GSEB Class 12 Maths Solutions Chapter 6 Application of Derivatives Exercise 6.4

Question 1. Using differentials, find the approximate value of each of the following upto 3 places of decimal :
(i) \( \sqrt{25.3} \)
(ii) \( \sqrt{49.5} \)
(iii) \( \sqrt{0.6} \)
(iv) \( (0.009)^{\frac { 1 }{ 3 }} \)
(v) \( (0.999)^{\frac { 1 }{ 10 }} \)
(vi) \( (15)^{\frac { 1 }{ 4 }} \)
(vii) \( (26)^{\frac { 1 }{ 3 }} \)
(viii) \( (255)^{\frac { 1 }{ 4 }} \)
(ix) \( (82)^{\frac { 1 }{ 4 }} \)
(x) \( (401)^{\frac { 1 }{ 2 }} \)
(xi) \( (0.0037)^{\frac { 1 }{ 2 }} \)
(xii) \( (26.57)^{\frac { 1 }{ 3 }} \)
(xiii) \( (81.5)^{\frac { 1 }{ 4 }} \)
(xiv) \( (3.968)^{\frac { 3 }{ 2 }} \)
(xv) \( (32.15)^{\frac { 1 }{ 5 }} \)
Answer:
(i) Let \( y = \sqrt{x} \). Given \( x = 25 \) and \( \Delta x = 0.3 \).
So, \( \Delta y = \sqrt{x+\Delta x} - \sqrt{x} = \sqrt{25.3} - \sqrt{25} = \sqrt{25.3} - 5 \).
This gives \( \sqrt{25.3} = 5 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x = \frac{1}{2\sqrt{25}} \times (0.3) = \frac{1}{10} \times 0.3 = 0.03 \).
Therefore, \( dy = \Delta y = 0.03 \).
Putting this value of \( \Delta y \) in (1), we get
\( \sqrt{25.3} = 5 + 0.03 = 5.03 \).
(ii) Let \( y = \sqrt{x} \). Given \( x = 49 \) and \( \Delta x = 0.5 \).
So, \( \Delta y = \sqrt{x+\Delta x} - \sqrt{x} = \sqrt{49.5} - \sqrt{49} = \sqrt{49.5} - 7 \).
This gives \( \sqrt{49.5} = 7 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x = \frac{1}{2\sqrt{49}} \times 0.5 = \frac{1}{14} \times 0.5 = 0.036 \).
Therefore, \( dy = \Delta y = 0.036 \).
Putting this value of \( \Delta y \) in (1), we get
\( \sqrt{49.5} = 7 + 0.036 = 7.036 \).
(iii) Let \( y = \sqrt{x} \). Given \( x = 0.64 \) and \( \Delta x = -0.04 \).
So, \( \Delta y = \sqrt{x+\Delta x} - \sqrt{x} = \sqrt{0.60} - \sqrt{0.64} = \sqrt{0.60} - 0.8 \).
This gives \( \sqrt{0.60} = 0.8 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x = \frac{1}{2\sqrt{0.64}} \times (-0.04) = \frac{1}{2 \times 0.8} \times (-0.04) = \frac{-0.04}{1.6} = -0.025 \).
Therefore, \( dy = \Delta y = -0.025 \).
Putting this value of \( \Delta y \) in (1), we get
\( \sqrt{0.60} = 0.8 - 0.025 = 0.775 \).
(iv) Let \( y = x^{\frac{1}{3}} \). Given \( x = 0.008 \) and \( \Delta x = 0.001 \). So, \( x + \Delta x = 0.009 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (0.009)^{\frac{1}{3}} - (0.008)^{\frac{1}{3}} = (0.009)^{\frac{1}{3}} - 0.2 \).
This gives \( (0.009)^{\frac{1}{3}} = 0.2 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{3x^{\frac{2}{3}}} \times \Delta x = \frac{1}{3(0.008)^{\frac{2}{3}}} \times 0.001 = \frac{1}{3(0.2)^2} \times 0.001 = \frac{0.001}{3 \times 0.04} = \frac{0.001}{0.12} = 0.00833 \).
Therefore, \( \Delta y = dy = 0.00833 \).
Putting this value of \( \Delta y \) in (1), we get
\( (0.009)^{\frac{1}{3}} = 0.2 + 0.00833 = 0.20833 \) (approx.).
(v) Let \( y = x^{\frac{1}{10}} \). Given \( x = 1 \) and \( \Delta x = -0.001 \). So, \( x + \Delta x = 0.999 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{10}} - x^{\frac{1}{10}} = (0.999)^{\frac{1}{10}} - (1)^{\frac{1}{10}} = (0.999)^{\frac{1}{10}} - 1 \).
This gives \( (0.999)^{\frac{1}{10}} = 1 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{10}x^{\frac{1}{10}-1} = \frac{1}{10}x^{-\frac{9}{10}} = \frac{1}{10x^{\frac{9}{10}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{10x^{\frac{9}{10}}} \times \Delta x = \frac{1}{10(1)^{\frac{9}{10}}} \times (-0.001) = \frac{1}{10} \times (-0.001) = -0.0001 \).
Therefore, \( \Delta y = dy = -0.0001 \).
Putting this value of \( \Delta y \) in (1), we get
\( (0.999)^{\frac{1}{10}} = 1 - 0.0001 = 0.9999 \).
(vi) Let \( y = x^{\frac{1}{4}} \). Given \( x = 16 \) and \( \Delta x = -1 \). So, \( x + \Delta x = 15 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (15)^{\frac{1}{4}} - (16)^{\frac{1}{4}} = (15)^{\frac{1}{4}} - 2 \).
This gives \( (15)^{\frac{1}{4}} = 2 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}-1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{4x^{\frac{3}{4}}} \times \Delta x = \frac{1}{4(16)^{\frac{3}{4}}} \times (-1) = \frac{1}{4(2^4)^{\frac{3}{4}}} \times (-1) = \frac{1}{4 \times 2^3} \times (-1) = \frac{-1}{32} = -0.03125 \).
Therefore, \( \Delta y = dy = -0.03125 \).
Putting this value of \( \Delta y \) in (1), we get
\( (15)^{\frac{1}{4}} = 2 - 0.03125 = 1.96875 \) (approx.).
(vii) Let \( y = x^{\frac{1}{3}} \). Given \( x = 27 \) and \( \Delta x = -1 \). So, \( x + \Delta x = 26 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (26)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (26)^{\frac{1}{3}} - 3 \).
This gives \( (26)^{\frac{1}{3}} = 3 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{3x^{\frac{2}{3}}} \times \Delta x = \frac{1}{3(27)^{\frac{2}{3}}} \times (-1) = \frac{1}{3(3^3)^{\frac{2}{3}}} \times (-1) = \frac{1}{3 \times 3^2} \times (-1) = \frac{-1}{27} = -0.037037 \).
Therefore, \( \Delta y = dy = -0.037037 \).
Putting this value of \( \Delta y \) in (1), we get
\( (26)^{\frac{1}{3}} = 3 - 0.037037 = 2.962963 \) (approx.).
(viii) Let \( y = x^{\frac{1}{4}} \). Given \( x = 256 \) and \( \Delta x = -1 \). So, \( x + \Delta x = 255 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (255)^{\frac{1}{4}} - (256)^{\frac{1}{4}} = (255)^{\frac{1}{4}} - 4 \).
This gives \( (255)^{\frac{1}{4}} = 4 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}-1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{4x^{\frac{3}{4}}} \times \Delta x = \frac{1}{4(256)^{\frac{3}{4}}} \times (-1) = \frac{1}{4(4^4)^{\frac{3}{4}}} \times (-1) = \frac{1}{4 \times 4^3} \times (-1) = \frac{-1}{256} = -0.003906 \).
Therefore, \( \Delta y = dy = -0.003906 \).
Putting this value of \( \Delta y \) in (1), we get
\( (255)^{\frac{1}{4}} = 4 - 0.003906 = 3.996094 \) (approx.).
(ix) Let \( y = x^{\frac{1}{4}} \). Given \( x = 81 \) and \( \Delta x = 1 \). So, \( x + \Delta x = 82 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (82)^{\frac{1}{4}} - (81)^{\frac{1}{4}} = (82)^{\frac{1}{4}} - 3 \).
This gives \( (82)^{\frac{1}{4}} = 3 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}-1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{4x^{\frac{3}{4}}} \times \Delta x = \frac{1}{4(81)^{\frac{3}{4}}} \times (1) = \frac{1}{4(3^4)^{\frac{3}{4}}} \times (1) = \frac{1}{4 \times 3^3} \times (1) = \frac{1}{108} = 0.009259 \).
Therefore, \( \Delta y = dy = 0.009259 \).
Putting this value of \( \Delta y \) in (1), we get
\( (82)^{\frac{1}{4}} = 3 + 0.009259 = 3.009259 \) (approx.).
(x) Let \( y = x^{\frac{1}{2}} \). Given \( x = 400 \) and \( \Delta x = 1 \). So, \( x + \Delta x = 401 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{2}} - x^{\frac{1}{2}} = (401)^{\frac{1}{2}} - (400)^{\frac{1}{2}} = (401)^{\frac{1}{2}} - 20 \).
This gives \( (401)^{\frac{1}{2}} = 20 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x = \frac{1}{2\sqrt{400}} \times (1) = \frac{1}{2 \times 20} \times (1) = \frac{1}{40} = 0.025 \).
Therefore, \( \Delta y = dy = 0.025 \).
Putting this value of \( \Delta y \) in (1), we get
\( (401)^{\frac{1}{2}} = 20 + 0.025 = 20.025 \).
(xi) Let \( y = \sqrt{x} \). Given \( x = 0.0036 \) and \( \Delta x = 0.0001 \). So, \( x + \Delta x = 0.0037 \).
So, \( \Delta y = \sqrt{x+\Delta x} - \sqrt{x} = \sqrt{0.0037} - \sqrt{0.0036} = \sqrt{0.0037} - 0.06 \).
This gives \( \sqrt{0.0037} = 0.06 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x = \frac{1}{2\sqrt{0.0036}} \times 0.0001 = \frac{1}{2 \times 0.06} \times 0.0001 = \frac{0.0001}{0.12} = 0.000833 \).
Therefore, \( \Delta y = dy = 0.000833 \).
Putting this value of \( \Delta y \) in (1), we get
\( \sqrt{0.0037} = 0.06 + 0.000833 = 0.060833 \) (approx.).
(xii) Let \( y = x^{\frac{1}{3}} \). Given \( x = 27 \) and \( \Delta x = -0.43 \). So, \( x + \Delta x = 26.57 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (26.57)^{\frac{1}{3}} - 3 \).
This gives \( (26.57)^{\frac{1}{3}} = 3 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{3x^{\frac{2}{3}}} \times \Delta x = \frac{1}{3(27)^{\frac{2}{3}}} \times (-0.43) = \frac{1}{3(3^3)^{\frac{2}{3}}} \times (-0.43) = \frac{1}{3 \times 3^2} \times (-0.43) = \frac{-0.43}{27} = -0.015926 \).
Therefore, \( \Delta y = dy = -0.015926 \).
Putting this value of \( \Delta y \) in (1), we get
\( (26.57)^{\frac{1}{3}} = 3 - 0.015926 = 2.984074 \) (approx.).
(xiii) Let \( y = x^{\frac{1}{4}} \). Given \( x = 81 \) and \( \Delta x = 0.5 \). So, \( x + \Delta x = 81.5 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - (81)^{\frac{1}{4}} = (81.5)^{\frac{1}{4}} - 3 \).
This gives \( (81.5)^{\frac{1}{4}} = 3 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}-1} = \frac{1}{4}x^{-\frac{3}{4}} = \frac{1}{4x^{\frac{3}{4}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{4x^{\frac{3}{4}}} \times \Delta x = \frac{1}{4(81)^{\frac{3}{4}}} \times (0.5) = \frac{1}{4(3^4)^{\frac{3}{4}}} \times (0.5) = \frac{1}{4 \times 3^3} \times (0.5) = \frac{0.5}{108} = 0.0046296 \).
Therefore, \( \Delta y = dy = 0.0046296 \).
Putting this value of \( \Delta y \) in (1), we get
\( (81.5)^{\frac{1}{4}} = 3 + 0.0046296 = 3.0046296 \) (approx.).
(xiv) Let \( y = x^{\frac{3}{2}} \). Given \( x = 4 \) and \( \Delta x = -0.032 \). So, \( x + \Delta x = 3.968 \).
So, \( \Delta y = (x+\Delta x)^{\frac{3}{2}} - x^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - (4)^{\frac{3}{2}} = (3.968)^{\frac{3}{2}} - 8 \).
This gives \( (3.968)^{\frac{3}{2}} = 8 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{3}{2}x^{\frac{3}{2}-1} = \frac{3}{2}x^{\frac{1}{2}} = \frac{3}{2}\sqrt{x} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{3}{2}\sqrt{x} \times \Delta x = \frac{3}{2}\sqrt{4} \times (-0.032) = \frac{3}{2} \times 2 \times (-0.032) = 3 \times (-0.032) = -0.096 \).
Therefore, \( \Delta y = dy = -0.096 \).
Putting this value of \( \Delta y \) in (1), we get
\( (3.968)^{\frac{3}{2}} = 8 - 0.096 = 7.904 \) (approx.).
(xv) Let \( y = x^{\frac{1}{5}} \). Given \( x = 32 \) and \( \Delta x = 0.15 \). So, \( x + \Delta x = 32.15 \).
So, \( \Delta y = (x+\Delta x)^{\frac{1}{5}} - x^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - (32)^{\frac{1}{5}} = (32.15)^{\frac{1}{5}} - 2 \).
This gives \( (32.15)^{\frac{1}{5}} = 2 + \Delta y \) ... (1)
Now, \( dy \) is approximately equal to \( \Delta y \).
Also, \( \frac{dy}{dx} = \frac{1}{5}x^{\frac{1}{5}-1} = \frac{1}{5}x^{-\frac{4}{5}} = \frac{1}{5x^{\frac{4}{5}}} \).
So, \( dy = \left(\frac{dy}{dx}\right) \Delta x = \frac{1}{5x^{\frac{4}{5}}} \times \Delta x = \frac{1}{5(32)^{\frac{4}{5}}} \times (0.15) = \frac{1}{5(2^5)^{\frac{4}{5}}} \times (0.15) = \frac{1}{5 \times 2^4} \times (0.15) = \frac{0.15}{80} = 0.001875 \).
Therefore, \( \Delta y = dy = 0.001875 \).
Putting this value of \( \Delta y \) in (1), we get
\( (32.15)^{\frac{1}{5}} = 2 + 0.001875 = 2.001875 \) (approx.).
In simple words: To find the approximate value, we use differentials. First, identify a close number \( x \) whose root or power is easy to calculate, and determine the small change \( \Delta x \). Then, calculate \( \Delta y \) using the differential \( dy = (\frac{dy}{dx})\Delta x \) and add it to the initial value.

Exam Tip: Remember to choose \( x \) such that it is close to the given number and its root/power is easily computable. Also, pay close attention to the sign of \( \Delta x \).

Question 2. Find the approximate value of f(2.01), where \( f(x) = 4x^2 + 5x +2 \).
Answer: Let \( x = 2 \) and \( \Delta x = 0.01 \). So, \( x + \Delta x = 2.01 \).
We know that \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).
First, calculate \( f(x) \): \( f(2) = 4(2)^2 + 5(2) + 2 = 4(4) + 10 + 2 = 16 + 10 + 2 = 28 \).
Next, find the derivative \( f'(x) \): \( f'(x) = \frac{d}{dx}(4x^2 + 5x + 2) = 8x + 5 \).
Then, calculate \( f'(2) \): \( f'(2) = 8(2) + 5 = 16 + 5 = 21 \).
Now, substitute these values into the approximation formula:
\( f(2.01) \approx f(2) + f'(2) \Delta x = 28 + (21)(0.01) = 28 + 0.21 = 28.21 \).
Thus, the approximate value of \( f(2.01) \) is \( 28.21 \).
In simple words: To estimate the function's value at a point slightly different from a known one, we find the function's value and its slope at the known point. Then, we multiply the slope by the small change and add it to the original function value.

Exam Tip: Remember the formula for approximation: \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \). Clearly identify \( x \) and \( \Delta x \) and compute \( f(x) \) and \( f'(x) \) accurately.

Question 3. Find the approximate value of f(5.001), where \( f(x) = x^3 – 7x^2 +15 \).
Answer: Let \( x = 5 \) and \( \Delta x = 0.001 \). So, \( x + \Delta x = 5.001 \).
We know that \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).
First, calculate \( f(x) \): \( f(5) = (5)^3 - 7(5)^2 + 15 = 125 - 7(25) + 15 = 125 - 175 + 15 = -35 \).
Next, find the derivative \( f'(x) \): \( f'(x) = \frac{d}{dx}(x^3 - 7x^2 + 15) = 3x^2 - 14x \).
Then, calculate \( f'(5) \): \( f'(5) = 3(5)^2 - 14(5) = 3(25) - 70 = 75 - 70 = 5 \).
Now, substitute these values into the approximation formula:
\( f(5.001) \approx f(5) + f'(5) \Delta x = -35 + (5)(0.001) = -35 + 0.005 = -34.995 \).
Thus, the approximate value of \( f(5.001) \) is \( -34.995 \).
In simple words: We find the function value and its rate of change at a simple point. Then, we use the rate of change multiplied by the tiny difference and add it to the original function value to get an estimate.

Exam Tip: Ensure precise calculation of the derivative and correct substitution of \( x \) and \( \Delta x \) values into the approximation formula. Be careful with signs, especially when \( f(x) \) is negative.

Question 4. Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.
Answer: Let the side of the cube be \( x \) metres.
The volume of the cube is given by \( V = x^3 \).
The increase in the side is \( 1\% \) of \( x \), which means \( \Delta x = 0.01x \).
The approximate change in volume, \( \Delta V \), is given by \( dV = (\frac{dV}{dx}) \Delta x \).
First, find the derivative of \( V \) with respect to \( x \): \( \frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2 \).
Now, substitute the values into the formula for \( dV \):
\( dV = (3x^2)(0.01x) = 0.03x^3 \).
Therefore, the approximate change in the volume is \( 0.03x^3 \) cubic metres.
In simple words: If a cube's side grows a little, its volume changes. We find how much the volume changes when the side changes by 1%, by using a special math tool called a derivative.

Exam Tip: For percentage change problems, convert the percentage into a decimal for \( \Delta x \). Always clearly state the formula for approximate change \( \Delta V = (\frac{dV}{dx})\Delta x \) and perform differentiation correctly.

Question 5. Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Answer: Let the side of the cube be \( x \) metres.
The surface area of the cube is given by \( S = 6x^2 \).
The decrease in the side is \( 1\% \) of \( x \), which means \( \Delta x = -0.01x \) (negative because it's a decrease).
The approximate change in surface area, \( \Delta S \), is given by \( dS = (\frac{dS}{dx}) \Delta x \).
First, find the derivative of \( S \) with respect to \( x \): \( \frac{dS}{dx} = \frac{d}{dx}(6x^2) = 12x \).
Now, substitute the values into the formula for \( dS \):
\( dS = (12x)(-0.01x) = -0.12x^2 \).
Therefore, the approximate change in the surface area is \( -0.12x^2 \) square metres. The negative sign indicates a decrease in surface area.
In simple words: When a cube's side shrinks by a small percentage, its surface area also changes. We calculate this approximate change by multiplying the rate of change of surface area by the small decrease in the side.

Exam Tip: Remember to use a negative sign for \( \Delta x \) when there is a decrease in the dimension. Clearly define the surface area formula for a cube and its derivative.

Question 6. If the radius of a sphere measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.
Answer: Let the radius of the sphere be \( r = 7 \) m.
The error in the measurement of the radius is given as \( \Delta r = 0.02 \) m.
The volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \).
The approximate error in calculating the volume, \( \Delta V \), is given by \( dV = (\frac{dV}{dr}) \Delta r \).
First, find the derivative of \( V \) with respect to \( r \): \( \frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi (3r^2) = 4\pi r^2 \).
Now, substitute the values of \( r \) and \( \Delta r \) into the formula for \( dV \):
\( dV = (4\pi r^2) \Delta r = 4\pi (7)^2 (0.02) = 4\pi (49)(0.02) = 196\pi(0.02) = 3.92\pi \).
Therefore, the approximate error in calculating the volume is \( 3.92\pi \) cubic metres.
In simple words: We have a sphere whose radius was measured with a small mistake. We use a method called differentials to figure out how much this small mistake in the radius affects the calculated volume of the sphere.

Exam Tip: Write down the correct formula for the volume of a sphere. Differentiate it correctly with respect to the radius and then substitute the given values for \( r \) and \( \Delta r \).

Question 7. If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Answer: Let the radius of the sphere be \( r = 9 \) m.
The error in the measurement of the radius is given as \( \Delta r = 0.03 \) m.
The surface area of a sphere is given by the formula \( S = 4\pi r^2 \).
The approximate error in calculating the surface area, \( \Delta S \), is given by \( dS = (\frac{dS}{dr}) \Delta r \).
First, find the derivative of \( S \) with respect to \( r \): \( \frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 4\pi (2r) = 8\pi r \).
Now, substitute the values of \( r \) and \( \Delta r \) into the formula for \( dS \):
\( dS = (8\pi r) \Delta r = 8\pi (9)(0.03) = 72\pi(0.03) = 2.16\pi \).
Therefore, the approximate error in calculating the surface area is \( 2.16\pi \) square metres.
In simple words: When a sphere's radius has a small error in its measurement, we use differentiation to calculate how much this error might affect the total surface area of the sphere.

Exam Tip: Be sure to recall the formula for the surface area of a sphere. Differentiate correctly and substitute the values of \( r \) and \( \Delta r \) into the error formula.

Question 8. 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
Let \( x = 3 \) and \( \Delta x = 0.02 \). So, \( x + \Delta x = 3.02 \).
We know that \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).
First, calculate \( f(x) \): \( f(3) = 3(3)^2 + 15(3) + 5 = 3(9) + 45 + 5 = 27 + 45 + 5 = 77 \).
Next, find the derivative \( f'(x) \): \( f'(x) = \frac{d}{dx}(3x^2 + 15x + 5) = 6x + 15 \).
Then, calculate \( f'(3) \): \( f'(3) = 6(3) + 15 = 18 + 15 = 33 \).
Now, substitute these values into the approximation formula:
\( f(3.02) \approx f(3) + f'(3) \Delta x = 77 + (33)(0.02) = 77 + 0.66 = 77.66 \).
Therefore, the approximate value of \( f(3.02) \) is \( 77.66 \).
In simple words: We're finding an estimated value of the function near a known point. We calculate the function's value and its rate of change at the simple point, then use those to project the value at the slightly different point.

Exam Tip: For MCQs, work through the approximation steps carefully to avoid errors. Double-check your derivative calculation and arithmetic to select the correct option.

Question 9. The approximate change in volume of a cube of side x metres caused by increasing the side by 3% is
(A) \( 0.06 x^3 m^3 \)
(B) \( 0.6 x^3 m^3 \)
(C) \( 0.09 x^3 m^3 \)
(D) \( 0.9 x^3 m^3 \)
Answer: (C) \( 0.09 x^3 m^3 \)
Let the side of the cube be \( x \) metres.
The volume of the cube is given by \( V = x^3 \).
The increase in the side is \( 3\% \) of \( x \), which means \( \Delta x = 0.03x \).
The approximate change in volume, \( \Delta V \), is given by \( dV = (\frac{dV}{dx}) \Delta x \).
First, find the derivative of \( V \) with respect to \( x \): \( \frac{dV}{dx} = \frac{d}{dx}(x^3) = 3x^2 \).
Now, substitute the values into the formula for \( dV \):
\( dV = (3x^2)(0.03x) = 0.09x^3 \).
Therefore, the approximate change in the volume is \( 0.09x^3 \) cubic metres.
In simple words: We calculate how much a cube's volume changes when its side length increases by a small percentage. We use the derivative of the volume formula to find this change.

Exam Tip: Convert percentage changes to decimal form (e.g., 3% to 0.03) for \( \Delta x \). The volume of a cube is \( x^3 \), so its derivative with respect to \( x \) is \( 3x^2 \).