Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.1

Question 1. Let \( f(x) = \sqrt[3]{x} \). Find the linear approximation at \( x = 27 \). Use the linear approximation to approximate \( \sqrt[3]{27.2} \)
Answer: We need to find the linear approximation for \( f(x) = \sqrt[3]{x} \) at \( x_0 = 27 \).
First, find the function value at \( x_0 \):
\( f(x_0) = f(27) = \sqrt[3]{27} = 3 \)
Next, find the derivative of \( f(x) \):
\( f(x) = x^{1/3} \)
\( f'(x) = \frac{1}{3}x^{(1/3) - 1} = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \)
Now, evaluate the derivative at \( x_0 = 27 \):
\( f'(27) = \frac{1}{3\sqrt[3]{27^2}} = \frac{1}{3 \times (\sqrt[3]{27})^2} = \frac{1}{3 \times 3^2} = \frac{1}{3 \times 9} = \frac{1}{27} \)
We want to approximate \( \sqrt[3]{27.2} \). Here, \( x = 27.2 \), so \( \Delta x = x - x_0 = 27.2 - 27 = 0.2 \).
The formula for linear approximation is: \( f(x_0 + \Delta x) \approx f(x_0) + f'(x_0) \Delta x \)
Substitute the values:
\( f(27.2) \approx f(27) + f'(27) (0.2) \)
\( f(27.2) \approx 3 + \frac{1}{27} (0.2) \)
\( f(27.2) \approx 3 + \frac{0.2}{27} \)
\( f(27.2) \approx 3 + 0.007407... \)
\( f(27.2) \approx 3.0074 \)
Therefore, the approximate value of \( \sqrt[3]{27.2} \) is \( 3.0074 \). Linear approximation is very useful for estimating values of complex functions near a point where the function value is easily calculated.
In simple words: We found a line that closely matches the curve of the function at \( x=27 \). Then, we used this line to estimate the value of the function at a slightly different point, \( 27.2 \), which gave us about \( 3.0074 \).

🎯 Exam Tip: Remember to clearly identify \( x_0 \) and \( \Delta x \) for the approximation. Always calculate both \( f(x_0) \) and \( f'(x_0) \) carefully before applying the linear approximation formula.

Question 2. Use the linear approximation to find approximate value of
(i) \( (123)^{2/3} \)
(ii) \( \sqrt[4]{15} \)
(iii) \( \sqrt[3]{26} \)
Answer: We will use the linear approximation formula \( f(x_0 + \Delta x) \approx f(x_0) + f'(x_0) \Delta x \) for each part.

(i) For \( (123)^{2/3} \):
Let \( f(x) = x^{2/3} \). We choose \( x_0 = 125 \) because it is close to 123 and \( 125^{1/3} = 5 \).
So, \( \Delta x = 123 - 125 = -2 \).
First, find \( f(x_0) \):
\( f(125) = (125)^{2/3} = (125^{1/3})^2 = 5^2 = 25 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = \frac{2}{3}x^{(2/3)-1} = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} \)
Evaluate \( f'(x_0) \):
\( f'(125) = \frac{2}{3\sqrt[3]{125}} = \frac{2}{3 \times 5} = \frac{2}{15} \)
Now apply the approximation formula:
\( f(123) \approx f(125) + f'(125)(-2) \)
\( f(123) \approx 25 + \frac{2}{15}(-2) \)
\( f(123) \approx 25 - \frac{4}{15} \)
\( f(123) \approx 25 - 0.2666... \)
\( f(123)^{2/3} \approx 24.7334 \)

(ii) For \( \sqrt[4]{15} \):
Let \( f(x) = x^{1/4} \). We choose \( x_0 = 16 \) because it is close to 15 and \( 16^{1/4} = 2 \).
So, \( \Delta x = 15 - 16 = -1 \).
First, find \( f(x_0) \):
\( f(16) = (16)^{1/4} = 2 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = \frac{1}{4}x^{(1/4)-1} = \frac{1}{4}x^{-3/4} = \frac{1}{4\sqrt[4]{x^3}} \)
Evaluate \( f'(x_0) \):
\( f'(16) = \frac{1}{4\sqrt[4]{16^3}} = \frac{1}{4 \times (\sqrt[4]{16})^3} = \frac{1}{4 \times 2^3} = \frac{1}{4 \times 8} = \frac{1}{32} \)
Now apply the approximation formula:
\( f(15) \approx f(16) + f'(16)(-1) \)
\( f(15) \approx 2 + \frac{1}{32}(-1) \)
\( f(15) \approx 2 - \frac{1}{32} \)
\( f(15) \approx 2 - 0.03125 \)
\( \sqrt[4]{15} \approx 1.96875 \)

(iii) For \( \sqrt[3]{26} \):
Let \( f(x) = x^{1/3} \). We choose \( x_0 = 27 \) because it is close to 26 and \( 27^{1/3} = 3 \).
So, \( \Delta x = 26 - 27 = -1 \).
First, find \( f(x_0) \):
\( f(27) = (27)^{1/3} = 3 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = \frac{1}{3}x^{(1/3)-1} = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \)
Evaluate \( f'(x_0) \):
\( f'(27) = \frac{1}{3\sqrt[3]{27^2}} = \frac{1}{3 \times (\sqrt[3]{27})^2} = \frac{1}{3 \times 3^2} = \frac{1}{27} \)
Now apply the approximation formula:
\( f(26) \approx f(27) + f'(27)(-1) \)
\( f(26) \approx 3 + \frac{1}{27}(-1) \)
\( f(26) \approx 3 - \frac{1}{27} \)
\( f(26) \approx 3 - 0.0370... \)
\( \sqrt[3]{26} \approx 2.963 \)
In simple words: For each number, we picked another very close number that was easy to calculate the root for. Then, we used a special formula with the function's slope at that easy number to guess the value for the original number. This method helps us find approximate values quickly.

🎯 Exam Tip: Always choose \( x_0 \) to be a value near the given number for which \( f(x_0) \) and \( f'(x_0) \) can be easily computed, often a perfect power (square, cube, etc.).

Question 3. Find a linear approximation for the following functions at the indicated points
(i) \( f(x) = x^3 - 5x + 12 \), \( x_0 = 2 \)
(ii) \( g(x) = \sqrt{x^2+9} \), \( x_0 = -4 \)
(iii) \( h(x) = \frac{x}{x+1} \), \( x_0 = 1 \)
Answer: We will find the linear approximation \( L(x) = f(x_0) + f'(x_0)(x - x_0) \) for each function at the given point.

(i) For \( f(x) = x^3 - 5x + 12 \) at \( x_0 = 2 \):
First, find \( f(x_0) \):
\( f(2) = (2)^3 - 5(2) + 12 = 8 - 10 + 12 = 10 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 3x^2 - 5 \)
Evaluate \( f'(x_0) \):
\( f'(2) = 3(2)^2 - 5 = 3(4) - 5 = 12 - 5 = 7 \)
Now, substitute these into the linear approximation formula:
\( L(x) = f(2) + f'(2)(x - 2) \)
\( L(x) = 10 + 7(x - 2) \)
\( L(x) = 10 + 7x - 14 \)
\( L(x) = 7x - 4 \)

(ii) For \( g(x) = \sqrt{x^2+9} \) at \( x_0 = -4 \):
First, find \( g(x_0) \):
\( g(-4) = \sqrt{(-4)^2+9} = \sqrt{16+9} = \sqrt{25} = 5 \)
Next, find the derivative \( g'(x) \):
\( g(x) = (x^2+9)^{1/2} \)
\( g'(x) = \frac{1}{2}(x^2+9)^{-1/2}(2x) = \frac{x}{\sqrt{x^2+9}} \)
Evaluate \( g'(x_0) \):
\( g'(-4) = \frac{-4}{\sqrt{(-4)^2+9}} = \frac{-4}{\sqrt{16+9}} = \frac{-4}{\sqrt{25}} = \frac{-4}{5} \)
Now, substitute these into the linear approximation formula:
\( L(x) = g(-4) + g'(-4)(x - (-4)) \)
\( L(x) = 5 + \left(\frac{-4}{5}\right)(x + 4) \)
\( L(x) = 5 - \frac{4x+16}{5} \)
\( L(x) = \frac{25 - (4x+16)}{5} \)
\( L(x) = \frac{25 - 4x - 16}{5} \)
\( L(x) = \frac{9 - 4x}{5} \)

(iii) For \( h(x) = \frac{x}{x+1} \) at \( x_0 = 1 \):
First, find \( h(x_0) \):
\( h(1) = \frac{1}{1+1} = \frac{1}{2} \)
Next, find the derivative \( h'(x) \) using the quotient rule:
\( h'(x) = \frac{(x+1)(1) - (x)(1)}{(x+1)^2} = \frac{x+1-x}{(x+1)^2} = \frac{1}{(x+1)^2} \)
Evaluate \( h'(x_0) \):
\( h'(1) = \frac{1}{(1+1)^2} = \frac{1}{2^2} = \frac{1}{4} \)
Now, substitute these into the linear approximation formula:
\( L(x) = h(1) + h'(1)(x - 1) \)
\( L(x) = \frac{1}{2} + \frac{1}{4}(x - 1) \)
\( L(x) = \frac{2}{4} + \frac{x - 1}{4} \)
\( L(x) = \frac{2 + x - 1}{4} \)
\( L(x) = \frac{x + 1}{4} \)
In simple words: For each function, we found the equation of the straight line that touches the function's curve at the given point. This line is called the linear approximation. It helps us estimate the function's values for inputs very close to that point.

🎯 Exam Tip: Remember that a linear approximation is essentially the equation of the tangent line to the curve at the given point. Practice finding derivatives correctly, as an error there will affect the entire approximation.

Question 4. The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm find the following in calculating the area of the circular plate:
(i) Absolute error
(ii) Relative error
(iii) Percentage error
Answer:
Given:
Actual radius \( r = 12.5 \) cm
Measured radius \( r_{measured} = 12.65 \) cm
The error in radius is \( dr = r_{measured} - r = 12.65 - 12.5 = 0.15 \) cm.

The area of a circular plate is given by \( A = \pi r^2 \).
The differential of the area is \( dA = \frac{dA}{dr} dr = 2\pi r dr \).
Substitute the actual radius and the error in radius to find the approximate change in area:
\( dA = 2\pi (12.5)(0.15) \)
\( dA = 3.75\pi \)
This value, \( dA = 3.75\pi \), represents the approximate change in area due to the small error in radius measurement.

Now, let's calculate the exact change in area:
Exact change in area \( = A(12.65) - A(12.5) \)
\( = \pi (12.65)^2 - \pi (12.5)^2 \)
\( = \pi (160.0225 - 156.25) \)
\( = \pi (3.7725) \)
\( = 3.7725\pi \)

(i) **Absolute error:**
The absolute error is the difference between the exact change and the approximate change.
Absolute error \( = |3.7725\pi - 3.75\pi| \)
\( = 0.0225\pi \) cm\(^2 \)

(ii) **Relative error:**
Relative error \( = \frac{\text{Absolute error}}{\text{Actual value of dA}} \)
In this context, the "actual value of dA" refers to the exact change in area or the calculated approximate change, depending on the definition used. Following the source calculation, it is taken as the ratio of absolute error to the exact change in area.
Relative error \( = \frac{0.0225\pi}{3.7725\pi} \)
\( = 0.00596 \)

(iii) **Percentage error:**
Percentage error \( = \text{Relative error} \times 100 \)
\( = 0.00596 \times 100 \)
\( = 0.596\% \)
Rounded to one decimal place, this is \( 0.6\% \).
In simple words: When we measure the radius of a circular plate a little bit wrong, it causes mistakes in calculating its area. We found three types of mistakes: absolute error (the actual size of the mistake), relative error (the mistake compared to the true value), and percentage error (the relative error shown as a percentage). Even a small error in radius can lead to a slight error in the area.

🎯 Exam Tip: Clearly differentiate between the actual value, measured value, and the error in measurement (dr). For area calculations, remember that the differential \( dA \) gives the approximate change in area, which is crucial for error analysis.

Question 5. A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:
(i) Change in the volume.
(ii) Change in the surface area
Answer:
Given: Initial radius \( r = 10 \) cm.
Final radius \( r_{final} = 9.8 \) cm.
The change in radius is \( dr = r_{final} - r = 9.8 - 10 = -0.2 \) cm.

(i) **Change in the volume:**
The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).
To find the approximate change in volume, we use its differential \( dV \).
\( dV = \frac{dV}{dr} dr \)
\( dV = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) dr \)
\( dV = 4\pi r^2 dr \)
Substitute the values: \( r = 10 \) cm and \( dr = -0.2 \) cm.
\( dV = 4\pi (10)^2 (-0.2) \)
\( dV = 4\pi (100) (-0.2) \)
\( dV = -80\pi \) cm\(^3 \)
The negative sign indicates a decrease in volume, so the change in volume is approximately \( 80\pi \) cm\(^3 \).

(ii) **Change in the surface area:**
The surface area of a sphere is given by \( S = 4\pi r^2 \).
To find the approximate change in surface area, we use its differential \( dS \).
\( dS = \frac{dS}{dr} dr \)
\( dS = \frac{d}{dr}(4\pi r^2) dr \)
\( dS = 8\pi r dr \)
Substitute the values: \( r = 10 \) cm and \( dr = -0.2 \) cm.
\( dS = 8\pi (10) (-0.2) \)
\( dS = -16\pi \) cm\(^2 \)
The negative sign indicates a decrease in surface area, so the change in surface area is approximately \( 16\pi \) cm\(^2 \). Differential changes help estimate how quantities like volume and surface area respond to small variations in radius.
In simple words: When the ice sphere's radius shrinks a little, both its total space (volume) and its outer covering (surface area) also decrease. We used a special math method called differentials to guess how much these values changed due to the small change in radius.

🎯 Exam Tip: Remember the formulas for volume and surface area of a sphere. When calculating changes, ensure you correctly determine \( dr \) (change in radius) and use the derivative to find \( dV \) and \( dS \).

Question 6. The time T, taken for a complete oscillation of a single pendulum with length I, is \( T = 2\pi\sqrt{\frac{l}{g}} \) where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of I.
Answer:
Given the formula for the time period of a pendulum: \( T = 2\pi\sqrt{\frac{l}{g}} \)
We can rewrite this as: \( T = 2\pi l^{1/2} g^{-1/2} \)
To find the percentage error, it's often helpful to take the logarithm of both sides. This simplifies expressions with products and powers.
Taking the natural logarithm on both sides:
\( \ln T = \ln(2\pi l^{1/2} g^{-1/2}) \)
Using logarithm properties \( \ln(ab) = \ln a + \ln b \) and \( \ln(a^b) = b \ln a \):
\( \ln T = \ln(2\pi) + \ln(l^{1/2}) + \ln(g^{-1/2}) \)
\( \ln T = \ln(2\pi) + \frac{1}{2}\ln l - \frac{1}{2}\ln g \)
Now, differentiate both sides with respect to \( l \) to find the relative change. Since \( g \) and \( 2\pi \) are constants, their derivatives are zero.
\( \frac{1}{T} \frac{dT}{dl} = 0 + \frac{1}{2} \frac{1}{l} - 0 \)
\( \frac{1}{T} \frac{dT}{dl} = \frac{1}{2l} \)
This gives us the relationship between the relative change in T and the relative change in l.
Multiplying both sides by \( \Delta l \), we get the approximate relative change in T:
\( \frac{\Delta T}{T} \approx \frac{1}{2} \frac{\Delta l}{l} \)
To find the percentage error, multiply by 100 on both sides:
\( \frac{\Delta T}{T} \times 100\% = \frac{1}{2} \left(\frac{\Delta l}{l} \times 100\%\right) \)
We are given that the percentage error in the value of \( l \) is 2 percent, which means \( \left(\frac{\Delta l}{l} \times 100\%\right) = 2\% \).
Substitute this value:
\( \frac{\Delta T}{T} \times 100\% = \frac{1}{2} (2\%) \)
\( \frac{\Delta T}{T} \times 100\% = 1\% \)
Therefore, the approximate percentage error in the calculated value of \( T \) is 1%. This method of using logarithms is very efficient for error propagation problems involving powers.
In simple words: The time a pendulum takes to swing depends on its length. If there is a small mistake of 2% in measuring the pendulum's length, we used a math trick with logarithms and derivatives to figure out how much this mistake affects the calculated swing time. It turns out the swing time will have an error of about 1%.

🎯 Exam Tip: For problems involving percentage errors with powers or products, taking logarithms before differentiating is a very effective technique. Remember that \( \frac{\Delta y}{y} \times 100\% \) represents the percentage error in \( y \).

Question 7. Show that the percentage error in the nth root of a number is approximately \( \frac{1}{n} \) times the percentage error in the number.
Answer:
Let \( x \) be the number and \( y \) be its nth root. So, we can write:
\( y = x^{1/n} \)
To relate the percentage errors, we take the natural logarithm of both sides:
\( \ln y = \ln(x^{1/n}) \)
Using the logarithm property \( \ln(a^b) = b \ln a \):
\( \ln y = \frac{1}{n} \ln x \)
Now, differentiate both sides. This gives us the relative change in \( y \) with respect to \( x \).
\( \frac{1}{y} dy = \frac{1}{n} \frac{1}{x} dx \)
This equation shows the relationship between small changes \( dy \) and \( dx \). We can approximate these small changes using \( \Delta y \) and \( \Delta x \).
So, \( \frac{\Delta y}{y} \approx \frac{1}{n} \frac{\Delta x}{x} \)
To express this in terms of percentage error, we multiply both sides by 100:
\( \frac{\Delta y}{y} \times 100\% = \frac{1}{n} \left(\frac{\Delta x}{x} \times 100\%\right) \)
Here, \( \left(\frac{\Delta y}{y} \times 100\%\right) \) is the percentage error in \( y \) (the nth root), and \( \left(\frac{\Delta x}{x} \times 100\%\right) \) is the percentage error in \( x \) (the original number).
Therefore, the percentage error in the nth root of a number is approximately \( \frac{1}{n} \) times the percentage error in the number itself. This formula is useful for understanding how errors propagate through power functions.
In simple words: If you have a number and there's a small mistake in it, and you then find the 'nth root' of that number (like a square root or cube root), the mistake in the root will be smaller. Specifically, the percentage mistake in the root will be the original percentage mistake divided by 'n'. For example, if there's a 10% error in a number, its square root will only have a 5% error.

🎯 Exam Tip: This derivation highlights a key principle of error propagation: fractional and percentage errors are easier to manage using logarithmic differentiation when dealing with powers and roots.