right) \)
\( \frac{dV}{dt} = \frac{\pi}{4} \left( 2x \frac{dx}{dt} h + x^2 \frac{dh}{dt} \right) \)
Since \( \frac{dV}{dt} = 0 \):
\( 0 = \frac{\pi}{4} \left( 2x h \frac{dx}{dt} + x^2 \frac{dh}{dt} \right) \)
Since \( \frac{\pi}{4} \neq 0 \), we can write:
\( 2x h \frac{dx}{dt} + x^2 \frac{dh}{dt} = 0 \)
We are given:
Instantaneous diameter \( x = 10 \) cm
Instantaneous altitude \( h = 20 \) cm
Rate of increase of diameter \( \frac{dx}{dt} = 2 \text{ cm/sec} \)
Substitute these values into the equation:
\( 2(10)(20)(2) + (10)^2 \frac{dh}{dt} = 0 \)
\( 800 + 100 \frac{dh}{dt} = 0 \)
\( 100 \frac{dh}{dt} = -800 \)
\( \frac{dh}{dt} = -\frac{800}{100} \)
\( \frac{dh}{dt} = -8 \text{ cm/sec} \)
The negative sign indicates that the altitude is decreasing. Thus, the altitude must decrease at the rate of \( 8 \text{ cm/sec} \) to keep the volume constant. This shows an inverse relationship between changes in diameter and height to maintain a fixed volume.
In simple words: A cylinder's volume needs to stay the same. If its width (diameter) is growing, its height (altitude) must shrink to balance it out. We calculate how fast the height needs to shrink, given the cylinder's current size and how fast its diameter is expanding.

🎯 Exam Tip: When the volume needs to remain constant, its rate of change \( \frac{dV}{dt} \) is zero. Remember to apply the product rule correctly when differentiating the volume formula with two changing variables (diameter and height).

Question 16. The radius of the base of a certain cone is increasing, at the rate of 3 cm per minute and the altitude is decreasing at the rate of 4 cm per minute. Find the rate of change of total surface of the cone when the radius is 7 cm and the altitude is 24 cm.
Answer:
Let r be the radius of the base and h be the altitude (height) of the cone.
The total surface area (S) of a cone is given by: \( S = \pi r^2 + \pi r l \), where l is the slant height.
The slant height \( l = \sqrt{r^2 + h^2} \).
So, \( S = \pi r^2 + \pi r \sqrt{r^2 + h^2} \) ...(i)
We are given:
\( \frac{dr}{dt} = 3 \text{ cm/min} \)
\( \frac{dh}{dt} = -4 \text{ cm/min} \) (negative because altitude is decreasing)
We need to find \( \frac{dS}{dt} \) when \( r = 7 \) cm and \( h = 24 \) cm.
First, calculate the slant height l at this instant:
\( l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \) cm.
Now, differentiate S with respect to time (t):
\( \frac{dS}{dt} = \frac{d}{dt}(\pi r^2) + \frac{d}{dt}(\pi r \sqrt{r^2 + h^2}) \)
\( \frac{dS}{dt} = 2\pi r \frac{dr}{dt} + \pi \left( \frac{dr}{dt}\sqrt{r^2 + h^2} + r \frac{d}{dt}(\sqrt{r^2 + h^2}) \right) \)
Let's differentiate \( \sqrt{r^2 + h^2} \) separately:
\( \frac{d}{dt}(\sqrt{r^2 + h^2}) = \frac{1}{2\sqrt{r^2 + h^2}} \left( 2r \frac{dr}{dt} + 2h \frac{dh}{dt} \right) = \frac{r \frac{dr}{dt} + h \frac{dh}{dt}}{\sqrt{r^2 + h^2}} \)
Substitute this back into the \( \frac{dS}{dt} \) equation:
\( \frac{dS}{dt} = 2\pi r \frac{dr}{dt} + \pi \left( \sqrt{r^2 + h^2} \frac{dr}{dt} + r \frac{r \frac{dr}{dt} + h \frac{dh}{dt}}{\sqrt{r^2 + h^2}} \right) \)
Substitute \( r = 7, h = 24, l = 25, \frac{dr}{dt} = 3, \frac{dh}{dt} = -4 \):
\( \frac{dS}{dt} = 2\pi(7)(3) + \pi \left( (25)(3) + 7 \frac{7(3) + 24(-4)}{25} \right) \)
\( \frac{dS}{dt} = 42\pi + \pi \left( 75 + 7 \frac{21 - 96}{25} \right) \)
\( \frac{dS}{dt} = 42\pi + \pi \left( 75 + 7 \frac{-75}{25} \right) \)
\( \frac{dS}{dt} = 42\pi + \pi \left( 75 + 7(-3) \right) \)
\( \frac{dS}{dt} = 42\pi + \pi (75 - 21) \)
\( \frac{dS}{dt} = 42\pi + 54\pi \)
\( \frac{dS}{dt} = 96\pi \text{ cm}^2/\text{min} \)
Thus, the total surface area of the cone is increasing at the rate of \( 96\pi \text{ cm}^2/\text{min} \). This calculation reveals how multiple changes in dimensions affect the overall surface area of a 3D shape.
In simple words: The base of a cone is getting wider, but its height is getting shorter. We want to find out how quickly its total outside area is changing when the cone has specific measurements. We use a formula that connects the radius, height, and slant height to find this rate.

🎯 Exam Tip: This type of problem requires careful application of the chain rule and product rule. Always calculate the slant height first, as it's often a necessary intermediate step.

Question 17. Water is dripping out at the steady rate of 1 c.c. per second through a tiny hole at the vertex of a conical vessel whose axis is vertical. When the slant height of the water in the filter is 4 cm, find the rate of decrease of (i) the slant height of water, (ii) the area of the water surface, given that the vertical angle of the vessaf is 60°.
Answer:
Let r be the radius, h be the height, and l be the slant height of the water.
The vertical angle of the funnel is \( 60^\circ \), so the semi-vertical angle is \( 30^\circ \).
From a right-angled triangle, we have:
\( r = l \sin 30^\circ = \frac{l}{2} \)
\( h = l \cos 30^\circ = \frac{\sqrt{3}l}{2} \)
The volume of water (V) is \( V = \frac{1}{3}\pi r^2 h \). Substitute r and h in terms of l:
\( V = \frac{1}{3}\pi \left(\frac{l}{2}\right)^2 \left(\frac{\sqrt{3}l}{2}\right) = \frac{1}{3}\pi \frac{l^2}{4} \frac{\sqrt{3}l}{2} = \frac{\pi \sqrt{3}}{24}l^3 \)
We are given \( \frac{dV}{dt} = -1 \text{ cm}^3/\text{s} \) (negative because water is dripping out).
Differentiate V with respect to time (t):
\( \frac{dV}{dt} = \frac{\pi \sqrt{3}}{24} \cdot 3l^2 \frac{dl}{dt} = \frac{\pi \sqrt{3}}{8}l^2 \frac{dl}{dt} \)

(i) Find the rate of decrease of the slant height of water \( \left(\frac{dl}{dt}\right) \) when \( l = 4 \) cm:
Substitute \( \frac{dV}{dt} = -1 \) and \( l = 4 \):
\( -1 = \frac{\pi \sqrt{3}}{8}(4)^2 \frac{dl}{dt} \)
\( -1 = \frac{\pi \sqrt{3}}{8}(16) \frac{dl}{dt} \)
\( -1 = 2\pi \sqrt{3} \frac{dl}{dt} \)
\( \frac{dl}{dt} = -\frac{1}{2\pi \sqrt{3}} \text{ cm/s} \)
The slant height is decreasing at the rate of \( \frac{1}{2\pi \sqrt{3}} \text{ cm/s} \).

(ii) Find the rate of decrease of the area of the water surface \( \left(\frac{dA}{dt}\right) \) when \( l = 4 \) cm:
The area of the water surface (A) is a circle: \( A = \pi r^2 \)
Substitute \( r = \frac{l}{2} \): \( A = \pi \left(\frac{l}{2}\right)^2 = \frac{\pi}{4}l^2 \)
Differentiate A with respect to time (t):
\( \frac{dA}{dt} = \frac{\pi}{4} \cdot 2l \frac{dl}{dt} = \frac{\pi}{2}l \frac{dl}{dt} \)
Substitute \( l = 4 \) cm and \( \frac{dl}{dt} = -\frac{1}{2\pi \sqrt{3}} \text{ cm/s} \):
\( \frac{dA}{dt} = \frac{\pi}{2}(4) \left(-\frac{1}{2\pi \sqrt{3}}\right) \)
\( \frac{dA}{dt} = 2\pi \left(-\frac{1}{2\pi \sqrt{3}}\right) \)
\( \frac{dA}{dt} = -\frac{1}{\sqrt{3}} \text{ cm}^2/\text{s} \)
The area of the water surface is decreasing at the rate of \( \frac{1}{\sqrt{3}} \text{ cm}^2/\text{s} \). These calculations help us understand how quickly different aspects of the water's shape change as it drains.
In simple words: Water is leaking from a cone, so everything is shrinking. We calculate how fast the slant height (the sloping side of the water level) is decreasing and then how fast the circular surface area of the water is shrinking at a particular moment.

🎯 Exam Tip: Break down complex problems into smaller parts. First find the rate of change of the primary variable (slant height in this case), then use it to find the rates of change of other related quantities.

Question 18. If the area of circle increases at a uniform rate, show that the rate of increase of the perimeter varies inversely as the radius.
Answer:
Let r be the radius of the circle.
The area of the circle is \( A = \pi r^2 \).
The perimeter (circumference) of the circle is \( P = 2\pi r \).
We are given that the area of the circle increases at a uniform rate, meaning \( \frac{dA}{dt} = k \), where k is a constant.
Differentiate A with respect to time (t):
\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Since \( \frac{dA}{dt} = k \):
\( k = 2\pi r \frac{dr}{dt} \)
From this, we can express \( \frac{dr}{dt} \):
\( \frac{dr}{dt} = \frac{k}{2\pi r} \) ...(1)
Now, differentiate the perimeter P with respect to time (t):
\( \frac{dP}{dt} = \frac{d}{dt}(2\pi r) \)
\( \frac{dP}{dt} = 2\pi \frac{dr}{dt} \)
Substitute the expression for \( \frac{dr}{dt} \) from (1) into this equation:
\( \frac{dP}{dt} = 2\pi \left(\frac{k}{2\pi r}\right) \)
\( \frac{dP}{dt} = \frac{k}{r} \)
Since k is a constant, this equation shows that \( \frac{dP}{dt} \) is inversely proportional to r (i.e., \( \frac{dP}{dt} \propto \frac{1}{r} \)). Thus, the rate of increase of the perimeter of the circle varies inversely as the radius. This means a larger circle will have its perimeter grow slower even if its area is growing at a steady pace.
In simple words: If a circle's area always grows at the same speed, we need to show that its perimeter (the distance around it) grows slower as the circle gets bigger. We use math to link how the area, radius, and perimeter change over time to prove this.

🎯 Exam Tip: To prove inverse proportionality, ensure your final derived rate is expressed as a constant divided by the variable in question. Clearly define your constants and show each differentiation step.

Question 18. If the area of circle increases at a uniform rate, show that the rate of increase of the perimeter varies inversely as the radius.
Answer: Let \( r \) be the radius of the circle, \( A \) be its area, and \( P \) be its perimeter.
The area of a circle is given by \( A = \pi r^2 \).
The perimeter of a circle is given by \( P = 2 \pi r \).
We differentiate both equations with respect to time \( t \):
For the area: \( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \implies \frac{dA}{dt} = 2 \pi r \frac{dr}{dt} \) (Equation 1)
For the perimeter: \( \frac{dP}{dt} = \frac{d}{dt}(2 \pi r) \)
\( \implies \frac{dP}{dt} = 2 \pi \frac{dr}{dt} \) (Equation 2)
We are given that the area increases at a uniform rate, so \( \frac{dA}{dt} = k \), where \( k \) is a constant.
From Equation 1, we can express \( \frac{dr}{dt} \):
\( k = 2 \pi r \frac{dr}{dt} \)
\( \implies \frac{dr}{dt} = \frac{k}{2 \pi r} \)
Now, substitute this expression for \( \frac{dr}{dt} \) into Equation 2:
\( \frac{dP}{dt} = 2 \pi \left( \frac{k}{2 \pi r} \right) \)
\( \implies \frac{dP}{dt} = \frac{k}{r} \)
Since \( k \) is a constant, \( 2\pi \) is also a constant, the term \( \frac{k}{r} \) means that \( \frac{dP}{dt} \) is directly proportional to \( \frac{1}{r} \). This shows that the rate of increase of the perimeter varies inversely as the radius. This relationship highlights how the rate of change of one geometric property can be linked to another.
In simple words: When a circle's area grows steadily, its outer edge (perimeter) grows faster when the circle is small, and slower when it is big. This means the speed at which the perimeter grows depends on the opposite of its size.

🎯 Exam Tip: When proving relationships involving rates of change, always start by defining your variables and writing down the basic formulas for area, perimeter, or volume. Then, differentiate these formulas with respect to time and substitute known rates to establish the desired relationship.