GSEB Class 12 Maths Solutions Chapter 6 Application of Derivatives Exercise 6.1

Question 1. Find the rate of change of the area of a circle with respect to its radius, when
(a) r = 3 cm (b) r = 4 cm
Answer: Let \( A \) represent the area of the circle and \( r \) be its radius. The formula for the area of a circle is given by \( A = \pi r^2 \).
To find the rate of change of area with respect to radius, we differentiate \( A \) with respect to \( r \):
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) \)
\( \implies \frac{dA}{dr} = 2\pi r \)
(a) When \( r = 3 \) cm:
\( \frac{dA}{dr} = 2\pi (3) = 6\pi \text{ cm}^2/\text{cm} \)
(b) When \( r = 4 \) cm:
\( \frac{dA}{dr} = 2\pi (4) = 8\pi \text{ cm}^2/\text{cm} \)
In simple words: We calculate how quickly the circle's area grows as its radius increases. This means taking the derivative of the area formula. Then we put in the given radius values to find the exact rate for each.

Exam Tip: Remember to differentiate the area formula with respect to the radius and substitute the given radius values to find the specific rates of change. Pay attention to the units.

Question 2. The volume of a cube is increasing at the rate of 8 cm³/sec. How fast is the surface area increasing, when the length of an edge is 12 cm?
Answer: Let \( x \) be the length of an edge of the cube. The volume \( V \) of the cube is \( V = x^3 \), and its surface area \( S \) is \( S = 6x^2 \).
We are given that the rate of change of volume is \( \frac{dV}{dt} = 8 \text{ cm}^3/\text{sec} \).
We need to find the rate of change of surface area, \( \frac{dS}{dt} \).
Differentiating \( V \) with respect to time \( t \):
\( \frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt} \)
Since \( \frac{dV}{dt} = 8 \):
\( 8 = 3x^2 \frac{dx}{dt} \)
\( \implies \frac{dx}{dt} = \frac{8}{3x^2} \) ...(1)
Now, differentiating \( S \) with respect to time \( t \):
\( \frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt} \)
Substitute \( \frac{dx}{dt} \) from equation (1) into this equation:
\( \frac{dS}{dt} = 12x \left(\frac{8}{3x^2}\right) \)
\( \implies \frac{dS}{dt} = \frac{96x}{3x^2} = \frac{32}{x} \)
When the length of the edge \( x = 12 \) cm:
\( \frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{sec} \)
In simple words: The volume of a cube is growing at a certain speed. We need to figure out how fast its outer surface is growing when the edge is 12 cm. First, we find a formula for how the edge length changes over time, then use that to find the surface area's change rate.

Exam Tip: For related rates problems, first identify the given rates and the rate you need to find. Then establish relationships between the variables using appropriate formulas (like volume and surface area). Finally, differentiate with respect to time and substitute the known values.

Question 3. The radius of a circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing, when the radius is 10 cm.
Answer: Let \( r \) be the radius of the circle and \( A \) be its area. We are given that the radius is increasing at a uniform rate of \( \frac{dr}{dt} = 3 \text{ cm/sec} \).
The area of a circle is given by \( A = \pi r^2 \).
To find the rate at which the area is increasing, we differentiate \( A \) with respect to time \( t \):
\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Substitute the given value \( \frac{dr}{dt} = 3 \text{ cm/sec} \):
\( \frac{dA}{dt} = 2\pi r (3) \)
\( \implies \frac{dA}{dt} = 6\pi r \)
Now, we need to find this rate when the radius \( r = 10 \) cm:
\( \frac{dA}{dt} = 6\pi (10) = 60\pi \text{ cm}^2/\text{sec} \)
In simple words: We know how fast a circle's radius is getting bigger. We need to calculate how quickly the circle's total area is growing when the radius reaches 10 cm. We use the area formula, differentiate it, and then put in the numbers.

Exam Tip: When solving related rates problems, ensure you differentiate the formula relating the variables with respect to time (using the chain rule) before substituting specific values. This ensures you account for how each variable changes over time.

Question 4. 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 10 cm long?
Answer: Let \( x \) be the length of the edge of the cube. The rate of change of the edge is given as \( \frac{dx}{dt} = 3 \text{ cm/sec} \).
The volume \( V \) of a cube with edge length \( x \) is \( V = x^3 \).
We need to find the rate of increase of volume, \( \frac{dV}{dt} \), when \( x = 10 \) cm.
Differentiate the volume formula with respect to time \( t \):
\( \frac{dV}{dt} = \frac{d}{dt}(x^3) \)
\( \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \)
Substitute the given value \( \frac{dx}{dt} = 3 \text{ cm/sec} \):
\( \frac{dV}{dt} = 3x^2 (3) \)
\( \implies \frac{dV}{dt} = 9x^2 \)
Now, we evaluate this rate when the edge length \( x = 10 \) cm:
\( \frac{dV}{dt} = 9(10)^2 = 9(100) = 900 \text{ cm}^3/\text{sec} \)
In simple words: If a cube's side length is growing at a constant speed, we want to know how fast its total volume is expanding when its side length reaches 10 cm. We use the volume formula for a cube, find its derivative with respect to time, and then plug in the numbers.

Exam Tip: Pay close attention to whether the question asks for the rate of change of an edge, area, or volume. Ensure you use the correct formula and differentiate it correctly using the chain rule with respect to time.

Question 5. A stone is dropped into a quiet lake and waves move in circles at the rate of 5 cm/sec. At the instant, when radius of the circular wave is 8 cm, how fast is the enclosed area increasing ?
Answer: Let \( r \) be the radius of the circular wave and \( A \) be the area enclosed by the wave.
We are given that the radius of the circular wave is increasing at the rate of \( \frac{dr}{dt} = 5 \text{ cm/sec} \).
The area of a circle is given by \( A = \pi r^2 \).
We need to find the rate at which the enclosed area is increasing, \( \frac{dA}{dt} \), when \( r = 8 \) cm.
Differentiate the area formula with respect to time \( t \):
\( \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \)
\( \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)
Substitute the given value \( \frac{dr}{dt} = 5 \text{ cm/sec} \):
\( \frac{dA}{dt} = 2\pi r (5) \)
\( \implies \frac{dA}{dt} = 10\pi r \)
Now, we evaluate this rate when the radius \( r = 8 \) cm:
\( \frac{dA}{dt} = 10\pi (8) = 80\pi \text{ cm}^2/\text{sec} \)
In simple words: When a stone makes ripples, the circle gets bigger. We know how fast the radius grows. We want to find out how quickly the area inside that circle is expanding when the radius is 8 cm. We will use the area formula, find its rate of change, and then plug in the numbers.

Exam Tip: Remember that "rate of change" implies differentiation with respect to time. Always write down the given information and the quantity you need to find. Use the chain rule for differentiation if the variable is dependent on time.

Question 6. The radius of a circle is increasing at 0.7 cm/s. What is the rate of increase of its circumference ?
Answer: Let \( r \) be the radius of the circle and \( C \) be its circumference.
We are given that the radius is increasing at the rate of \( \frac{dr}{dt} = 0.7 \text{ cm/sec} \).
The circumference of a circle is given by \( C = 2\pi r \).
We need to find the rate of increase of its circumference, \( \frac{dC}{dt} \).
Differentiate the circumference formula with respect to time \( t \):
\( \frac{dC}{dt} = \frac{d}{dt}(2\pi r) \)
\( \implies \frac{dC}{dt} = 2\pi \frac{dr}{dt} \)
Substitute the given value \( \frac{dr}{dt} = 0.7 \text{ cm/sec} \):
\( \frac{dC}{dt} = 2\pi (0.7) \)
\( \implies \frac{dC}{dt} = 1.4\pi \text{ cm/sec} \)
In simple words: If a circle's radius is growing at a certain speed, we need to find how fast its perimeter (circumference) is growing. We use the circumference formula, differentiate it with respect to time, and then insert the given rate of radius increase.

Exam Tip: Distinguish between area and circumference formulas. When a rate of change is asked, always remember to differentiate with respect to time and apply the chain rule correctly. Units are important to include in your final answer.

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, then y = 6 cm. Find the rate of change of
(a) the perimeter and
(b) the area of the rectangle.
Answer: Let \( x \) be the length and \( y \) be the width of the rectangle.
Given rates:
\( \frac{dx}{dt} = -5 \text{ cm/min} \) (decreasing length)
\( \frac{dy}{dt} = 4 \text{ cm/min} \) (increasing width)
Given dimensions at a specific instant: \( x = 8 \) cm and \( y = 6 \) cm.

(a) **Rate of change of the perimeter**
The perimeter \( P \) of a rectangle is \( P = 2(x + y) \).
Differentiate \( P \) with respect to time \( t \):
\( \frac{dP}{dt} = \frac{d}{dt}(2(x + y)) \)
\( \implies \frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} \right) \)
Substitute the given rates:
\( \frac{dP}{dt} = 2(-5 + 4) \)
\( \implies \frac{dP}{dt} = 2(-1) = -2 \text{ cm/min} \)
This means the perimeter is decreasing at the rate of \( 2 \text{ cm/min} \).

(b) **Rate of change of the area**
The area \( A \) of a rectangle is \( A = xy \).
Differentiate \( A \) with respect to time \( t \) using the product rule:
\( \frac{dA}{dt} = \frac{d}{dt}(xy) \)
\( \implies \frac{dA}{dt} = \frac{dx}{dt} y + x \frac{dy}{dt} \)
Substitute the given values for \( x, y, \frac{dx}{dt}, \) and \( \frac{dy}{dt} \):
\( \frac{dA}{dt} = (-5)(6) + (8)(4) \)
\( \implies \frac{dA}{dt} = -30 + 32 \)
\( \implies \frac{dA}{dt} = 2 \text{ cm}^2/\text{min} \)
This means the area is increasing at the rate of \( 2 \text{ cm}^2/\text{min} \).
In simple words: The length of a rectangle is getting smaller while its width is getting larger. We need to find out how fast both its total outside length (perimeter) and its inside space (area) are changing at a specific moment when we know the length and width. We will differentiate the formulas for perimeter and area and then plug in the rates and dimensions.

Exam Tip: For problems involving rates of change of geometric figures, clearly define variables and write down all given and required rates. Remember to use the product rule for differentiating area if both length and width are changing. Negative rates indicate a decrease.

Question 8. A variable cube 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 increases, when the radius is 15 cm.
Answer: Let \( r \) be the radius of the spherical balloon and \( V \) be its volume.
We are given that gas is pumped into the balloon at a rate of \( \frac{dV}{dt} = 900 \text{ cm}^3/\text{sec} \).
The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
We need to find the rate at which the radius increases, \( \frac{dr}{dt} \), when \( r = 15 \) cm.
Differentiate the volume formula with respect to time \( t \):
\( \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \)
\( \implies \frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \)
\( \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)
Substitute the given value \( \frac{dV}{dt} = 900 \text{ cm}^3/\text{sec} \):
\( 900 = 4\pi r^2 \frac{dr}{dt} \)
We want to find \( \frac{dr}{dt} \) when \( r = 15 \) cm:
\( 900 = 4\pi (15)^2 \frac{dr}{dt} \)
\( 900 = 4\pi (225) \frac{dr}{dt} \)
\( 900 = 900\pi \frac{dr}{dt} \)
Divide both sides by \( 900\pi \):
\( \frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \text{ cm/sec} \)
In simple words: A balloon is getting air pumped into it, making its volume grow. We know how fast the air goes in. We need to figure out how fast the balloon's size (radius) is expanding when its radius is 15 cm. We use the volume formula for a sphere, differentiate it over time, and then solve for the rate of radius change.

Exam Tip: Clearly identify which quantity is changing and at what rate. For spherical objects, remember the volume formula \( V = \frac{4}{3}\pi r^3 \) and its derivative with respect to time for related rates problems. Always isolate the unknown rate after differentiation.

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.
Answer: Let \( r \) be the variable radius of the spherical balloon and \( V \) be its volume.
The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
We need to find the rate at which its volume is increasing *with respect to the radius*, which is \( \frac{dV}{dr} \), when \( r = 10 \) cm.
Differentiate the volume formula with respect to \( r \):
\( \frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \)
\( \implies \frac{dV}{dr} = \frac{4}{3}\pi (3r^2) \)
\( \implies \frac{dV}{dr} = 4\pi r^2 \)
Now, we evaluate this rate when the radius \( r = 10 \) cm:
\( \frac{dV}{dr} = 4\pi (10)^2 \)
\( \implies \frac{dV}{dr} = 4\pi (100) = 400\pi \text{ cm}^3/\text{cm} \)
In simple words: We have a balloon that's always round, and its size can change. We want to find out how quickly the balloon's total space (volume) increases as its size (radius) gets bigger, specifically when the radius is 10 cm. We use the volume formula for a sphere and find its derivative concerning the radius.

Exam Tip: Read carefully whether the rate of change is with respect to time (\( \frac{dV}{dt} \)) or with respect to a spatial dimension like radius (\( \frac{dV}{dr} \)). The differentiation variable changes accordingly, and the chain rule is only applied if differentiating with respect to time.

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?
Answer: Let \( L \) be the length of the ladder, \( x \) be the distance of the foot of the ladder from the wall, and \( y \) be the height of the top of the ladder on the wall.
Given: Ladder length \( L = 5 \) m.
The bottom is pulled away from the wall at \( \frac{dx}{dt} = 2 \text{ cm/s} = 0.02 \text{ m/s} \).
We need to find \( \frac{dy}{dt} \) when \( x = 4 \) m.
Using the Pythagorean theorem for the right-angled triangle formed by the wall, ground, and ladder:
\( x^2 + y^2 = L^2 \)
\( x^2 + y^2 = 5^2 \)
\( x^2 + y^2 = 25 \)
Differentiate this equation with respect to time \( t \):
\( \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \)
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)
\( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \) ...(1)
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, substitute \( x = 4 \) m, \( y = 3 \) m, and \( \frac{dx}{dt} = 0.02 \text{ m/s} \) into equation (1):
\( (4)(0.02) + (3) \frac{dy}{dt} = 0 \)
\( 0.08 + 3 \frac{dy}{dt} = 0 \)
\( 3 \frac{dy}{dt} = -0.08 \)
\( \frac{dy}{dt} = -\frac{0.08}{3} = -\frac{8}{300} = -\frac{2}{75} \text{ m/s} \)
The negative sign shows that the height is decreasing.
To convert to cm/s:
\( -\frac{2}{75} \text{ m/s} = -\frac{2}{75} \times 100 \text{ cm/s} = -\frac{200}{75} \text{ cm/s} = -\frac{8}{3} \text{ cm/s} \)
Hence, the height of the ladder on the wall is decreasing at the rate of \( \frac{2}{75} \) m/s or \( \frac{8}{3} \) cm/s.
In simple words: A 5-meter ladder is leaning on a wall. Its bottom is being pulled away from the wall at 2 cm/s. We want to know how quickly the top of the ladder is sliding down the wall when the bottom is 4 meters away. We use the Pythagorean theorem to link the ladder's position, differentiate it with respect to time, and then put in the given values.

Exam Tip: In ladder problems, always draw a diagram to visualize the setup. Use the Pythagorean theorem to relate the sides of the right triangle formed. Remember to convert all units to be consistent (e.g., cm/s to m/s or vice versa) before solving. A negative rate indicates a decreasing quantity.

Question 11. A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Answer: The equation of the curve is \( 6y = x^3 + 2 \) ...(1)
We are given that the y-coordinate is changing 8 times as fast as the x-coordinate, which can be written as \( \frac{dy}{dt} = 8 \frac{dx}{dt} \) ...(2)
Differentiate equation (1) with respect to time \( t \):
\( \frac{d}{dt}(6y) = \frac{d}{dt}(x^3 + 2) \)
\( 6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt} \)
Substitute \( \frac{dy}{dt} = 8 \frac{dx}{dt} \) from equation (2) into this differentiated equation:
\( 6 \left( 8 \frac{dx}{dt} \right) = 3x^2 \frac{dx}{dt} \)
\( 48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt} \)
Since \( \frac{dx}{dt} \) cannot be zero (otherwise there would be no movement), we can divide both sides by \( \frac{dx}{dt} \):
\( 48 = 3x^2 \)
\( x^2 = \frac{48}{3} \)
\( x^2 = 16 \)
\( \implies x = \pm 4 \)
Now, find the corresponding y-coordinates using equation (1):
**Case 1: When \( x = 4 \)**
\( 6y = (4)^3 + 2 \)
\( 6y = 64 + 2 \)
\( 6y = 66 \)
\( y = \frac{66}{6} = 11 \)
So, one point is \( (4, 11) \).
**Case 2: When \( x = -4 \)**
\( 6y = (-4)^3 + 2 \)
\( 6y = -64 + 2 \)
\( 6y = -62 \)
\( y = \frac{-62}{6} = -\frac{31}{3} \)
So, the other point is \( \left(-4, -\frac{31}{3}\right) \).
The required points on the curve are \( (4, 11) \) and \( \left(-4, -\frac{31}{3}\right) \).
In simple words: A particle moves along a specific curved path. We need to locate the exact spots on this path where its vertical movement is 8 times faster than its horizontal movement. We will differentiate the curve's equation with respect to time and then use the given relationship between the rates to solve for x and y.

Exam Tip: When given a relationship between rates like \( \frac{dy}{dt} = k \frac{dx}{dt} \), substitute this directly into the differentiated equation of the curve. Be careful with algebra and solving for \( x \) and \( y \), remembering to consider both positive and negative roots for \( x^2 \).

Question 12. The radius of an air bubble is increasing at the rate of \( \frac{1}{2} \) cm per second. At what rate is the volume of the bubble increasing, when the radius is 1 cm?
Answer: Let \( r \) be the radius of the air bubble and \( V \) be its volume. We are given that the radius is increasing at the rate of \( \frac{dr}{dt} = \frac{1}{2} \text{ cm/sec} \).
The volume of a sphere (which an air bubble approximates) is given by \( V = \frac{4}{3}\pi r^3 \).
We need to find the rate at which the volume is increasing, \( \frac{dV}{dt} \), when the radius \( r = 1 \) cm.
Differentiate the volume formula with respect to time \( t \):
\( \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \)
\( \implies \frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \)
\( \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)
Substitute the given value \( \frac{dr}{dt} = \frac{1}{2} \text{ cm/sec} \):
\( \frac{dV}{dt} = 4\pi r^2 \left(\frac{1}{2}\right) \)
\( \implies \frac{dV}{dt} = 2\pi r^2 \)
Now, we evaluate this rate when the radius \( r = 1 \) cm:
\( \frac{dV}{dt} = 2\pi (1)^2 \)
\( \implies \frac{dV}{dt} = 2\pi \text{ cm}^3/\text{sec} \)
In simple words: An air bubble is growing, and we know how fast its radius is expanding. We want to find out how quickly its total space (volume) is increasing when the radius is exactly 1 cm. We will use the volume formula for a sphere, differentiate it with respect to time, and then plug in the known rates.

Exam Tip: Ensure you use the correct geometric formula for the shape involved. Differentiate carefully with respect to time, remembering the chain rule for variables dependent on time. Always include appropriate units for your answer.

Question 13. A balloon which always remains spherical has a variable diameter \( \frac{3}{2}(2x + 1) \). Find the rate of change of its volume with respect to x.
Answer: Let \( D \) be the diameter of the spherical balloon. Given \( D = \frac{3}{2}(2x + 1) \).
The radius \( r \) of the sphere is half of its diameter:
\( r = \frac{D}{2} = \frac{1}{2} \times \frac{3}{2}(2x + 1) = \frac{3}{4}(2x + 1) \)
The volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
Substitute the expression for \( r \) into the volume formula:
\( V = \frac{4}{3}\pi \left(\frac{3}{4}(2x + 1)\right)^3 \)
\( V = \frac{4}{3}\pi \left(\frac{27}{64}(2x + 1)^3\right) \)
\( V = \frac{4 \times 27}{3 \times 64}\pi (2x + 1)^3 \)
\( V = \frac{9}{16}\pi (2x + 1)^3 \)
We need to find the rate of change of its volume with respect to \( x \), which is \( \frac{dV}{dx} \).
Differentiate \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{d}{dx}\left(\frac{9}{16}\pi (2x + 1)^3\right) \)
\( \implies \frac{dV}{dx} = \frac{9}{16}\pi \times 3(2x + 1)^2 \times \frac{d}{dx}(2x + 1) \)
\( \implies \frac{dV}{dx} = \frac{9}{16}\pi \times 3(2x + 1)^2 \times 2 \)
\( \implies \frac{dV}{dx} = \frac{54}{16}\pi (2x + 1)^2 \)
\( \implies \frac{dV}{dx} = \frac{27}{8}\pi (2x + 1)^2 \)
In simple words: A round balloon's size (diameter) is changing based on a variable 'x'. We want to figure out how quickly its total space (volume) changes as 'x' changes. We will first find the volume in terms of 'x' and then differentiate that formula with respect to 'x'.

Exam Tip: When the radius or diameter is given as a function of another variable, substitute that function into the volume (or area) formula *before* differentiating. Use the chain rule carefully when differentiating a composite function like \( (2x+1)^3 \).

Question 14. Sand is pouring from a pipe at the rate of 12 cm³/sec. 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?
Answer: Let \( V \) be the volume of the sand cone, \( r \) be its base radius, and \( h \) be its height.
We are given that sand is pouring from a pipe at the rate of \( \frac{dV}{dt} = 12 \text{ cm}^3/\text{sec} \).
We are also given the relationship between height and radius: \( h = \frac{1}{6}r \), which means \( r = 6h \).
The volume of a cone is given by \( V = \frac{1}{3}\pi r^2 h \).
To express \( V \) solely in terms of \( h \), substitute \( r = 6h \) into the volume formula:
\( V = \frac{1}{3}\pi (6h)^2 h \)
\( V = \frac{1}{3}\pi (36h^2) h \)
\( V = 12\pi h^3 \)
We need to find the rate at which the height is increasing, \( \frac{dh}{dt} \), when \( h = 4 \) cm.
Differentiate \( V \) with respect to time \( t \):
\( \frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) \)
\( \implies \frac{dV}{dt} = 12\pi (3h^2) \frac{dh}{dt} \)
\( \implies \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \)
Substitute the given value \( \frac{dV}{dt} = 12 \text{ cm}^3/\text{sec} \):
\( 12 = 36\pi h^2 \frac{dh}{dt} \)
We want to find \( \frac{dh}{dt} \) when \( h = 4 \) cm:
\( 12 = 36\pi (4)^2 \frac{dh}{dt} \)
\( 12 = 36\pi (16) \frac{dh}{dt} \)
\( 12 = 576\pi \frac{dh}{dt} \)
\( \frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \text{ cm/sec} \)
In simple words: Sand is falling from a pipe, creating a cone. We know how fast the sand volume grows, and that the cone's height is always one-sixth of its base radius. We need to find how fast the cone's height is getting taller when it reaches 4 cm. We will write the cone's volume formula using only height, differentiate it over time, and then solve for the rate of height change.

Exam Tip: For problems with related dimensions (like height and radius in a cone), always substitute the given relationship into the volume (or area) formula *before* differentiating. This helps reduce the number of variables and simplifies the differentiation process.

Question 15. The total cost C(x) associated with production of x units of an item is given by C(x) = \( 0.007x^3 - 0.003 x^2 + 15x + 4000 \). Find the marginal cost, when 17 units are produced.
Answer: The total cost function is given by \( C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000 \).
Marginal cost (MC) is the instantaneous rate of change of total cost with respect to the number of units produced. It is found by differentiating the total cost function with respect to \( x \):
\( MC = \frac{dC}{dx} \)
\( \frac{dC}{dx} = \frac{d}{dx}(0.007x^3 - 0.003x^2 + 15x + 4000) \)
\( \implies \frac{dC}{dx} = 0.007(3x^2) - 0.003(2x) + 15(1) + 0 \)
\( \implies \frac{dC}{dx} = 0.021x^2 - 0.006x + 15 \)
Now, we need to find the marginal cost when \( x = 17 \) units are produced. Substitute \( x = 17 \) into the marginal cost function:
\( MC(17) = 0.021(17)^2 - 0.006(17) + 15 \)
\( MC(17) = 0.021(289) - 0.102 + 15 \)
\( MC(17) = 6.069 - 0.102 + 15 \)
\( MC(17) = 5.967 + 15 \)
\( MC(17) = 20.967 \)
Rounding to two decimal places, the marginal cost is Rs. 20.97.
In simple words: We have a formula that tells us the total cost to make 'x' items. We want to find the extra cost to make just one more item (the marginal cost) when we are already making 17 items. We do this by finding the derivative of the cost formula and then putting in 17 for 'x'.

Exam Tip: Remember that marginal cost is the derivative of the total cost function. Carefully differentiate each term and then substitute the given number of units to calculate the specific marginal cost. Pay attention to decimal places in calculations.

Question 16. The total revenue received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue, when x = 7.
Answer: The total revenue function is given by \( R(x) = 13x^2 + 26x + 15 \).
Marginal Revenue (MR) is the rate of change of total revenue with respect to the number of items sold. It is found by differentiating the total revenue function with respect to \( x \):
\( MR = \frac{dR}{dx} \)
\( \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15) \)
\( \implies \frac{dR}{dx} = 13(2x) + 26(1) + 0 \)
\( \implies \frac{dR}{dx} = 26x + 26 \)
\( \implies \frac{dR}{dx} = 26(x + 1) \)
Now, we need to find the marginal revenue when \( x = 7 \) units are sold. Substitute \( x = 7 \) into the marginal revenue function:
\( MR(7) = 26(7 + 1) \)
\( MR(7) = 26(8) \)
\( MR(7) = 208 \)
Hence, the marginal revenue when \( x = 7 \) is Rs. 208.
In simple words: We have a formula showing how much money we make from selling 'x' items. We want to find the extra money we get from selling just one more item (marginal revenue) when we have already sold 7 items. We do this by taking the derivative of the revenue formula and then putting in 7 for 'x'.

Exam Tip: Marginal revenue is calculated as the derivative of the total revenue function. Ensure you differentiate correctly and then substitute the specified number of units to obtain the exact marginal revenue.

Question 17. The rate of change of area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Answer: (B) 12π
Let \( A \) be the area of a circle with radius \( r \).
The area formula is \( A = \pi r^2 \).
The rate of change of area with respect to its radius \( r \) is found by differentiating \( A \) with respect to \( r \):
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) \)
\( \implies \frac{dA}{dr} = 2\pi r \)
Now, we evaluate this rate when \( r = 6 \) cm:
\( \frac{dA}{dr} = 2\pi (6) = 12\pi \text{ cm}^2/\text{cm} \)
Comparing this with the given options, option (B) is the correct answer.
In simple words: To find how fast a circle's area changes as its radius changes, we differentiate the area formula. Then, we plug in the specific radius value to get the exact rate.

Exam Tip: For MCQs, first solve the problem completely to get your answer, then compare it with the given options. Be careful to differentiate with respect to the correct variable (radius, in this case) and not time, as the question specifies.

Question 18. The total revenue (in Rupees) received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue at x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
Answer: (D) 126
The total revenue function is given by \( R(x) = 3x^2 + 36x + 5 \).
Marginal Revenue (MR) is the derivative of the total revenue function with respect to \( x \):
\( MR = \frac{dR}{dx} \)
\( \frac{dR}{dx} = \frac{d}{dx}(3x^2 + 36x + 5) \)
\( \implies \frac{dR}{dx} = 3(2x) + 36(1) + 0 \)
\( \implies \frac{dR}{dx} = 6x + 36 \)
Now, we need to find the marginal revenue when \( x = 15 \) units are sold. Substitute \( x = 15 \) into the marginal revenue function:
\( MR(15) = 6(15) + 36 \)
\( MR(15) = 90 + 36 \)
\( MR(15) = 126 \)
Comparing this with the given options, option (D) is the correct answer.
In simple words: We have a formula for total money earned from selling 'x' items. We want to find the extra money gained from selling one more item (marginal revenue) when 15 items have already been sold. We achieve this by differentiating the revenue formula and then substituting the value for 'x'.

Exam Tip: For MCQs on marginal revenue, compute the derivative of the revenue function and then substitute the specific value of x. Avoid calculation errors, especially when dealing with polynomials.