RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives More Questions

Question 1. If radius and height of a cylinder is r and A, then find the rate of change of surface area with respect to its radius.
Answer: Let the radius of the cylinder be \( r \) and the height be \( h \).
The surface area of a cylinder is given by \( S = 2\pi r^2 + 2\pi rh \).
To find the rate of change of surface area with respect to its radius, we need to differentiate \( S \) with respect to \( r \):
\( \frac{dS}{dr} = \frac{d}{dr}(2\pi r^2 + 2\pi rh) \)
\( \implies \frac{dS}{dr} = 2\pi \frac{d}{dr}(r^2) + 2\pi h \frac{d}{dr}(r) \)
\( \implies \frac{dS}{dr} = 2\pi (2r) + 2\pi h (1) \)
\( \implies \frac{dS}{dr} = 4\pi r + 2\pi h \)
This means that as the radius changes, the surface area changes at a rate determined by both the radius and the height. The surface area changes faster for larger radii and heights.
In simple words: We find how quickly the cylinder's outer skin changes when we change its radius. This is done by a math process called differentiation. The answer shows that the change depends on the radius and the height of the cylinder.

🎯 Exam Tip: Remember the formula for the surface area of a cylinder and know how to differentiate with respect to a specific variable while treating others as constants.

Question 2. Diff w.r.t. t,
\( \frac { dy }{ dt } = 3x^2 \frac {dx}{ dt } .....(i) \)
According to question
\( \frac {dy }{ dt } = 3 \frac { dx }{ dt } .....(ii) \)
From equation (i) and (ii)
\( 3x^2 = 3 \)
\( \implies x = \pm 1 \)
In equation \( y = x^3 + 21 \)
Putting \( x = 1 \)
\( y = 1^3 + 21 \)
\( = 1 + 21 =22 \)
Putting \( x = -1 \)
\( y = (-1)^3 + 21 = -1 + 21 = 20 \)
So, \( x = \pm 1 \) and \( y = 22, 20 \)
Answer: The problem asks us to find the values of \( x \) and \( y \) given two differential equations and a relationship between \( y \) and \( x \).
We have:
1. \( \frac { dy }{ dt } = 3x^2 \frac {dx}{ dt } \) (Given as equation (i))
2. \( \frac {dy }{ dt } = 3 \frac { dx }{ dt } \) (Given as equation (ii))
We also know:
3. \( y = x^3 + 21 \)

First, we compare equations (i) and (ii):
\( 3x^2 \frac {dx}{ dt } = 3 \frac { dx }{ dt } \)
Since \( \frac {dx}{ dt } \) is likely not zero (otherwise \( y \) would not change), we can divide both sides by \( 3 \frac {dx}{ dt } \):
\( x^2 = 1 \)
\( \implies x = \pm 1 \)

Next, we use the equation \( y = x^3 + 21 \) to find the corresponding \( y \) values for each \( x \).
Case 1: When \( x = 1 \)
\( y = (1)^3 + 21 \)
\( y = 1 + 21 \)
\( y = 22 \)

Case 2: When \( x = -1 \)
\( y = (-1)^3 + 21 \)
\( y = -1 + 21 \)
\( y = 20 \)

So, the possible pairs of \( (x, y) \) are \( (1, 22) \) and \( (-1, 20) \). This method helps us find unknown values by using different rates of change.
In simple words: We are given how \( y \) changes with time in two ways, and how \( y \) is related to \( x \). By comparing the ways \( y \) changes, we find possible values for \( x \). Then, we use these \( x \) values to find the matching \( y \) values.

🎯 Exam Tip: When solving equations involving derivatives, always ensure you consider all possible solutions for the variables, including both positive and negative roots.

Question 3. Prove that exponential function (\( e^x \)) is an increasing function.
Answer: To prove that an exponential function \( y = e^x \) is an increasing function, we need to show that its derivative with respect to \( x \) is always positive.
Let the function be \( y = e^x \).
We differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(e^x) \)
\( \implies \frac{dy}{dx} = e^x \)
For any real number \( x \), the value of \( e^x \) is always positive. For example, \( e^0=1 \), \( e^1 \approx 2.718 \), \( e^{-1} \approx 0.368 \).
Since \( e^x > 0 \) for all \( x \in R \), the derivative \( \frac{dy}{dx} \) is always positive.
When the derivative of a function is always positive, it means the function is always increasing. This shows that the graph of \( y = e^x \) always goes upwards as \( x \) increases.
In simple words: An increasing function is one that always goes up as you move from left to right on a graph. For \( e^x \), its "rate of change" (derivative) is always positive, which means it is always getting bigger, so it is an increasing function.

🎯 Exam Tip: To prove a function is increasing, always show that its first derivative is greater than zero for all values in its domain. For \( e^x \), remember that \( e^x \) is always positive.

Question 4. Prove that \( f(x) = \log (\sin x) \) is increasing in \( (0,\frac {\pi }{2}) \) and decreasing in \( (\frac {\pi }{2},\pi) \)
Answer: To prove whether a function is increasing or decreasing, we examine the sign of its first derivative.
Given the function \( f(x) = \log (\sin x) \).
First, we find the derivative of \( f(x) \) with respect to \( x \):
\( f'(x) = \frac{d}{dx}(\log (\sin x)) \)
Using the chain rule, \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \), where \( u = \sin x \).
\( \implies f'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) \)
\( \implies f'(x) = \frac{1}{\sin x} \cdot \cos x \)
\( \implies f'(x) = \cot x \)

Now we need to check the sign of \( f'(x) = \cot x \) in the given intervals.
In the interval \( (0, \frac{\pi}{2}) \):
This is the first quadrant. In the first quadrant, all trigonometric ratios (including \( \cot x \)) are positive.
So, \( \cot x > 0 \) for \( x \in (0, \frac{\pi}{2}) \).
Since \( f'(x) > 0 \), the function \( f(x) = \log (\sin x) \) is increasing in \( (0, \frac{\pi}{2}) \).

In the interval \( (\frac{\pi}{2}, \pi) \):
This is the second quadrant. In the second quadrant, \( \sin x \) is positive, but \( \cos x \) is negative. Therefore, \( \cot x = \frac{\cos x}{\sin x} \) is negative.
So, \( \cot x < 0 \) for \( x \in (\frac{\pi}{2}, \pi) \).
Since \( f'(x) < 0 \), the function \( f(x) = \log (\sin x) \) is decreasing in \( (\frac{\pi}{2}, \pi) \).
This shows how the slope of the function changes across different parts of its domain.
In simple words: We found the "slope" of the function \( f(x) = \log(\sin x) \), which is \( \cot x \). In the first quarter of the circle (0 to 90 degrees), \( \cot x \) is positive, so the function is going up. In the second quarter (90 to 180 degrees), \( \cot x \) is negative, so the function is going down.

🎯 Exam Tip: To determine if a function is increasing or decreasing, always calculate its first derivative. Then, analyze the sign of the derivative in the specified intervals, recalling the signs of trigonometric functions in different quadrants.

Question 5. Equation \( \sqrt { x } - \sqrt { y } = \sqrt {a} .....(i) \)
Let, tangent cut the axis OX and OY at point P and Q, point on curve contact with tangent is (h, k).
Since, (h, k) situated on curve.
So, \( \sqrt {h} - \sqrt {k} = \sqrt {a} .....(ii) \)
Diff. (i) w.r.t. x,
\( \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
\( \implies \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} \)
\( \therefore \) Slope of tangent at point (h, k) of curve.
\( m = [\frac{dy}{dx}]_{(h, k)} = \frac{\sqrt{k}}{\sqrt{h}} \)
So, from \( y – y_1 = m(x – x_1) \) equation of tangent at the point (h, k).
Answer: The problem asks to find the equation of the tangent to the curve \( \sqrt{x} - \sqrt{y} = \sqrt{a} \) at a point \( (h, k) \) and then to find the intercepts.
Given curve: \( \sqrt{x} - \sqrt{y} = \sqrt{a} \) (Equation i)
Since the point \( (h, k) \) lies on the curve, it satisfies the equation:
\( \sqrt{h} - \sqrt{k} = \sqrt{a} \) (Equation ii)

First, we find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent. Differentiating equation (i) with respect to \( x \):
\( \frac{d}{dx}(\sqrt{x}) - \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(\sqrt{a}) \)
\( \implies \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \)
Now, we solve for \( \frac{dy}{dx} \):
\( \frac{1}{2\sqrt{y}} \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{2\sqrt{y}}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} \)

The slope of the tangent \( m \) at the point \( (h, k) \) is found by substituting \( x=h \) and \( y=k \) into the derivative:
\( m = [\frac{dy}{dx}]_{(h, k)} = \frac{\sqrt{k}}{\sqrt{h}} \)

Now, we write the equation of the tangent line using the point-slope form \( y - y_1 = m(x - x_1) \), with \( (x_1, y_1) = (h, k) \) and \( m = \frac{\sqrt{k}}{\sqrt{h}} \):
\( y - k = \frac{\sqrt{k}}{\sqrt{h}}(x - h) \)
To make it simpler, we multiply by \( \sqrt{h} \):
\( \sqrt{h}(y - k) = \sqrt{k}(x - h) \)
\( \implies y\sqrt{h} - k\sqrt{h} = x\sqrt{k} - h\sqrt{k} \)
Rearrange the terms to get the standard form of a line:
\( x\sqrt{k} - y\sqrt{h} = h\sqrt{k} - k\sqrt{h} \)
Now, to find the x-intercept (Point P, where \( y=0 \)) and y-intercept (Point Q, where \( x=0 \)):
For x-intercept (P): Set \( y=0 \)
\( x\sqrt{k} = h\sqrt{k} - k\sqrt{h} \)
\( x = \frac{h\sqrt{k} - k\sqrt{h}}{\sqrt{k}} \)
\( x = h - \frac{k\sqrt{h}}{\sqrt{k}} = h - \sqrt{k}\sqrt{h} = h - \sqrt{hk} \)
So, \( P = (h - \sqrt{hk}, 0) \).
For y-intercept (Q): Set \( x=0 \)
\( -y\sqrt{h} = h\sqrt{k} - k\sqrt{h} \)
\( y = -\frac{h\sqrt{k} - k\sqrt{h}}{\sqrt{h}} \)
\( y = -( \frac{h\sqrt{k}}{\sqrt{h}} - k ) = -( \sqrt{h}\sqrt{k} - k ) = k - \sqrt{hk} \)
So, \( Q = (0, k - \sqrt{hk}) \).
The intercepts are determined by where the tangent line crosses the x-axis and y-axis. The derivation shows how the slope helps define the tangent line.
In simple words: We first find the slope of the curve at a specific point \( (h,k) \). Then, we use this slope and the point to write the equation of the straight line (tangent) that just touches the curve there. Finally, we find where this tangent line cuts the x-axis and y-axis.

🎯 Exam Tip: Remember to differentiate correctly, especially with square root functions. The point-slope form of a line is crucial for finding the tangent equation, and setting \( x=0 \) or \( y=0 \) will give the intercepts.

Question 6. Find equation of tangent of curve \( y = \cos (x + y) \), \( x \in [- 2\pi, 2\pi] \) which is parallel to the line \( x + 2y = 0 \)
Answer: We need to find the equation of the tangent to the curve \( y = \cos(x+y) \) that is parallel to the line \( x + 2y = 0 \).
First, find the slope of the given line \( x + 2y = 0 \).
\( 2y = -x \)
\( y = -\frac{1}{2}x \)
The slope of this line is \( -\frac{1}{2} \). Since the tangent line is parallel to this line, its slope must also be \( -\frac{1}{2} \).

Next, we find the derivative \( \frac{dy}{dx} \) of the curve \( y = \cos(x+y) \). We use implicit differentiation:
\( \frac{d}{dx}(y) = \frac{d}{dx}(\cos(x+y)) \)
\( \implies \frac{dy}{dx} = -\sin(x+y) \cdot (1 + \frac{dy}{dx}) \)
\( \implies \frac{dy}{dx} = -\sin(x+y) - \sin(x+y)\frac{dy}{dx} \)
Move all \( \frac{dy}{dx} \) terms to one side:
\( \frac{dy}{dx} + \sin(x+y)\frac{dy}{dx} = -\sin(x+y) \)
Factor out \( \frac{dy}{dx} \):
\( \frac{dy}{dx}(1 + \sin(x+y)) = -\sin(x+y) \)
\( \implies \frac{dy}{dx} = \frac{-\sin(x+y)}{1 + \sin(x+y)} \)

Now, we set the slope of the tangent equal to \( -\frac{1}{2} \):
\( \frac{-\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2} \)
\( \implies 2\sin(x+y) = 1 + \sin(x+y) \)
\( \implies \sin(x+y) = 1 \)

For \( \sin \theta = 1 \), the general solution is \( \theta = n\pi + (-1)^n \frac{\pi}{2} \).
So, \( x + y = n\pi + (-1)^n \frac{\pi}{2} \), where \( n \in Z \).
Since \( x \in [-2\pi, 2\pi] \), we need to find values of \( x+y \) that are within a reasonable range for \( \sin(x+y)=1 \).
Possible values for \( x+y \) are \( \frac{\pi}{2}, -\frac{3\pi}{2} \) etc.

From the original curve equation, \( y = \cos(x+y) \). If \( \sin(x+y) = 1 \), then \( x+y \) must be of the form \( \frac{\pi}{2} + 2k\pi \) for some integer \( k \).
For such values, \( \cos(x+y) = \cos(\frac{\pi}{2} + 2k\pi) = 0 \).
So, from \( y = \cos(x+y) \), we get \( y = 0 \).

If \( y = 0 \), then \( x+y = x \). So we have \( \sin x = 1 \).
For \( x \in [-2\pi, 2\pi] \), the values of \( x \) for which \( \sin x = 1 \) are:
\( x = \frac{\pi}{2} \) (when \( n=0 \) in \( x = n\pi + (-1)^n \frac{\pi}{2} \))
\( x = -\frac{3\pi}{2} \) (when \( n=-2 \) in \( x = n\pi + (-1)^n \frac{\pi}{2} \))
So the tangent points are \( (\frac{\pi}{2}, 0) \) and \( (-\frac{3\pi}{2}, 0) \).

Now we write the equations of the tangent lines using \( y-y_1 = m(x-x_1) \) with \( m = -\frac{1}{2} \):
For point \( (\frac{\pi}{2}, 0) \):
\( y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \)
\( 2y = -(x - \frac{\pi}{2}) \)
\( 2y = -x + \frac{\pi}{2} \)
\( x + 2y - \frac{\pi}{2} = 0 \)
Multiplying by 2 to clear the fraction:
\( 2x + 4y - \pi = 0 \) (Equation ii)

For point \( (-\frac{3\pi}{2}, 0) \):
\( y - 0 = -\frac{1}{2}(x - (-\frac{3\pi}{2})) \)
\( y = -\frac{1}{2}(x + \frac{3\pi}{2}) \)
\( 2y = -(x + \frac{3\pi}{2}) \)
\( 2y = -x - \frac{3\pi}{2} \)
\( x + 2y + \frac{3\pi}{2} = 0 \)
Multiplying by 2:
\( 2x + 4y + 3\pi = 0 \) (Equation i)
These equations represent the straight lines that touch the curve and are parallel to the given line.
In simple words: We want to find a line that touches the curve \( y = \cos(x+y) \) and runs side-by-side (parallel) to another line \( x+2y=0 \). First, we find the "steepness" (slope) of the given line. Then, we find the "steepness" of our curve by taking its derivative. We make these slopes equal to find the points where the tangent lines touch the curve. Finally, we write the equations for these tangent lines.

🎯 Exam Tip: When dealing with parallel lines, remember their slopes are equal. For implicit differentiation, differentiate both sides with respect to \( x \) and remember the chain rule for terms involving \( y \).

Question 7. Find the percentage error in calculating the volume of a cubical box if an error of 5% is made in measuring its edge.
Answer: Let \( V \) be the volume of the cubical box and \( x \) be the length of its edge.
The formula for the volume of a cube is \( V = x^3 \).

We need to find the percentage error in volume, which is \( \frac{\Delta V}{V} \times 100 \).
First, find the differential of \( V \) with respect to \( x \):
\( \frac{dV}{dx} = 3x^2 \)
The approximate change in volume, \( \Delta V \), can be related to the change in edge length, \( \Delta x \), by:
\( \Delta V \approx \frac{dV}{dx} \cdot \Delta x \)
\( \implies \Delta V = 3x^2 \cdot \Delta x \)

Now, we find the relative error in volume, \( \frac{\Delta V}{V} \):
\( \frac{\Delta V}{V} = \frac{3x^2 \Delta x}{x^3} \)
\( \implies \frac{\Delta V}{V} = 3 \frac{\Delta x}{x} \)

The problem states that there is an error of 5% in measuring the edge. This means the percentage error in \( x \) is 5%.
\( \frac{\Delta x}{x} \times 100 = 5\% \)

To find the percentage error in volume, we multiply the relative error by 100:
\( \frac{\Delta V}{V} \times 100 = 3 \left( \frac{\Delta x}{x} \times 100 \right) \)
Substitute the given percentage error in \( x \):
\( \frac{\Delta V}{V} \times 100 = 3 \times 5\% \)
\( \implies \frac{\Delta V}{V} \times 100 = 15\% \)
So, if there is a 5% error in measuring the side of a cube, the volume calculation will have a 15% error. This illustrates how errors in measurement can affect calculated quantities.
In simple words: If you make a small mistake measuring the side of a cube (like 5% off), that mistake gets bigger when you calculate its volume. For a cube, the volume error will be three times the side measurement error. So, a 5% error in the side leads to a 15% error in the volume.

🎯 Exam Tip: Remember the formula for relative error \( \frac{\Delta y}{y} = \frac{dy}{dx} \frac{\Delta x}{y} \) and percentage error \( \frac{\Delta y}{y} \times 100 \). For power functions like \( V=x^n \), the percentage error in \( V \) is \( n \) times the percentage error in \( x \).

Question 8. A circular metal plate expands under heating so that its radius increased by 2%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm.
Answer: Let \( r \) be the radius of the circular metal plate and \( S \) be its area.
The formula for the area of a circular plate is \( S = \pi r^2 \).

We are given that the radius increases by 2%. This means the percentage change in radius is 2%.
\( \frac{\Delta r}{r} \times 100 = 2\% \)
We can write this as \( \frac{\Delta r}{r} = 0.02 \). So, \( \Delta r = 0.02r \).

The initial radius is given as \( r = 10 \) cm.

To find the approximate increase in area, \( \Delta S \), we first find the derivative of \( S \) with respect to \( r \):
\( \frac{dS}{dr} = \frac{d}{dr}(\pi r^2) \)
\( \implies \frac{dS}{dr} = 2\pi r \)

The approximate change in area, \( \Delta S \), is given by:
\( \Delta S \approx \frac{dS}{dr} \cdot \Delta r \)
Substitute \( \frac{dS}{dr} = 2\pi r \) and \( \Delta r = 0.02r \):
\( \Delta S = (2\pi r) \cdot (0.02r) \)
\( \implies \Delta S = 0.04\pi r^2 \)

Now, substitute the initial radius \( r = 10 \) cm into the equation for \( \Delta S \):
\( \Delta S = 0.04\pi (10)^2 \)
\( \implies \Delta S = 0.04\pi (100) \)
\( \implies \Delta S = 4\pi \)

So, the approximate increase in the area of the plate is \( 4\pi \) cm\(^2\). This calculation helps in understanding how small changes in dimensions affect the overall size.
In simple words: A round metal plate gets bigger when heated, its radius grows by 2%. We want to know how much its total flat surface area increases. We use a math tool to find the change in area based on the change in radius. If the radius was 10 cm, the area increases by about \( 4\pi \) square centimeters.

🎯 Exam Tip: For approximate changes, remember to use differentials: \( \Delta y \approx \frac{dy}{dx} \Delta x \). Pay attention to whether the given percentage is for the actual value or for the change, and convert it to a decimal before calculating.

Question 9. O is the centre of circle. So, \( OA = OB = OQ = R \) (radius)
Again \( PQ=x \implies OP = (R - x) \)
then \( AP = AO + OP \)
\( \implies AP = R + R - x = 2R-x \)
Height of cone, \( (h) = 2R - x \)
Radius of cone \( (r). PB = \sqrt{OB^2-OP^2} \)
\( = \sqrt{R^2-(R-x)^2} \)
\( = \sqrt{R^2-(R^2-2Rx+x^2)} \)
\( = \sqrt{R^2-R^2+2Rx-x^2} \)
\( r = \sqrt{2Rx-x^2} \)
Let, value of cone is V, then
\( V = \frac{1}{3}\pi r^2 h \)
\( V = \frac{1}{3}\pi (\sqrt{2Rx-x^2})^2 (2R - x) \)
\( V = \frac{1}{3}\pi (2Rx-x^2) (2R - x) \)
Diff. w.r.t. x,
\( \implies \frac{1}{3}\pi [(2R-x)(2R-3x)] = 0 \)
\( \implies (2R-x) [(2R-2x)-x] = 0 \)
\( \implies (2R-x) (2R-3x) = 0 \)
\( x \neq 2R \), then \( x = \frac{2R}{3} \)
Diff. again to eq. (i) w.r.t. x
\( \frac{d^2V}{dx^2} = 2\pi (x-R) \)
For \( x = \frac{2R}{3} \)
\( \frac{d^2V}{dx^2} = 2\pi (\frac{2R}{3} - R) \)
\( \frac{d^2V}{dx^2} = 2\pi (\frac{2R-3R}{3}) = 2\pi (-\frac{R}{3}) \)
\( \frac{d^2V}{dx^2} = -\frac{2\pi R}{3} \)
Since \( R \) is a positive radius, \( -\frac{2\pi R}{3} \) is negative. This means at \( x=\frac{2R}{3} \), the volume \( V \) is maximum.

Again, height of cone
\( (h) = 2R-x = 2R - \frac{2R}{3} = \frac{6R-2R}{3} = \frac{4R}{3} \)
Radius of cone
\( (r) = \sqrt{2Rx-x^2} = \sqrt{2R(\frac{2R}{3}) - (\frac{2R}{3})^2} \)
\( = \sqrt{\frac{4R^2}{3} - \frac{4R^2}{9}} = \sqrt{\frac{12R^2-4R^2}{9}} = \sqrt{\frac{8R^2}{9}} = \frac{2\sqrt{2}R}{3} \)
Volume of cone
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\frac{2\sqrt{2}R}{3})^2 (\frac{4R}{3}) \)
\( V = \frac{1}{3}\pi (\frac{8R^2}{9}) (\frac{4R}{3}) = \frac{32\pi R^3}{81} \)
This maximum volume is also \( \frac{8}{27} \) of the volume of a sphere with radius R.
\( \frac{8}{27} \times (\frac{4}{3}\pi R^3) = \frac{32\pi R^3}{81} \)
In simple words: This problem is about finding the largest possible volume of a cone that can fit inside a sphere. We use calculus by taking derivatives to find the specific dimensions (height and radius) of the cone that give the maximum volume. The steps involve setting the first derivative to zero and checking the second derivative to confirm it's a maximum.

🎯 Exam Tip: For optimization problems, always define the quantity to be maximized or minimized (here, volume V) in terms of one variable. Then, differentiate, set the derivative to zero, and use the second derivative test to confirm maximum or minimum.

Question 10. Prove that semivertical angle of a cylindrical cone of given surface area and maximum volume is \( \sin^{-1}(\frac{1}{3}) \).
Answer: Let \( r \) be the radius and \( l \) be the slant height of the cone. Let \( h \) be the height and \( \alpha \) be the semivertical angle.
The given surface area \( S \) of the cone is:
\( S = \pi r^2 + \pi r l \) (where \( \pi r^2 \) is the area of the base and \( \pi r l \) is the curved surface area)
From this, we can express \( l \) in terms of \( S \) and \( r \):
\( \pi r l = S - \pi r^2 \)
\( \implies l = \frac{S - \pi r^2}{\pi r} \) (Equation i)

The volume \( V \) of the cone is \( V = \frac{1}{3}\pi r^2 h \).
We know that \( l^2 = r^2 + h^2 \) (from the Pythagorean theorem for the cone).
So, \( h^2 = l^2 - r^2 \), which means \( h = \sqrt{l^2 - r^2} \).
Substitute \( h \) into the volume formula:
\( V = \frac{1}{3}\pi r^2 \sqrt{l^2 - r^2} \)

To maximize \( V \), it is easier to maximize \( V^2 \) (since \( V > 0 \)). Let \( Z = V^2 \).
\( Z = (\frac{1}{3}\pi r^2 \sqrt{l^2 - r^2})^2 \)
\( \implies Z = \frac{1}{9}\pi^2 r^4 (l^2 - r^2) \)
Now, substitute the expression for \( l \) from Equation (i) into \( Z \):
\( Z = \frac{1}{9}\pi^2 r^4 \left( \left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2 \right) \)
\( \implies Z = \frac{1}{9}\pi^2 r^4 \left( \frac{(S - \pi r^2)^2}{\pi^2 r^2} - r^2 \right) \)
\( \implies Z = \frac{1}{9}\pi^2 r^4 \left( \frac{S^2 - 2S\pi r^2 + \pi^2 r^4 - \pi^2 r^4}{\pi^2 r^2} \right) \)
\( \implies Z = \frac{1}{9}\pi^2 r^4 \left( \frac{S^2 - 2S\pi r^2}{\pi^2 r^2} \right) \)
\( \implies Z = \frac{1}{9} r^2 (S^2 - 2S\pi r^2) \)
\( \implies Z = \frac{1}{9} (S^2 r^2 - 2S\pi r^4) \)

To find the maximum volume, we differentiate \( Z \) with respect to \( r \) and set it to zero:
\( \frac{dZ}{dr} = \frac{1}{9} \frac{d}{dr}(S^2 r^2 - 2S\pi r^4) \)
\( \implies \frac{dZ}{dr} = \frac{1}{9} (S^2 (2r) - 2S\pi (4r^3)) \)
\( \implies \frac{dZ}{dr} = \frac{1}{9} (2S^2 r - 8S\pi r^3) \)
Set \( \frac{dZ}{dr} = 0 \):
\( \frac{1}{9} (2S^2 r - 8S\pi r^3) = 0 \)
\( \implies 2S^2 r - 8S\pi r^3 = 0 \)
Factor out \( 2Sr \):
\( 2Sr (S - 4\pi r^2) = 0 \)
Since \( S \neq 0 \) and \( r \neq 0 \) (for a cone to exist), we must have:
\( S - 4\pi r^2 = 0 \)
\( \implies S = 4\pi r^2 \)

Now, we check the second derivative to confirm this gives a maximum:
\( \frac{d^2Z}{dr^2} = \frac{1}{9} \frac{d}{dr}(2S^2 r - 8S\pi r^3) \)
\( \implies \frac{d^2Z}{dr^2} = \frac{1}{9} (2S^2 - 8S\pi (3r^2)) \)
\( \implies \frac{d^2Z}{dr^2} = \frac{1}{9} (2S^2 - 24S\pi r^2) \)
Substitute \( S = 4\pi r^2 \) into the second derivative:
\( \frac{d^2Z}{dr^2} = \frac{1}{9} (2(4\pi r^2)^2 - 24(4\pi r^2)\pi r^2) \)
\( \implies \frac{d^2Z}{dr^2} = \frac{1}{9} (2(16\pi^2 r^4) - 96\pi^2 r^4) \)
\( \implies \frac{d^2Z}{dr^2} = \frac{1}{9} (32\pi^2 r^4 - 96\pi^2 r^4) \)
\( \implies \frac{d^2Z}{dr^2} = \frac{1}{9} (-64\pi^2 r^4) \)
Since \( \pi, r \) are positive, \( \frac{d^2Z}{dr^2} \) is negative, confirming that \( S = 4\pi r^2 \) corresponds to a maximum volume.

Now we need to find the semivertical angle \( \alpha \). From the geometry of the cone, we know that \( \sin \alpha = \frac{r}{l} \).
We have \( S = 4\pi r^2 \). We also know \( S = \pi r^2 + \pi r l \).
Equating the two expressions for \( S \):
\( 4\pi r^2 = \pi r^2 + \pi r l \)
Subtract \( \pi r^2 \) from both sides:
\( 3\pi r^2 = \pi r l \)
Divide both sides by \( \pi r \) (since \( r \neq 0 \)):
\( 3r = l \)

Now substitute this relationship into the formula for \( \sin \alpha \):
\( \sin \alpha = \frac{r}{l} = \frac{r}{3r} \)
\( \implies \sin \alpha = \frac{1}{3} \)
Therefore, \( \alpha = \sin^{-1}(\frac{1}{3}) \).
This demonstrates how a specific geometric relationship emerges when optimizing a cone's volume under a fixed surface area.
In simple words: We want to find the angle at the tip of a cone that gives the biggest possible volume when its total outside area is fixed. We write equations for the volume and surface area, then use calculus (differentiation) to find the conditions for maximum volume. This leads us to a relationship between the cone's radius and slant height, which then tells us the sine of the semivertical angle is \( \frac{1}{3} \).

🎯 Exam Tip: When maximizing or minimizing a volume or area, often it's easier to work with the square of the quantity (like \( V^2 \) or \( A^2 \)) to avoid square roots in differentiation. Always relate geometric variables like \( r, h, l \) to the angle using trigonometric ratios.