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

Question 1. Using differentials, find the approximate value of each of the following upto 3 places of decimal.
(i) \( \sqrt{37} \)
(ii) \( (15)^{1/4} \)
(iii) \( (255)^{1/4} \)
(v) \( (0.0037)^{1/2} \)
(vi) \( (0.6)^{1/2} \)
(vii) \( (26.57)^{1/3} \)
(viii) \( (0.009)^{1/3} \)
(ix) \( \log_e = 4.04 \), it being given that \( \log_{10}4 = 0.6021 \) and \( \log_{10} e = 0.4343 \)
(x) \( \log_e = 10.02 \), it being given that \( \log_e 10 = 2.3026 \)
Answer:
(i) Let \( y = f(x) = \sqrt{x} \).
Take \( x = 36 \), so \( x + \Delta x = 37 \). This means \( \Delta x = 37 - 36 = 1 \).
When \( x = 36 \), then \( y = \sqrt{36} = 6 \).
Let \( \Delta x = dx = 1 \).
Now, \( y = \sqrt{x} \). So, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
At \( x = 36 \), \( \frac{dy}{dx} = \frac{1}{2\sqrt{36}} = \frac{1}{2 \times 6} = \frac{1}{12} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{12} \times 1 = \frac{1}{12} \).
Hence, \( \sqrt{37} = y + \Delta y = 6 + \frac{1}{12} = 6 + 0.0833 = 6.083 \).

(ii) Let \( y = f(x) = x^{1/4} \).
Take \( x = 16 \), so \( x + \Delta x = 15 \). This means \( \Delta x = 15 - 16 = -1 \).
When \( x = 16 \), then \( y = (16)^{1/4} = 2 \).
Let \( \Delta x = dx = -1 \).
Now, \( y = x^{1/4} \). So, \( \frac{dy}{dx} = \frac{1}{4}x^{-3/4} \).
At \( x = 16 \), \( \frac{dy}{dx} = \frac{1}{4}(16)^{-3/4} = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{32} \times (-1) = -\frac{1}{32} \).
Hence, \( (15)^{1/4} = y + \Delta y = 2 - \frac{1}{32} = \frac{64-1}{32} = \frac{63}{32} = 1.969 \).

(iii) Let \( y = f(x) = x^{1/4} \).
Take \( x = 256 \), so \( x + \Delta x = 255 \). This means \( \Delta x = -1 \).
When \( x = 256 \), then \( y = (256)^{1/4} = 4 \).
Let \( dx = \Delta x = -1 \).
Now, \( y = x^{1/4} \). So, \( \frac{dy}{dx} = \frac{1}{4}x^{-3/4} \).
At \( x = 256 \), \( \frac{dy}{dx} = \frac{1}{4}(256)^{-3/4} = \frac{1}{4} \times \frac{1}{64} = \frac{1}{256} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{256} \times (-1) = -\frac{1}{256} \).
Hence, \( (255)^{1/4} = y + \Delta y = 4 - \frac{1}{256} = \frac{1024-1}{256} = \frac{1023}{256} = 3.996 \).

(iv) Let \( y = f(x) = x^{1/4} \).
Take \( x = 81 \), so \( x + \Delta x = 81.5 \). This means \( \Delta x = 0.5 \).
When \( x = 81 \), then \( y = (81)^{1/4} = 3 \).
Let \( dx = \Delta x = 0.5 \).
Now, \( y = x^{1/4} \). So, \( \frac{dy}{dx} = \frac{1}{4}x^{-3/4} \).
At \( x = 81 \), \( \frac{dy}{dx} = \frac{1}{4}(81)^{-3/4} = \frac{1}{4} \times \frac{1}{27} = \frac{1}{108} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{108} \times 0.5 = \frac{0.5}{108} = 0.0046 \).
Hence, \( (81.5)^{1/4} = y + \Delta y = 3 + 0.0046 = 3.005 \).

(v) Let \( y = f(x) = \sqrt{x} \).
Let \( x = 0.0036 \) and \( x + \delta x = 0.0037 \).
So, \( \delta x = 0.0037 - 0.0036 = 0.0001 \).
From the definition of differentials, \( y + \delta y = \sqrt{x+\delta x} \).
Therefore, \( \delta y = \sqrt{x+\delta x} - y \).
We need to find \( (0.0037)^{1/2} = \sqrt{0.0037} \).
Since \( y = \sqrt{x} \), we have \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
Then \( dy = \frac{dy}{dx} \delta x = \frac{1}{2\sqrt{x}} \delta x \).
Substituting \( x = 0.0036 \) and \( \delta x = 0.0001 \):
\( dy = \frac{1}{2\sqrt{0.0036}} \times 0.0001 = \frac{1}{2 \times 0.06} \times 0.0001 = \frac{1}{0.12} \times 0.0001 \).
\( dy = \frac{0.0001}{0.12} = \frac{1}{1200} \approx 0.000833 \).
Since \( \Delta y \approx dy \), the approximate value of \( \sqrt{0.0037} \) is \( y + \Delta y = \sqrt{0.0036} + 0.000833 = 0.06 + 0.000833 = 0.06083 \).

(vi) Let \( y = f(x) = \sqrt{x} \).
Let \( x = 0.64 \) and \( x + \delta x = 0.6 \).
So, \( \delta x = 0.6 - 0.64 = -0.04 \).
When \( x = 0.64 \), \( y = \sqrt{0.64} = 0.8 \).
The derivative is \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
At \( x = 0.64 \), \( \frac{dy}{dx} = \frac{1}{2\sqrt{0.64}} = \frac{1}{2 \times 0.8} = \frac{1}{1.6} \).
Now, \( \Delta y = dy = \frac{dy}{dx} \delta x = \frac{1}{1.6} \times (-0.04) = -0.025 \).
So, \( \sqrt{0.6} = y + \Delta y = 0.8 - 0.025 = 0.775 \).

(vii) Let \( y = f(x) = x^{1/3} \).
Take \( x = 27 \), so \( x + \Delta x = 26.57 \). This means \( \Delta x = 26.57 - 27 = -0.43 \).
When \( x = 27 \), then \( y = (27)^{1/3} = 3 \).
Let \( dx = \Delta x = -0.43 \).
Now, \( y = x^{1/3} \). So, \( \frac{dy}{dx} = \frac{1}{3}x^{-2/3} \).
At \( x = 27 \), \( \frac{dy}{dx} = \frac{1}{3}(27)^{-2/3} = \frac{1}{3} \times \frac{1}{9} = \frac{1}{27} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{27} \times (-0.43) = -\frac{0.43}{27} \approx -0.0159 \).
Hence, \( (26.57)^{1/3} = y + \Delta y = 3 - 0.0159 = 2.984 \).

(viii) Let \( y = f(x) = x^{1/3} \).
Take \( x = 0.008 \), so \( x + \Delta x = 0.009 \). This means \( \Delta x = 0.001 \).
When \( x = 0.008 \), then \( y = (0.008)^{1/3} = 0.2 \).
Let \( dx = \Delta x = 0.001 \).
Now, \( y = x^{1/3} \). So, \( \frac{dy}{dx} = \frac{1}{3}x^{-2/3} \).
At \( x = 0.008 \), \( \frac{dy}{dx} = \frac{1}{3}(0.008)^{-2/3} = \frac{1}{3} \times (0.2)^{-2} = \frac{1}{3} \times \frac{1}{0.04} = \frac{1}{0.12} \).
We know that \( \Delta y = dy \).
So, \( \Delta y = \frac{dy}{dx} dx = \frac{1}{0.12} \times 0.001 = \frac{0.001}{0.12} = \frac{1}{120} \approx 0.00833 \).
Hence, \( (0.009)^{1/3} = y + \Delta y = 0.2 + 0.00833 = 0.208 \).

(ix) Let us consider the function \( y = \log_e x = f(x) \).
Let \( x = 4 \) and \( x + \Delta x = 4.04 \). This means \( \Delta x = dx = 4.04 - 4 = 0.04 \).
When \( x = 4 \), \( y = \log_e 4 \). We are given \( \log_{10} 4 = 0.6021 \).
We know \( \log_e 4 = \frac{\log_{10} 4}{\log_{10} e} = \frac{0.6021}{0.4343} \approx 1.386 \).
The derivative of \( y = \log_e x \) is \( \frac{dy}{dx} = \frac{1}{x} \).
At \( x = 4 \), \( \frac{dy}{dx} = \frac{1}{4} \).
Now, \( \Delta y = dy = \frac{dy}{dx} dx = \frac{1}{4} \times 0.04 = 0.01 \).
So, \( \log_e 4.04 = y + \Delta y = \log_e 4 + 0.01 \approx 1.386 + 0.01 = 1.396 \).

(x) Let us consider the function \( y = \log_e x = f(x) \).
Let \( x = 10 \) and \( x + \Delta x = 10.02 \). This means \( \Delta x = dx = 10.02 - 10 = 0.02 \).
When \( x = 10 \), \( y = \log_e 10 \). We are given \( \log_e 10 = 2.3026 \).
The derivative of \( y = \log_e x \) is \( \frac{dy}{dx} = \frac{1}{x} \).
At \( x = 10 \), \( \frac{dy}{dx} = \frac{1}{10} \).
Now, \( \Delta y = dy = \frac{dy}{dx} dx = \frac{1}{10} \times 0.02 = 0.002 \).
So, \( \log_e 10.02 = y + \Delta y = \log_e 10 + 0.002 = 2.3026 + 0.002 = 2.3046 \).
In simple words: To find the approximate value of a slightly changed number using differentials, we break the number into a known part (x) and a very small change (Δx). Then, we use the derivative of the function to find the small change in the function's value (Δy) for that small change in x. Adding the original function value (y) and this small change (Δy) gives us the approximate answer.

🎯 Exam Tip: Remember that \( \Delta x \) and \( dx \) are considered approximately equal for small changes. Make sure to choose 'x' as a value close to the given number for which the function is easy to calculate, and \( \Delta x \) represents the difference. Always carry out calculations to at least four decimal places before rounding the final answer to three places.

Question 2. If \( f(x) = 3x^2 + 15x + 5 \), then the approximate value of \( f(3.02) \) is
(a) 47.66
(b) 57.66
(c) 67.66
(d) 77.66
Answer: (d) 77.66
In simple words: To find the approximate value of the function at a slightly changed input, we first find the function's value and its rate of change (derivative) at the nearby whole number. Then we multiply the rate of change by the small difference in the input and add it to the original function value.

🎯 Exam Tip: When dealing with polynomials, differentiation is straightforward. Remember that \( \Delta y \approx \frac{dy}{dx} \times \Delta x \) is the core formula for approximation. Be careful with signs when calculating \( \Delta x \).

Question 3. The volume of a sphere, radius r cm is \( \frac{4}{3}\pi r^3 \) cm\( ^3 \). Find the approximate decrease in volume of a sphere when the radius decreases from 3 to 2.98 cm.
Answer: Given the volume of a sphere, \( V = \frac{4}{3}\pi r^3 \).
To find the decrease in volume, we differentiate V with respect to r:
\( \frac{dV}{dr} = \frac{d}{dr} \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi (3r^2) = 4\pi r^2 \).
The radius decreases from \( r = 3 \) cm to \( 2.98 \) cm.
So, the initial radius is \( r = 3 \) cm.
The change in radius is \( \delta r = 2.98 - 3 = -0.02 \) cm.
The approximate change in volume, \( dV \), is given by \( dV = \frac{dV}{dr} \times \delta r \).
Substitute the values:
\( dV = (4\pi r^2) \times \delta r = 4\pi (3^2) \times (-0.02) \).
\( dV = 4\pi (9) \times (-0.02) = 36\pi \times (-0.02) \).
\( dV = -0.72\pi \) cm\( ^3 \).
The negative sign indicates a decrease in volume. Therefore, the approximate decrease in volume of the sphere is \( 0.72\pi \) cm\( ^3 \). This calculation helps in understanding how sensitive the volume is to small changes in radius.
In simple words: When a sphere's radius gets a little smaller, its volume also gets smaller. We can figure out how much smaller the volume becomes by using a special math trick with derivatives. We find the rate at which volume changes with radius, then multiply it by the small change in radius.

🎯 Exam Tip: Always pay attention to the sign of \( \delta r \). If the quantity is decreasing, \( \delta r \) (or \( \Delta x \)) will be negative, leading to a negative differential \( dV \) (or \( \Delta y \)), correctly showing a decrease. Clearly state your initial values and the change.

Question 4. Find the approximate change in \( 1/x \) when \( x = 1 \), \( \delta x = 0.2 \).
Answer: Let \( y = f(x) = \frac{1}{x} \).
We are given \( x = 1 \) and \( \delta x = 0.2 \).
First, find the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (x^{-1}) = -1x^{-2} = -\frac{1}{x^2} \).
Next, calculate the value of the derivative at \( x = 1 \):
\( \left(\frac{dy}{dx}\right)_{x=1} = -\frac{1}{1^2} = -1 \).
The approximate change in \( y \), denoted as \( \delta y \) or \( dy \), is given by:
\( \delta y = \frac{dy}{dx} \times \delta x \).
Substitute the values:
\( \delta y = (-1) \times (0.2) = -0.2 \).
So, the approximate change in \( 1/x \) when \( x \) changes from 1 by \( 0.2 \) is \( -0.2 \). This means the value of \( 1/x \) decreases by 0.2.
In simple words: We want to know how much a fraction like "1 divided by a number" changes if the number itself changes a little. We use the idea of a derivative to find how fast the fraction changes. Then, we multiply that rate by the small change in the number to get the approximate change in the fraction.

🎯 Exam Tip: For functions like \( 1/x \) or \( \sqrt{x} \), remember the power rule for differentiation \( \frac{d}{dx}(x^n) = nx^{n-1} \). Be careful with negative exponents and their derivatives. The sign of the approximate change indicates increase (+) or decrease (-).

Question 5. A sphere of radius 10 cm shrinks to radius 9.8 cm. Find approximately the decrease in (i) volume, and (ii) surface area.
Answer: Let \( r \) be the radius of the sphere.
Given initial radius \( r = 10 \) cm.
The radius shrinks to \( 9.8 \) cm, so the change in radius is \( \Delta r = 9.8 - 10 = -0.2 \) cm.
We can let \( dr = \Delta r = -0.2 \) cm.

(i) Volume of a sphere, \( V = \frac{4}{3}\pi r^3 \).
The derivative of volume with respect to radius is \( \frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \).
The approximate decrease in volume, \( dV \), is \( dV = \frac{dV}{dr} \times dr \).
Substitute \( r = 10 \) and \( dr = -0.2 \):
\( dV = 4\pi (10^2) \times (-0.2) = 4\pi (100) \times (-0.2) = -80\pi \) cm\( ^3 \).
The negative sign indicates a decrease. So, the approximate decrease in volume is \( 80\pi \) cm\( ^3 \). This shows how a small change in radius can lead to a significant change in volume for a sphere.

(ii) Surface area of a sphere, \( S = 4\pi r^2 \).
The derivative of surface area with respect to radius is \( \frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 8\pi r \).
The approximate decrease in surface area, \( dS \), is \( dS = \frac{dS}{dr} \times dr \).
Substitute \( r = 10 \) and \( dr = -0.2 \):
\( dS = 8\pi (10) \times (-0.2) = 80\pi \times (-0.2) = -16\pi \) cm\( ^2 \).
The negative sign indicates a decrease. So, the approximate decrease in surface area is \( 16\pi \) cm\( ^2 \).
In simple words: When a ball shrinks a little bit, both its inside space (volume) and its outside skin (surface area) get smaller. We use derivatives to calculate how much they reduce. We find how fast the volume or area changes with radius, then multiply that speed by the tiny amount the radius changed.

🎯 Exam Tip: Remember the formulas for volume and surface area of a sphere. When calculating the decrease, ensure \( \Delta r \) is negative to get the correct sign for the differential, indicating reduction.

Question 6. A circular plate expands when heated from a radius of 5 cm to 5.06 cm. Find the approximate increase in area.
Answer: Let \( A \) be the area of a circle with radius \( r \).
The formula for the area of a circle is \( A = \pi r^2 \).
The initial radius is \( r = 5 \) cm.
The radius expands to \( 5.06 \) cm.
The change in radius is \( \Delta r = 5.06 - 5 = 0.06 \) cm.
We can let \( dr = \Delta r = 0.06 \) cm.
First, find the derivative of the area with respect to the radius:
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r \).
Now, calculate the value of the derivative at the initial radius \( r = 5 \):
\( \left(\frac{dA}{dr}\right)_{r=5} = 2\pi (5) = 10\pi \).
The approximate increase in area, \( dA \), is given by \( dA = \frac{dA}{dr} \times dr \).
Substitute the values:
\( dA = (10\pi) \times (0.06) = 0.6\pi \) cm\( ^2 \).
Since \( \Delta r \) is positive, \( dA \) is also positive, indicating an increase. So, the approximate increase in area is \( 0.6\pi \) cm\( ^2 \). This shows how heat causes materials to expand, affecting their area.
In simple words: When a round plate gets hot, its radius grows a little, which makes its whole area bigger. To find out how much bigger the area gets, we use derivatives. We calculate how fast the area changes for each bit of radius increase, then multiply by the small change in radius.

🎯 Exam Tip: Remember the area formula for a circle. Ensure you calculate \( \Delta r \) correctly and use the initial radius when evaluating the derivative. The result will be positive for an increase and negative for a decrease.

Question 7. If the side of a cube is 10.01 cm, find approximately the volume of the cube.
Answer: Let \( x \) be the side of the cube.
The volume of a cube is \( V = x^3 \).
We are given the side length as \( 10.01 \) cm.
Let \( x = 10 \) cm and \( x + \Delta x = 10.01 \) cm.
So, the change in side length is \( \Delta x = 10.01 - 10 = 0.01 \) cm.
We can let \( dx = \Delta x = 0.01 \) cm.
First, calculate the volume for the known side \( x = 10 \):
\( V = (10)^3 = 1000 \) cm\( ^3 \).
Next, find the derivative of the volume with respect to the side length:
\( \frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2 \).
Evaluate the derivative at \( x = 10 \):
\( \left(\frac{dV}{dx}\right)_{x=10} = 3(10^2) = 3 \times 100 = 300 \).
The approximate change in volume, \( dV \), is given by \( dV = \frac{dV}{dx} \times dx \).
Substitute the values:
\( dV = (300) \times (0.01) = 3 \) cm\( ^3 \).
The approximate volume of the cube is \( V + dV \).
Required volume \( = 1000 + 3 = 1003 \) cm\( ^3 \). This method gives a very close estimate for volumes of objects with slight changes in dimensions.
In simple words: To find the approximate volume of a cube when its side is just a tiny bit bigger than a round number, we calculate the volume for the round number first. Then, we figure out how much the volume changes for a small increase in side length using calculus, and add that change to our first volume.

🎯 Exam Tip: For volume calculations, ensure you identify the base value 'x' for which the function is easily calculable and the small change 'dx'. The result of \( V + dV \) is the approximate value, and always include units for volume (cm\( ^3 \)).

Question 8. The shape of a bowl is such that \( V = h^3 + 3h^2 + 11h \) where \( V \) cm\( ^3 \) is the volume of water in the bowl and \( h \) cm the depth of the water. If, when \( h = 7 \), an additional small volume \( \delta V \) cm\( ^3 \) of water is poured into the bowl, prove that the level of the water rises approximately \( \frac{\delta V}{200} \) cm.
Answer: Given the volume of water in the bowl as \( V = h^3 + 3h^2 + 11h \).
To find how the height \( h \) changes with volume \( V \), we first differentiate \( V \) with respect to \( h \):
\( \frac{dV}{dh} = \frac{d}{dh}(h^3 + 3h^2 + 11h) = 3h^2 + 6h + 11 \).
We are interested in the situation when the depth of water is \( h = 7 \) cm.
Calculate \( \frac{dV}{dh} \) at \( h = 7 \):
\( \left(\frac{dV}{dh}\right)_{h=7} = 3(7^2) + 6(7) + 11 = 3(49) + 42 + 11 = 147 + 42 + 11 = 200 \).
We know that for small changes, \( \delta V \approx dV \) and \( \delta h \approx dh \).
Also, \( dV = \frac{dV}{dh} \times dh \).
So, \( \delta V = \left(\frac{dV}{dh}\right)_{h=7} \times \delta h \).
Substituting the value of the derivative, we get:
\( \delta V = 200 \times \delta h \).
To find the rise in water level \( \delta h \), we rearrange the equation:
\( \delta h = \frac{\delta V}{200} \) cm.
This proves that the level of the water rises approximately \( \frac{\delta V}{200} \) cm when an additional small volume \( \delta V \) is poured in. This is a practical application of derivatives to understand how water levels change in containers of complex shapes.
In simple words: We have a bowl shaped so that its water volume depends on its depth. If we add a tiny bit more water, we want to know how much the water level goes up. By using a derivative, we find how fast the volume changes with depth. Then, we use this rate to link the extra water volume to the small rise in water level.

🎯 Exam Tip: When proving a relationship involving differentials, clearly state the given function and its derivative. Remember that \( \frac{dV}{dh} \) represents the rate of change of volume with respect to height, which is then used to relate small changes \( \delta V \) and \( \delta h \).

Question 9. A closed circular cylinder has height 16 cm, and radius \( r \) cm. The total surface area is \( A \) cm\( ^2 \). Prove that \( \frac{dA}{dr} = 4\pi(r + 8) \). Hence, calculate an approximate increase in area if the radius increases from 4 to 4.02 cm, the height remaining constant.
Answer: For a closed circular cylinder, the height \( h = 16 \) cm and radius \( r \) cm.
The total surface area \( A \) is given by the formula:
\( A = 2\pi r^2 + 2\pi rh \).
Since \( h = 16 \), substitute this into the area formula:
\( A = 2\pi r^2 + 2\pi r(16) \).
\( A = 2\pi r^2 + 32\pi r \).
Now, differentiate \( A \) with respect to \( r \) to find \( \frac{dA}{dr} \):
\( \frac{dA}{dr} = \frac{d}{dr}(2\pi r^2 + 32\pi r) = 2\pi(2r) + 32\pi(1) = 4\pi r + 32\pi \).
Factor out \( 4\pi \):
\( \frac{dA}{dr} = 4\pi(r + 8) \). This proves the first part of the question.

For the second part, calculate the approximate increase in area.
The radius increases from \( r = 4 \) cm to \( 4.02 \) cm.
So, the initial radius is \( r = 4 \) cm.
The change in radius is \( \Delta r = 4.02 - 4 = 0.02 \) cm.
We can let \( dr = \Delta r = 0.02 \) cm.
First, evaluate \( \frac{dA}{dr} \) at \( r = 4 \):
\( \left(\frac{dA}{dr}\right)_{r=4} = 4\pi(4 + 8) = 4\pi(12) = 48\pi \).
The approximate increase in area, \( dA \), is given by \( dA = \frac{dA}{dr} \times dr \).
Substitute the values:
\( dA = (48\pi) \times (0.02) = 0.96\pi \) cm\( ^2 \).
So, the approximate increase in surface area is \( 0.96\pi \) cm\( ^2 \). This helps to calculate how much material might be needed if the cylinder's dimensions change slightly.
In simple words: For a cylinder of a fixed height, we first write down how its total outside area depends on its radius. Then, we use a derivative to see how fast this area changes when the radius changes. If the radius grows a tiny bit, we multiply this "speed of change" by the small increase in radius to find out how much the area increases.

🎯 Exam Tip: Clearly write down the formula for the total surface area of a closed cylinder. Remember to substitute the constant height before differentiating. The derivative \( \frac{dA}{dr} \) shows the rate of change of area per unit change in radius.