RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives Exercise 8.1

Question 1. Find the rate of change of the area of a circle with respect to its radius r, when r = 3 cm and r = 4 cm.
Answer: Let \(A\) be the area of the circle and \(r\) be its radius. The formula for the area of a circle is \( A = \pi r^2 \).
To find the rate of change of the area with respect to the radius, we need to calculate the derivative of \(A\) with respect to \(r\).
\[ \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) \]
\[ \frac{dA}{dr} = 2\pi r \]
Now, we will find the rate of change at the given radius values:
When \( r = 3 \) cm:
\[ \frac{dA}{dr} = 2\pi (3) \]
\[ \frac{dA}{dr} = 6\pi \text{ cm}^2/\text{sec} \]
When \( r = 4 \) cm:
\[ \frac{dA}{dr} = 2\pi (4) \]
\[ \frac{dA}{dr} = 8\pi \text{ cm}^2/\text{sec} \]
In simple words: The area of a circle changes faster when its radius is bigger. We found out how quickly the area grows for every tiny increase in the radius, at two different sizes of the circle.

🎯 Exam Tip: Remember to include the correct units in your final answer for rates of change, which are typically a unit of area divided by a unit of time or length, depending on the context.

Question 2. A particle moves along \( y = \frac {2}{3} x^3 + 1 \). Find the points on the curve at which y – coordinate is changing twice as fast as the x – coordinate.
Answer: The equation of the curve is given as \( y = \frac {2}{3} x^3 + 1 \). Let's call this Equation (i).
The position of the particle at time 't' is denoted by \( (x, y) \), which means \( (x, y) \) is always on the curve (i).
We are told that the rate of change of the y-coordinate is twice as fast as the rate of change of the x-coordinate. In mathematical terms, this means:
\[ \frac{dy}{dt} = 2 \frac{dx}{dt} \]
Now, we need to differentiate Equation (i) with respect to time 't':
\[ \frac{dy}{dt} = \frac{d}{dt} \left( \frac {2}{3} x^3 + 1 \right) \]
\[ \frac{dy}{dt} = \frac {2}{3} \cdot 3x^2 \frac{dx}{dt} + 0 \]
\[ \frac{dy}{dt} = 2x^2 \frac{dx}{dt} \]
Now we can substitute the condition \( \frac{dy}{dt} = 2 \frac{dx}{dt} \) into this differentiated equation:
\[ 2 \frac{dx}{dt} = 2x^2 \frac{dx}{dt} \]
Assuming \( \frac{dx}{dt} \neq 0 \) (since the x-coordinate is changing), we can divide both sides by \( 2 \frac{dx}{dt} \):
\[ 1 = x^2 \]
So, \( x = \pm 1 \).
Now, we find the corresponding y-values using the original curve equation \( y = \frac {2}{3} x^3 + 1 \):
When \( x = 1 \):
\[ y = \frac {2}{3} (1)^3 + 1 \]
\[ y = \frac {2}{3} + 1 \]
\[ y = \frac {2+3}{3} = \frac {5}{3} \]
So, one point is \( \left(1, \frac {5}{3}\right) \).
When \( x = -1 \):
\[ y = \frac {2}{3} (-1)^3 + 1 \]
\[ y = \frac {2}{3} (-1) + 1 \]
\[ y = -\frac {2}{3} + 1 \]
\[ y = \frac {-2+3}{3} = \frac {1}{3} \]
So, the other point is \( \left(-1, \frac {1}{3}\right) \).
Therefore, the required points on the curve are \( \left(1, \frac {5}{3}\right) \) and \( \left(-1, \frac {1}{3}\right) \).
In simple words: We needed to find specific spots on the curve where the 'up-down' change is exactly twice as fast as the 'left-right' change. By using calculus to describe these changes and applying the given condition, we found two exact points where this happens.

🎯 Exam Tip: Always remember to find both the x and y coordinates for "points on the curve" questions. After finding x, substitute it back into the *original* equation of the curve to get the y-value.

Question 3. Let AB be a ladder, its end 'A' is x distance far from wall and upper end is at y from the ground. Let slope between ladder and ground is \( \theta \). Find the rate of change of the angle \( \theta \) when the base of the ladder is moving away from the wall at 1.5 m/s and it is 12m away from the wall.

Answer: Let the ladder AB have a length of 13 m. Let 'x' be the distance of the base of the ladder (end A) from the wall (point O), and 'y' be the height of the top of the ladder (end B) from the ground. The angle the ladder makes with the ground is \( \theta \).
From the right-angled triangle formed by the wall, ground, and ladder, we can write:
1. Using the Pythagorean theorem:
\[ x^2 + y^2 = (13)^2 \]
\[ x^2 + y^2 = 169 \] Now, differentiate this equation with respect to time 't':
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
This simplifies to:
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \quad \text{(Equation 1)} \]
2. Using trigonometry for the angle \( \theta \):
\[ \tan \theta = \frac{y}{x} \]
Now, differentiate this equation with respect to time 't':
\[ \sec^2 \theta \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \quad \text{(Equation 2)} \]
We are given that the base of the ladder is moving away from the wall at 1.5 m/s, so \( \frac{dx}{dt} = 1.5 \) m/s.
We are also given that \( x = 12 \) m.
First, find 'y' when \( x = 12 \) m using \( x^2 + y^2 = 169 \):
\[ (12)^2 + y^2 = 169 \]
\[ 144 + y^2 = 169 \]
\[ y^2 = 169 - 144 \]
\[ y^2 = 25 \]
\[ y = 5 \text{ m} \] Now, use Equation 1 to find \( \frac{dy}{dt} \):
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]
\[ 12 (1.5) + 5 \frac{dy}{dt} = 0 \]
\[ 18 + 5 \frac{dy}{dt} = 0 \]
\[ 5 \frac{dy}{dt} = -18 \]
\[ \frac{dy}{dt} = -\frac{18}{5} = -3.6 \text{ m/s} \]
The negative sign for \( \frac{dy}{dt} \) means the top of the ladder is sliding down the wall.
Next, we need \( \sec^2 \theta \). From the triangle, \( \cos \theta = \frac{x}{13} \), so \( \sec \theta = \frac{13}{x} \).
\[ \sec^2 \theta = \left(\frac{13}{x}\right)^2 = \left(\frac{13}{12}\right)^2 = \frac{169}{144} \]
Now substitute all values into Equation 2:
\[ \sec^2 \theta \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \]
\[ \frac{169}{144} \frac{d\theta}{dt} = \frac{12 (-3.6) - 5 (1.5)}{12^2} \]
\[ \frac{169}{144} \frac{d\theta}{dt} = \frac{-43.2 - 7.5}{144} \]
\[ \frac{169}{144} \frac{d\theta}{dt} = \frac{-50.7}{144} \]
\[ \frac{d\theta}{dt} = \frac{-50.7}{169} \]
\[ \frac{d\theta}{dt} = -\frac{507}{1690} \] Let's re-check the OCR provided solution, it seems to be using an alternative way to calculate \( \frac{d\theta}{dt} \) where it transforms \( \sec^2 \theta \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \) further. Given from OCR: \( \frac{d\theta}{dt} = \frac{-3(x^2 + y^2)}{2x^2y(1+\tan^2 \theta)} = \frac{-3(x^2 + y^2)}{2x^2y \sec^2 \theta} \). This is incorrect simplification from \( \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \) to \( \frac{-3(x^2+y^2)}{...} \). Let's use the source's approach to the extent possible, specifically the final calculation shown.
The source has simplified \( \sec^2\theta \frac{d\theta}{dt} = \frac{x(\frac{-3x}{2y}\frac{dx}{dt}) - y\frac{dx}{dt}}{x^2} \). Wait, the source's derivation from `tanθ = y/x` is confusing and seems to have errors in intermediate steps to reach the formula it uses. Let's follow the numeric values and final answer shown in the source, assuming the final expression `dθ/dt = -3/(2y√(13^2-x^2))` is the one intended. The source simplifies `(dθ/dt) = -3/(2y√(13^2-x^2))` which becomes `(dθ/dt) = -3/(2y * y)` since `y = √(13^2-x^2)`. So, `dθ/dt = -3/(2y^2)`. This looks like a different approach or simplified expression. Let's use the derivative of \( \cos \theta = \frac{x}{13} \). Differentiating with respect to t:
\[ -\sin \theta \frac{d\theta}{dt} = \frac{1}{13} \frac{dx}{dt} \]
We know \( \sin \theta = \frac{y}{13} \).
\[ -\frac{y}{13} \frac{d\theta}{dt} = \frac{1}{13} \frac{dx}{dt} \]
\[ -y \frac{d\theta}{dt} = \frac{dx}{dt} \]
\[ \frac{d\theta}{dt} = -\frac{1}{y} \frac{dx}{dt} \]
Given \( x=12 \), we found \( y=5 \). Also \( \frac{dx}{dt} = 1.5 \).
\[ \frac{d\theta}{dt} = -\frac{1}{5} (1.5) \]
\[ \frac{d\theta}{dt} = -\frac{1.5}{5} = -0.3 \text{ radian/s} \] This result, \( -0.3 \) rad/s, can be written as \( -\frac{3}{10} \) rad/s. The OCR-provided working has steps that lead to `dθ/dt = -3/10 radian/s` at `x=12`. Specifically, it reaches `dθ/dt = -3 / (2 * 5)` which gives `-3/10` when `x=12` and `y=5`. The formula used in the OCR: `dθ/dt = -3 / (2y√(13^2-x^2))` When `x = 12`, then `y = √(13^2 - 12^2) = √(169 - 144) = √25 = 5`. So, `dθ/dt = -3 / (2 * 5 * √(13^2 - 12^2))` `dθ/dt = -3 / (2 * 5 * 5)` `dθ/dt = -3 / (50)`. This does not match `-3/10`. Let's re-evaluate from the source's image, if `dx/dt = 1.5 m/s`. Then `dθ/dt = -1/y dx/dt` `dθ/dt = -1/5 * 1.5 = -0.3` which is `-3/10`. The provided source computation seems to follow `dθ/dt = -3 / (2√(13^2-x^2) * 2y)` for some reason, which simplifies to `dθ/dt = -3 / (2y * y)` (if `y = √(13^2-x^2)`) then `dθ/dt = -3 / (2y^2)`. At `x=12`, `y=5`. So `dθ/dt = -3 / (2 * 5^2) = -3/50`. This is a mismatch. However, the final conclusion of the source *is* `dθ/dt = -3/10 radian/s`. I will use the `dθ/dt = -1/y dx/dt` approach, as it directly yields the correct value shown in the end of the source. Solution reconstruction: Let `x` be the distance of the base of the ladder from the wall, `y` be the height of the ladder on the wall, and `L = 13` m be the length of the ladder. By Pythagorean theorem: \( x^2 + y^2 = L^2 = 13^2 = 169 \). The angle \( \theta \) between the ladder and the ground can be expressed as: \( \cos \theta = \frac{x}{L} = \frac{x}{13} \) Differentiate both sides with respect to time 't': \( -\sin \theta \frac{d\theta}{dt} = \frac{1}{13} \frac{dx}{dt} \) We know that \( \sin \theta = \frac{y}{L} = \frac{y}{13} \). Substitute this into the equation: \( -\frac{y}{13} \frac{d\theta}{dt} = \frac{1}{13} \frac{dx}{dt} \) Multiply by 13 on both sides:
\( -y \frac{d\theta}{dt} = \frac{dx}{dt} \) So, \( \frac{d\theta}{dt} = -\frac{1}{y} \frac{dx}{dt} \) We are given \( \frac{dx}{dt} = 1.5 \) m/s and we need to find \( \frac{d\theta}{dt} \) when \( x = 12 \) cm. First, find 'y' when \( x = 12 \) m:
\( 12^2 + y^2 = 13^2 \)
\( 144 + y^2 = 169 \)
\( y^2 = 169 - 144 = 25 \)
\( y = 5 \) m (since height cannot be negative). Now substitute the values of \( y \) and \( \frac{dx}{dt} \) into the equation for \( \frac{d\theta}{dt} \):
\( \frac{d\theta}{dt} = -\frac{1}{5} (1.5) \)
\( \frac{d\theta}{dt} = -0.3 \)
\( \frac{d\theta}{dt} = -\frac{3}{10} \) radian/s
The negative sign indicates that the angle \( \theta \) is decreasing as the base of the ladder moves away from the wall.
In simple words: When the bottom of the ladder slides away from the wall, the angle the ladder makes with the ground gets smaller. We calculated how fast this angle changes when the ladder is 12 meters away from the wall and moving at a certain speed.

🎯 Exam Tip: For related rates problems involving triangles, always draw a clear diagram, use the Pythagorean theorem or trigonometric relations to establish connections between variables, and then differentiate implicitly with respect to time 't'.

Question 4. An edge of a variable cube is increasing at the rate of 3 cm/sec. How is the volume of the cube increasing when the edge is 10 cm long?
Answer: Let \( x \) be the length of an edge of the cube and \( V \) be its volume.
The formula for the volume of a cube is \( V = x^3 \).
We are given that the edge is increasing at the rate of 3 cm/sec, which means \( \frac{dx}{dt} = 3 \) cm/sec.
We need to find the rate at which the volume of the cube is increasing, i.e., \( \frac{dV}{dt} \), when the edge is \( x = 10 \) cm.
Differentiate the volume formula \( V = x^3 \) with respect to time 't':
\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) \]
\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]
Now, substitute the given values:
\( x = 10 \) cm
\( \frac{dx}{dt} = 3 \) cm/sec
\[ \frac{dV}{dt} = 3(10)^2 (3) \]
\[ \frac{dV}{dt} = 3(100)(3) \]
\[ \frac{dV}{dt} = 900 \text{ cm}^3/\text{sec} \]
Thus, the volume of the cube is increasing at the rate of 900 cm³/sec.
In simple words: As the side of a cube gets longer, its volume grows much faster. When the side length is 10 cm and growing by 3 cm every second, the cube's total space is getting bigger by 900 cubic centimeters each second.

🎯 Exam Tip: Clearly define your variables (e.g., x for edge, V for volume) and state the given rates and the rate you need to find. Always remember to use the chain rule when differentiating with respect to time.

Question 5. A Balloon which is always spherical, is being inflated by pumping in 900 cm³ of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.
Answer: Let \( r \) be the radius of the spherical balloon and \( V \) be its volume at any time 't'.
The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \).
We are given that gas is being pumped in at a rate of 900 cm³/sec. This is the rate of change of volume with respect to time: \( \frac{dV}{dt} = 900 \) cm³/sec.
We need to find the rate at which the radius is increasing, \( \frac{dr}{dt} \), when the radius is \( r = 15 \) cm.
Differentiate the volume formula \( V = \frac{4}{3}\pi r^3 \) with respect to time 't':
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \]
\[ \frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \]
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Now, substitute the known values:
\( \frac{dV}{dt} = 900 \)
\( r = 15 \)
\[ 900 = 4\pi (15)^2 \frac{dr}{dt} \]
\[ 900 = 4\pi (225) \frac{dr}{dt} \]
\[ 900 = 900\pi \frac{dr}{dt} \]
To find \( \frac{dr}{dt} \), divide both sides by \( 900\pi \):
\[ \frac{dr}{dt} = \frac{900}{900\pi} \]
\[ \frac{dr}{dt} = \frac{1}{\pi} \text{ cm/s} \]
The rate of change of the radius of the balloon is \( \frac{1}{\pi} \) cm/s.
In simple words: We know how fast air is going into the balloon, making its volume bigger. We used this to figure out how quickly the balloon's skin stretches outwards, making its radius grow, especially when it's already 15 cm big.

🎯 Exam Tip: For problems involving spheres, ensure you use the correct formulas for volume (\( V = \frac{4}{3}\pi r^3 \)) and surface area (\( S = 4\pi r^2 \)). Be careful with calculations involving \( \pi \), leaving it in symbolic form unless a numerical approximation is requested.

Question 6. A balloon which is always spherical, has diameter \( \frac {3}{2} (2x+1) \). Find rate of change of its volume w.r.t. x.
Answer: Let \( V \) be the volume of the spherical balloon.
The diameter of the balloon is given as \( D = \frac {3}{2} (2x+1) \).
The radius \( r \) is half of the diameter, so:
\[ r = \frac{1}{2} D = \frac{1}{2} \cdot \frac{3}{2} (2x+1) = \frac{3}{4} (2x+1) \]
The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \).
Substitute the expression for \( r \) into the volume formula:
\[ V = \frac{4}{3}\pi \left( \frac{3}{4} (2x+1) \right)^3 \]
\[ V = \frac{4}{3}\pi \left( \frac{3^3}{4^3} (2x+1)^3 \right) \]
\[ V = \frac{4}{3}\pi \left( \frac{27}{64} (2x+1)^3 \right) \]
\[ V = \frac{4 \cdot 27}{3 \cdot 64} \pi (2x+1)^3 \]
\[ V = \frac{108}{192} \pi (2x+1)^3 \]
Simplify the fraction \( \frac{108}{192} \): divide both by 12, gives \( \frac{9}{16} \).
\[ V = \frac{9\pi}{16} (2x+1)^3 \]
Now, we need to find the rate of change of its volume with respect to \( x \), i.e., \( \frac{dV}{dx} \).
Differentiate \( V \) with respect to \( x \):
\[ \frac{dV}{dx} = \frac{d}{dx} \left( \frac{9\pi}{16} (2x+1)^3 \right) \]
Using the chain rule:
\[ \frac{dV}{dx} = \frac{9\pi}{16} \cdot 3 (2x+1)^{3-1} \cdot \frac{d}{dx}(2x+1) \]
\[ \frac{dV}{dx} = \frac{9\pi}{16} \cdot 3 (2x+1)^2 \cdot 2 \]
\[ \frac{dV}{dx} = \frac{9\pi \cdot 3 \cdot 2}{16} (2x+1)^2 \]
\[ \frac{dV}{dx} = \frac{54\pi}{16} (2x+1)^2 \]
Simplify the fraction \( \frac{54}{16} \): divide both by 2, gives \( \frac{27}{8} \).
\[ \frac{dV}{dx} = \frac{27\pi}{8} (2x+1)^2 \]
This is the rate of change of the volume of the balloon with respect to \( x \).
In simple words: The size of the balloon is linked to a value 'x'. We figured out how quickly the balloon's total space (volume) changes every time 'x' increases by a tiny bit.

🎯 Exam Tip: When dealing with composite functions, remember to apply the chain rule correctly. In this case, differentiating \( (2x+1)^3 \) with respect to \( x \) involves multiplying by the derivative of the inner function \( (2x+1) \), which is 2.

Question 7. The total cost C(x) associated with the production of x units of a product is given by \( C(x) = 0.005x^3 – 0.02x² + 30x + 5000 \). Find the marginal cost when 3 units are produced.
Answer: The total cost function for producing \( x \) units is given as \( C(x) = 0.005x^3 – 0.02x² + 30x + 5000 \).
Marginal cost (MC) is the rate of change of total cost with respect to the number of units produced. To find the marginal cost, we need to differentiate \( C(x) \) with respect to \( x \).
\[ MC = \frac{dC}{dx} = \frac{d}{dx}(0.005x^3 – 0.02x² + 30x + 5000) \]
\[ MC = 0.005(3x^2) – 0.02(2x) + 30(1) + 0 \]
\[ MC = 0.015x^2 – 0.04x + 30 \]
Now, we need to find the marginal cost when \( x = 3 \) units are produced. Substitute \( x = 3 \) into the marginal cost function:
\[ MC(3) = 0.015(3)^2 – 0.04(3) + 30 \]
\[ MC(3) = 0.015(9) – 0.12 + 30 \]
\[ MC(3) = 0.135 – 0.12 + 30 \]
\[ MC(3) = 0.015 + 30 \]
\[ MC(3) = 30.015 \]
Rounding to two decimal places, the marginal cost is approximately Rs. 30.02.
In simple words: Marginal cost tells us how much extra it costs to make one more item. We found that when 3 items are already being made, making the 4th item would add about Rs. 30.02 to the total cost.

🎯 Exam Tip: Remember that marginal cost is found by taking the first derivative of the total cost function. Pay close attention to decimal places and calculations when substituting values.

Question 8. The radius of a soap bubble is increasing at the rate of 0.2 cm/sec. At what rate is the surface area of a bubble increasing when the radius is 7 cm and at what rate is the volume of that bubble increasing when the radius is 5 cm.
Answer: Let \( r \) be the radius of the soap bubble.
We are given that the radius is increasing at the rate of 0.2 cm/sec, so \( \frac{dr}{dt} = 0.2 \) cm/sec.
**Part 1: Rate of change of surface area**
Let \( S \) be the surface area of the spherical soap bubble. The formula for the surface area of a sphere is \( S = 4\pi r^2 \).
We need to find \( \frac{dS}{dt} \) when \( r = 7 \) cm.
Differentiate \( S \) with respect to time 't':
\[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) \]
\[ \frac{dS}{dt} = 4\pi (2r) \frac{dr}{dt} \]
\[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \]
Substitute the values \( r = 7 \) cm and \( \frac{dr}{dt} = 0.2 \) cm/sec:
\[ \frac{dS}{dt} = 8\pi (7) (0.2) \]
\[ \frac{dS}{dt} = 56\pi (0.2) \]
\[ \frac{dS}{dt} = 11.2\pi \text{ cm}^2/\text{s} \]
So, the surface area of the bubble is increasing at the rate of \( 11.2\pi \) cm²/s when its radius is 7 cm.
**Part 2: Rate of change of volume**
Let \( V \) be the volume of the spherical soap bubble. The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \).
We need to find \( \frac{dV}{dt} \) when \( r = 5 \) cm.
Differentiate \( V \) with respect to time 't':
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \]
\[ \frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \]
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Substitute the values \( r = 5 \) cm and \( \frac{dr}{dt} = 0.2 \) cm/sec:
\[ \frac{dV}{dt} = 4\pi (5)^2 (0.2) \]
\[ \frac{dV}{dt} = 4\pi (25) (0.2) \]
\[ \frac{dV}{dt} = 100\pi (0.2) \]
\[ \frac{dV}{dt} = 20\pi \text{ cm}^3/\text{s} \]
So, the volume of the bubble is increasing at the rate of \( 20\pi \) cm³/s when its radius is 5 cm.
In simple words: We have a growing soap bubble and know how fast its size (radius) is changing. We used this to calculate two things: first, how quickly its outer skin (surface area) is getting bigger when it's 7 cm wide, and second, how quickly the space inside it (volume) is expanding when it's 5 cm wide.

🎯 Exam Tip: Always tackle multi-part questions by breaking them down into separate calculations. Ensure you use the correct formulas for surface area and volume of a sphere and substitute the given values accurately for each part of the problem.

Question 9. 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 height of the sand cone increasing when the height is 4 cm ?
Answer: Let \( V \) be the volume of the sand cone, \( h \) its height, and \( r \) the radius of its base at any time 't'.
The volume of a cone is given by \( V = \frac{1}{3}\pi r^2 h \).
We are given that sand is pouring from a pipe at the rate of 12 cm³/sec, so \( \frac{dV}{dt} = 12 \) cm³/sec.
We are also given a relationship between the height and radius: the height of the cone is always one-sixth of the radius of the base.
So, \( h = \frac{1}{6}r \). This means \( r = 6h \).
We need to find how fast the height is increasing, \( \frac{dh}{dt} \), when the height is \( h = 4 \) cm.
First, substitute \( r = 6h \) into the volume formula to express \( V \) solely in terms of \( h \):
\[ V = \frac{1}{3}\pi (6h)^2 h \]
\[ V = \frac{1}{3}\pi (36h^2) h \]
\[ V = \frac{36\pi}{3} h^3 \]
\[ V = 12\pi h^3 \]
Now, differentiate \( V \) with respect to time 't':
\[ \frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) \]
\[ \frac{dV}{dt} = 12\pi (3h^2) \frac{dh}{dt} \]
\[ \frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt} \]
Substitute the known values: \( \frac{dV}{dt} = 12 \) and \( h = 4 \):
\[ 12 = 36\pi (4)^2 \frac{dh}{dt} \]
\[ 12 = 36\pi (16) \frac{dh}{dt} \]
\[ 12 = 576\pi \frac{dh}{dt} \]
To find \( \frac{dh}{dt} \), divide both sides by \( 576\pi \):
\[ \frac{dh}{dt} = \frac{12}{576\pi} \]
Simplify the fraction \( \frac{12}{576} \). Both are divisible by 12: \( 576 \div 12 = 48 \).
\[ \frac{dh}{dt} = \frac{1}{48\pi} \text{ cm/s} \]
The height of the sand cone is increasing at the rate of \( \frac{1}{48\pi} \) cm/s when the height is 4 cm.
In simple words: Sand is falling and creating a cone shape. We know how fast the sand is piling up (volume change) and how the cone's height and base are related. We used this to figure out how fast the cone's height is growing taller when it's 4 cm high.

🎯 Exam Tip: In related rates problems, always try to express the primary changing quantity (volume, area) in terms of a single variable (radius, height) using the given relationships. This simplifies the differentiation process significantly.

Question 10. The total revenue received from the sale of x units of a product is given by \( R(x) = 13x² + 26x + 15 \). Find marginal cost when x = 15.
Answer: The total revenue function for the sale of \( x \) units is given as \( R(x) = 13x² + 26x + 15 \).
The question asks for marginal cost, but it provides a revenue function. It's highly probable that it actually means "marginal revenue." We will calculate the marginal revenue, as the subsequent steps in the source align with this calculation.
Marginal Revenue (MR) is the rate of change of total revenue with respect to the number of units sold. To find the marginal revenue, we differentiate \( R(x) \) with respect to \( x \).
\[ MR = \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15) \]
\[ MR = 13(2x) + 26(1) + 0 \]
\[ MR = 26x + 26 \]
Now, we need to find the marginal revenue when \( x = 15 \) units are sold. Substitute \( x = 15 \) into the marginal revenue function:
\[ MR(15) = 26(15) + 26 \]
\[ MR(15) = 390 + 26 \]
\[ MR(15) = 416 \]
Following the given solution's final statement, we state: Hence, Marginal cost = Rs. 416.
In simple words: When a company sells more items, its total money earned changes. We calculated that if 15 items are already sold, selling one more would bring in an extra Rs. 416. The question used the term 'marginal cost' but referred to the revenue function, so we calculated what would typically be called 'marginal revenue' from the provided formula.

🎯 Exam Tip: Understand the difference between cost, revenue, and profit functions. Marginal cost is \( C'(x) \), marginal revenue is \( R'(x) \), and marginal profit is \( P'(x) \). Always pay close attention to the function provided and the quantity asked, and be prepared to derive the relevant marginal function through differentiation.