Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.2

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.2

Question 1. Find the differential dy for each of the following functions.
(i) \( y = \frac { (1-2x)^3 }{ 3-4x } \)
(ii) \( y = (3 + \sin 2x)^{2/3} \)
(iii) \( y = e^{x^2 – 5x +7} \cos(x^2 – 1) \)
Answer:
(i) Given, \( y = \frac { (1-2x)^3 }{ 3-4x } \). We use the quotient rule for differentiation here.
\( \frac{dy}{dx} = \frac{(3-4x) \cdot 3(1-2x)^2 \cdot (-2) - (1-2x)^3 \cdot (-4)}{(3-4x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1-2x)^2 [ -6(3-4x) + 4(1-2x) ]}{(3-4x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1-2x)^2 [ -18 + 24x + 4 - 8x ]}{(3-4x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1-2x)^2 [ 16x - 14 ]}{(3-4x)^2} \)
\( \implies \frac{dy}{dx} = \frac{2(1-2x)^2 (8x - 7)}{(3-4x)^2} \)
So, the differential is \( dy = \frac{2(1-2x)^2 (8x - 7)}{(3-4x)^2} dx \). This method helps us find the rate of change of y with respect to x.
(ii) Given, \( y = (3 + \sin 2x)^{2/3} \). We apply the chain rule for this function.
\( \frac{dy}{dx} = \frac{2}{3} (3 + \sin 2x)^{(2/3) - 1} \cdot (\cos 2x) \cdot 2 \)
\( \implies \frac{dy}{dx} = \frac{4}{3} (3 + \sin 2x)^{-1/3} \cos 2x \)
\( \implies \frac{dy}{dx} = \frac{4 \cos 2x}{3 (3 + \sin 2x)^{1/3}} \)
Therefore, the differential is \( dy = \frac{4 \cos 2x}{3 (3 + \sin 2x)^{1/3}} dx \). The chain rule is essential when you have a function inside another function.
(iii) Given, \( y = e^{x^2 – 5x +7} \cos(x^2 – 1) \). This involves the product rule and chain rule.
\( \frac{dy}{dx} = e^{x^2 – 5x +7} \frac{d}{dx}(\cos(x^2 – 1)) + \cos(x^2 – 1) \frac{d}{dx}(e^{x^2 – 5x +7}) \)
\( \implies \frac{dy}{dx} = e^{x^2 – 5x +7} (-\sin(x^2 – 1) \cdot 2x) + \cos(x^2 – 1) (e^{x^2 – 5x +7} \cdot (2x - 5)) \)
\( \implies \frac{dy}{dx} = e^{x^2 – 5x +7} [ -2x \sin(x^2 – 1) + (2x - 5) \cos(x^2 – 1) ] \)
Thus, the differential is \( dy = e^{x^2 – 5x +7} [ -2x \sin(x^2 – 1) + (2x - 5) \cos(x^2 – 1) ] dx \). Differentiating composite functions properly is key to accuracy.
In simple words: For each function, we found its derivative using rules like the quotient rule, chain rule, and product rule. The differential \(dy\) is then found by multiplying the derivative by \(dx\). It represents a small change in \(y\) for a small change in \(x\).

🎯 Exam Tip: Remember to apply the correct differentiation rules (product, quotient, chain) for each part of the function. Pay close attention to negative signs and exponents.

Question 2. Find df for \( f(x) = x^2 + 3x \) and evaluate it for
(i) \( x = 2 \) and \( dx = 0.1 \)
(ii) \( x = 3 \) and \( dx = 0.02 \)
Answer:
Given the function \( f(x) = x^2 + 3x \). We first find the differential \( df \).
\( df = f'(x) dx \)
First, find the derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}(x^2 + 3x) = 2x + 3 \)
So, \( df = (2x + 3) dx \). This equation helps us estimate the change in the function's value.
(i) Now, we evaluate \( df \) for \( x = 2 \) and \( dx = 0.1 \):
\( df = (2(2) + 3) (0.1) \)
\( \implies df = (4 + 3) (0.1) \)
\( \implies df = 7 (0.1) \)
\( \implies df = 0.7 \)
(ii) Next, we evaluate \( df \) for \( x = 3 \) and \( dx = 0.02 \):
\( df = (2(3) + 3) (0.02) \)
\( \implies df = (6 + 3) (0.02) \)
\( \implies df = 9 (0.02) \)
\( \implies df = 0.18 \)
In simple words: First, we find the derivative of the function. Then, we multiply it by the small change in \(x\) (which is \(dx\)) to get the approximate change in the function, \(df\). We did this for two different sets of values.

🎯 Exam Tip: Remember that \( df = f'(x) dx \). Be careful with the substitution of \(x\) and \(dx\) values, especially with decimals.

Question 3. Find \( \Delta f \) and \( df \) for the function \( f \) for the indicated values of \( x, \Delta x \) and compare:
(i) \( f(x) = x^3 – 2x^2, x = 2, \Delta x = dx = 0.5 \)
(ii) \( f(x) = x^2 + 2x + 3, x = -0.5, \Delta x = dx = 0.1 \)
Answer:
(i) Given function: \( y = f(x) = x^3 – 2x^2 \). We need to find both the exact change \( \Delta f \) and the approximate change \( df \).
First, let's find \( df \):
\( df = f'(x) dx \)
\( f'(x) = \frac{d}{dx}(x^3 – 2x^2) = 3x^2 – 4x \)
So, \( df = (3x^2 – 4x) dx \). This is the formula for the differential.
Evaluate \( df \) when \( x = 2 \) and \( dx = 0.5 \):
\( df = (3(2)^2 – 4(2)) (0.5) \)
\( \implies df = (3 \cdot 4 – 8) (0.5) \)
\( \implies df = (12 – 8) (0.5) \)
\( \implies df = 4 (0.5) \)
\( \implies df = 2 \)
So, the approximate change is \( df = 2 \).
Next, let's find \( \Delta f \):
\( \Delta f = f(x + \Delta x) – f(x) \)
Here, \( x = 2 \) and \( \Delta x = 0.5 \). So, \( x + \Delta x = 2 + 0.5 = 2.5 \).
\( f(x + \Delta x) = f(2.5) = (2.5)^3 – 2(2.5)^2 \)
\( \implies f(2.5) = 15.625 – 2(6.25) \)
\( \implies f(2.5) = 15.625 – 12.5 \)
\( \implies f(2.5) = 3.125 \)
And \( f(x) = f(2) = (2)^3 – 2(2)^2 \)
\( \implies f(2) = 8 – 2(4) \)
\( \implies f(2) = 8 – 8 \)
\( \implies f(2) = 0 \)
So, \( \Delta f = 3.125 – 0 = 3.125 \). Comparing \( df = 2 \) and \( \Delta f = 3.125 \), we see they are close but not exactly the same, as expected for an approximation.
(ii) Given function: \( y = f(x) = x^2 + 2x + 3 \).
First, let's find \( df \):
\( df = f'(x) dx \)
\( f'(x) = \frac{d}{dx}(x^2 + 2x + 3) = 2x + 2 \)
So, \( df = (2x + 2) dx \).
Evaluate \( df \) when \( x = -0.5 \) and \( dx = 0.1 \):
\( df = (2(-0.5) + 2) (0.1) \)
\( \implies df = (-1 + 2) (0.1) \)
\( \implies df = 1 (0.1) \)
\( \implies df = 0.1 \)
So, the approximate change is \( df = 0.1 \).
Next, let's find \( \Delta f \):
\( \Delta f = f(x + \Delta x) – f(x) \)
Here, \( x = -0.5 \) and \( \Delta x = 0.1 \). So, \( x + \Delta x = -0.5 + 0.1 = -0.4 \).
\( f(x + \Delta x) = f(-0.4) = (-0.4)^2 + 2(-0.4) + 3 \)
\( \implies f(-0.4) = 0.16 – 0.8 + 3 \)
\( \implies f(-0.4) = 2.36 \)
And \( f(x) = f(-0.5) = (-0.5)^2 + 2(-0.5) + 3 \)
\( \implies f(-0.5) = 0.25 – 1 + 3 \)
\( \implies f(-0.5) = 2.25 \)
So, \( \Delta f = 2.36 – 2.25 = 0.11 \). Comparing \( df = 0.1 \) and \( \Delta f = 0.11 \), they are very close, showing that the differential provides a good approximation for small changes.
In simple words: We calculated two things for each function: \(df\) (the approximate change) and \( \Delta f \) (the actual change). \(df\) is found using the derivative, while \( \Delta f \) is found by calculating the function's value at the new point minus its value at the old point. We then compared these two values, which are usually close for small changes.

🎯 Exam Tip: Clearly distinguish between \( df \) (differential, an approximation) and \( \Delta f \) (actual change). \( \Delta f \) needs two function evaluations, while \( df \) uses the derivative at the initial point.

Question 4. Assuming \( \log_{10} e = 0.4343 \), find an approximate value of \( \log_{10} 1003 \).
Answer:
Let \( f(x) = \log_{10} x \). We want to find \( f(1003) \). We know \( f(1000) = \log_{10} 1000 = 3 \).
We can use the differential approximation: \( \Delta f \approx df = f'(x) \Delta x \).
First, find the derivative \( f'(x) \):
\( f(x) = \log_{10} x = \frac{\ln x}{\ln 10} \)
\( f'(x) = \frac{1}{x \ln 10} \). We know \( \ln 10 = \frac{1}{\log_{10} e} \).
So, \( f'(x) = \frac{1}{x} \log_{10} e \). This allows us to use the given value directly.
Here, let \( x = 1000 \) and \( \Delta x = 3 \) (since \( 1003 = 1000 + 3 \)).
\( \Delta f = f(1003) – f(1000) \)
\( df = f'(1000) \cdot \Delta x = \frac{1}{1000} \log_{10} e \cdot 3 \)
Substitute the given value \( \log_{10} e = 0.4343 \):
\( df = \frac{0.4343}{1000} \cdot 3 \)
\( \implies df = 0.0004343 \cdot 3 \)
\( \implies df = 0.0013029 \)
Now, \( f(1003) = f(1000) + df \)
\( \implies \log_{10} 1003 = \log_{10} 1000 + 0.0013029 \)
\( \implies \log_{10} 1003 = 3 + 0.0013029 \)
\( \implies \log_{10} 1003 = 3.0013029 \)
Therefore, the approximate value of \( \log_{10} 1003 \) is \( 3.0013029 \). Using known values and differentials makes calculations simpler.
In simple words: We want to find the value of \( \log_{10} 1003 \). We know the value of \( \log_{10} 1000 \) is 3. We use the idea that for a small change, the approximate change in the function is its derivative multiplied by the small change. We calculate this small change and add it to 3 to get our answer.

🎯 Exam Tip: When approximating function values, choose a nearby value for \(x\) where the function is easy to calculate. Remember the formula \( f(x + \Delta x) \approx f(x) + f'(x) \Delta x \).

Question 5. The trunk of a tree has a diameter of 30 cm. During the following year, the circumference grew 6 cm.
(i) How much did the tree diameter grow?
(ii) What is the percentage increase in the area of the cross-section of the tree?
Answer:
Given: Initial diameter \( D = 30 \) cm.
Change in circumference \( dS = 6 \) cm.
We know the circumference of a circle is \( S = \pi D \).
To find the change in diameter, we differentiate \( S \) with respect to \( D \): \( dS = \pi dD \). This shows how circumference and diameter changes relate.
(i) We are given \( dS = 6 \) cm.
So, \( 6 = \pi dD \).
Therefore, the increase in diameter \( dD = \frac{6}{\pi} \) cm.
(ii) We need to find the percentage increase in the area of the cross-section.
The area of a circular cross-section is \( A = \pi r^2 \). Since \( r = \frac{D}{2} \), we have \( A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \).
To find the change in area \( dA \), we differentiate \( A \) with respect to \( D \):
\( dA = \frac{\pi}{4} (2D) dD \)
\( \implies dA = \frac{\pi D}{2} dD \). This formula helps us estimate area change from diameter change.
Substitute the known values: \( D = 30 \) cm and \( dD = \frac{6}{\pi} \) cm.
\( dA = \frac{\pi (30)}{2} \left(\frac{6}{\pi}\right) \)
\( \implies dA = 15 \cdot 6 \)
\( \implies dA = 90 \) cm\(^2 \).
Percentage increase in area \( = \frac{dA}{A} \times 100\% \).
First, calculate the initial area \( A \):
\( A = \frac{\pi (30)^2}{4} = \frac{900\pi}{4} = 225\pi \) cm\(^2 \).
Now, calculate the percentage increase:
Percentage increase \( = \frac{90}{225\pi} \times 100 \)
\( = \frac{2}{5\pi} \times 100 \)
\( = \frac{200}{5\pi} \)
\( = \frac{40}{\pi} \% \). This tells us the relative increase in the tree's cross-section.
In simple words: The tree's circumference grew by 6 cm. Using the formula for circumference, we found how much the diameter grew. Then, using the formula for the area of a circle, we found the approximate increase in the cross-sectional area and calculated what percentage this increase was compared to the original area.

🎯 Exam Tip: Remember the basic formulas for circumference (\( S = \pi D \)) and area (\( A = \frac{\pi D^2}{4} \)) of a circle. Differentiation helps relate changes in these quantities.

Question 6. An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is 5 mm and the radius to the outside of the shell is 5.3 mm, find the volume of the shell approximately.
Answer:
The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \).
We need to find the approximate volume of the shell. This can be estimated using differentials.
Let \( r \) be the inside radius and \( dr \) be the thickness of the shell.
Given: inside radius \( r = 5 \) mm.
Outside radius \( = 5.3 \) mm.
So, the thickness \( dr = \text{outside radius} - \text{inside radius} = 5.3 - 5 = 0.3 \) mm. This is the change in radius.
The approximate change in volume \( dV \) is given by the differential of \( V \):
\( dV = \frac{d}{dr}\left(\frac{4}{3} \pi r^3\right) dr \)
\( \implies dV = \frac{4}{3} \pi (3r^2) dr \)
\( \implies dV = 4 \pi r^2 dr \). This formula gives the approximate volume of a thin shell.
Substitute \( r = 5 \) mm and \( dr = 0.3 \) mm:
\( dV = 4 \pi (5)^2 (0.3) \)
\( \implies dV = 4 \pi (25) (0.3) \)
\( \implies dV = 100 \pi (0.3) \)
\( \implies dV = 30 \pi \) mm\(^3 \).
Therefore, the approximate volume of the shell is \( 30 \pi \) mm\(^3 \). Differentials provide a quick estimate for such problems.
In simple words: An egg shell's volume is like a thin layer. We use the formula for the volume of a sphere and its derivative to find the approximate volume of this thin shell. We take the inside radius as \(r\) and the thickness as \(dr\), then calculate \(4 \pi r^2 dr\).

🎯 Exam Tip: For approximate shell volumes or changes, always use the differential \( dV = \frac{dV}{dr} dr \). Make sure to correctly identify \( r \) (the inner radius) and \( dr \) (the thickness).

Question 7. Assume that the cross-section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from 2 mm to 2.1 mm, how much is the cross-sectional area increased approximately?
Answer:
The cross-section of an artery is circular. The area of a circle is \( A = \pi r^2 \).
Given: Initial radius \( r = 2 \) mm.
New radius \( = 2.1 \) mm.
The change in radius is \( dr = 2.1 - 2 = 0.1 \) mm. This is the small change we consider.
To find the approximate increase in area, we use the differential \( dA \).
\( dA = \frac{dA}{dr} dr \)
First, find the derivative of \( A \) with respect to \( r \):
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2 \pi r \). This represents the rate at which area changes with radius.
Now, substitute the values into the differential formula:
\( dA = (2 \pi r) dr \)
\( \implies dA = 2 \pi (2) (0.1) \)
\( \implies dA = 4 \pi (0.1) \)
\( \implies dA = 0.4 \pi \) mm\(^2 \).
Therefore, the approximate increase in the cross-sectional area is \( 0.4 \pi \) mm\(^2 \). This small increase can significantly affect blood flow.
In simple words: The artery's area is a circle. When its radius gets a little bigger, we can find the approximate increase in its area using a simple calculation. We multiply the derivative of the area formula by the small change in the radius.

🎯 Exam Tip: Clearly identify the initial radius \( r \) and the change in radius \( dr \). Use the formula \( dA = 2 \pi r dr \) for the approximate change in the area of a circle.

Question 8. In a newly developed city, it is estimated that the voting population (in thousands) will increase according to \( V(t) =30 + 12t^2 – t^3 \), \( 0 \leq t \leq 8 \) where \( t \) is the time in years. Find the approximate change in voters for the time change from 4 to \( 4 \frac{1}{6} \) years.
Answer:
Given the voting population function \( V(t) = 30 + 12t^2 – t^3 \). The time \( t \) is in years.
We need to find the approximate change in voters for a time change from \( t = 4 \) years to \( t = 4 \frac{1}{6} \) years.
Here, the initial time is \( t = 4 \).
The change in time is \( dt = 4 \frac{1}{6} - 4 = \frac{1}{6} \) years. This represents our \( \Delta t \).
The approximate change in voters is \( dV = V'(t) dt \).
First, find the derivative \( V'(t) \):
\( V'(t) = \frac{d}{dt}(30 + 12t^2 – t^3) = 24t – 3t^2 \). This tells us the rate of change of voters.
Now, substitute \( t = 4 \) and \( dt = \frac{1}{6} \) into the \( dV \) formula:
\( dV = (24(4) – 3(4)^2) \left(\frac{1}{6}\right) \)
\( \implies dV = (96 – 3(16)) \left(\frac{1}{6}\right) \)
\( \implies dV = (96 – 48) \left(\frac{1}{6}\right) \)
\( \implies dV = 48 \left(\frac{1}{6}\right) \)
\( \implies dV = 8 \).
Since \( V(t) \) is in thousands, the approximate change in voters is \( 8 \times 1000 = 8000 \). This shows how differentials can predict population shifts.
In simple words: We have a formula for how the voting population changes over time. To find the approximate change in voters for a small period, we first find the rate at which voters change (the derivative). Then, we multiply this rate by the small change in time. Since the original population was given in thousands, we multiply our answer by 1000.

🎯 Exam Tip: Remember to express the fraction \( 4 \frac{1}{6} \) correctly as a change in time. Ensure the units (thousands in this case) are applied correctly at the end of the calculation.

Question 9. The relation between the number of words y a person learns in x hours is given by \( y = 52\sqrt{x}, 0 \leq x \leq 9 \). What is the approximate number of words learned when x changes from
(i) 1 to 1.1 hours?
(ii) 4 to 4.1 hours?
Answer:
Given the function \( y = 52\sqrt{x} \). This function describes how many words a person learns.
The approximate number of words learned is given by the differential \( dy = y'(x) dx \).
First, find the derivative \( y'(x) \):
\( y'(x) = \frac{d}{dx}(52x^{1/2}) = 52 \cdot \frac{1}{2} x^{(1/2) - 1} \)
\( \implies y'(x) = 26x^{-1/2} = \frac{26}{\sqrt{x}} \). This gives the rate of learning.
So, \( dy = \frac{26}{\sqrt{x}} dx \).
(i) When \( x \) changes from 1 to 1.1 hours:
Here, \( x = 1 \) and \( dx = 1.1 - 1 = 0.1 \).
\( dy = \frac{26}{\sqrt{1}} (0.1) \)
\( \implies dy = 26 (0.1) \)
\( \implies dy = 2.6 \)
Since the number of words must be an integer, we round this to 3 words. This indicates a small but measurable increase in words learned.
(ii) When \( x \) changes from 4 to 4.1 hours:
Here, \( x = 4 \) and \( dx = 4.1 - 4 = 0.1 \).
\( dy = \frac{26}{\sqrt{4}} (0.1) \)
\( \implies dy = \frac{26}{2} (0.1) \)
\( \implies dy = 13 (0.1) \)
\( \implies dy = 1.3 \)
Rounding to the nearest whole number, this is approximately 1 word. As study time increases, the rate of learning new words might slow down.
In simple words: We used a formula that tells us how many words are learned over time. For a small increase in study time, we found the rate of learning and multiplied it by the extra time. This gave us the approximate number of extra words learned, which we then rounded to a whole number.

🎯 Exam Tip: For problems involving counting discrete items like "words learned," it's often appropriate to round the final differential value to the nearest integer. Remember to correctly handle fractional exponents in differentiation.

Question 10. A circular plate expands uniformly under the influence of heat. If its radius increases from 10.5 cm to 10.75 cm, then find an approximate change in the area and the approximate percentage change in the area.
Answer:
The area of a circular plate is \( A = \pi r^2 \).
Given: Initial radius \( r = 10.5 \) cm.
New radius \( = 10.75 \) cm.
The change in radius \( dr = 10.75 - 10.5 = 0.25 \) cm. This is the small change in the radius.
First, find the approximate change in the area \( dA \).
\( dA = \frac{dA}{dr} dr \)
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2 \pi r \). This shows how the area expands as the radius increases.
Substitute \( r = 10.5 \) and \( dr = 0.25 \):
\( dA = 2 \pi (10.5) (0.25) \)
\( \implies dA = 21 \pi (0.25) \)
\( \implies dA = 5.25 \pi \) cm\(^2 \). This is the approximate increase in area.
Next, find the approximate percentage change in the area.
Percentage change in Area \( = \frac{dA}{A} \times 100\% \).
First, calculate the initial area \( A \):
\( A = \pi (10.5)^2 \)
\( \implies A = \pi (110.25) \)
\( \implies A = 110.25 \pi \) cm\(^2 \).
Now, calculate the percentage change:
Percentage change \( = \frac{5.25 \pi}{110.25 \pi} \times 100 \)
\( \implies \) Percentage change \( = \frac{5.25}{110.25} \times 100 \)
\( \implies \) Percentage change \( = 0.047619 \times 100 \)
\( \implies \) Percentage change \( = 4.7619 \% \). Rounding to two decimal places, this is approximately \( 4.76\% \). This indicates a significant expansion due to heat.
In simple words: A circular plate got warmer, and its radius grew a little. We found the approximate increase in its area by using calculus. Then, we calculated what percentage this area increase was compared to the original area of the plate.

🎯 Exam Tip: When calculating percentage change, always divide the change in quantity (\(dA\)) by the original quantity (\(A\)), then multiply by 100. Make sure to use the initial radius for calculating the original area \( A \).

Question 11. A coat of paint of thickness 0.2 cm is applied to the faces of cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint on this cube. Also calculate the exact amount of paint used to paint this cube.
Answer:
Let \( V \) be the volume of the cube and \( a \) be its edge length.
The volume of a cube is \( V = a^3 \).
Given: Initial edge length \( a = 10 \) cm.
Thickness of paint \( da = 0.2 \) cm. This represents the small change in the edge length.
First, find the approximate amount of paint using differentials:
\( dV = \frac{dV}{da} da \)
\( \frac{dV}{da} = \frac{d}{da}(a^3) = 3a^2 \). This is the rate at which volume changes with edge length.
Substitute \( a = 10 \) and \( da = 0.2 \):
\( dV = 3(10)^2 (0.2) \)
\( \implies dV = 3(100) (0.2) \)
\( \implies dV = 300 (0.2) \)
\( \implies dV = 60 \) cm\(^3 \).
So, the approximate amount of paint used is \( 60 \) cm\(^3 \).
Next, calculate the exact amount of paint used.
The paint adds to the thickness of the cube's edge. So, the new edge length \( a + \Delta a = 10 + 2(0.2) = 10 + 0.4 = 10.4 \) cm. (Note: paint is applied to *faces*, so it adds to the length on both sides of each original edge dimension, effectively increasing the overall cube's dimensions).
No, the paint is applied to the faces, so it increases the edge length on each side. The thickness of paint is \(0.2\) cm. So a 10cm cube becomes a \(10 + 2 \times 0.2 = 10.4\) cm cube. The exact volume of the painted cube is \( V_{new} = (10.4)^3 \).
\( V_{new} = 1124.864 \) cm\(^3 \).
The original volume of the cube is \( V_{original} = (10)^3 = 1000 \) cm\(^3 \).
The exact amount of paint used is \( V_{new} - V_{original} \):
Exact paint volume \( = 1124.864 - 1000 = 124.864 \) cm\(^3 \).
The differential \(dV = 60\) cm\(^3\) is an approximation, but the exact value is much larger. This difference arises because \( da \) is relatively large compared to the edge, making the approximation less accurate.
In simple words: We found the approximate amount of paint needed by using the derivative of the cube's volume. Then, we found the exact amount by calculating the volume of the cube with paint and subtracting the volume of the cube without paint. The exact amount was much higher, showing that approximation works best for very small changes.

🎯 Exam Tip: For paint thickness problems on a cube, remember that the total increase in edge length is twice the paint thickness (\(2 \times da\)) because paint is applied to both sides. Also, note that differentials provide approximations, and the accuracy decreases as the change (\(da\)) becomes larger.