RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives Exercise 8.4

 

Question 1. Approximate the value of \( (0.009)^{1/3} \).
Answer: Let the function be \( y = x^{1/3} \). We choose \( x = 0.008 \) because it's a perfect cube near 0.009, making calculations easier. So, \( \Delta x = 0.009 - 0.008 = 0.001 \). First, find \( y \) for \( x = 0.008 \): \( y = (0.008)^{1/3} = ( (0.2)^3 )^{1/3} = 0.2 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{1/3}) = \frac{1}{3} x^{(1/3) - 1} = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 0.008 \) is \( \frac{1}{3(0.008)^{2/3}} = \frac{1}{3( (0.2)^3 )^{2/3}} = \frac{1}{3(0.2)^2} = \frac{1}{3 \times 0.04} = \frac{1}{0.12} \). So, \( \Delta y = \frac{1}{0.12} \times 0.001 = \frac{0.001}{0.12} \approx 0.0083 \). The approximate value of \( (0.009)^{1/3} \) is \( y + \Delta y \): \( (0.009)^{1/3} \approx 0.2 + 0.0083 = 0.2083 \). Therefore, the approximate value of \( (0.009)^{1/3} \) is \( 0.2083 \).In simple words: We find a number close to 0.009 that is easy to cube root, which is 0.008. We then use a small change formula to find out how much the cube root changes when we go from 0.008 to 0.009. Adding this small change to the cube root of 0.008 gives us our final approximate answer.

🎯 Exam Tip: When using differentials for approximation, always choose an 'x' value that is very close to the given number and allows for easy calculation of the function and its derivative. Ensure \( \Delta x \) is small for accurate approximation.

 

Question 2. Approximate the value of \( (0.999)^{1/10} \).
Answer: Let the function be \( y = x^{1/10} \). We choose \( x = 1 \) because it is a convenient number near 0.999. So, \( \Delta x = 0.999 - 1 = -0.001 \). First, find \( y \) for \( x = 1 \): \( y = (1)^{1/10} = 1 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{1/10}) = \frac{1}{10} x^{(1/10) - 1} = \frac{1}{10} x^{-9/10} = \frac{1}{10x^{9/10}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 1 \) is \( \frac{1}{10(1)^{9/10}} = \frac{1}{10 \times 1} = \frac{1}{10} \). So, \( \Delta y = \frac{1}{10} \times (-0.001) = -0.0001 \). The approximate value of \( (0.999)^{1/10} \) is \( y + \Delta y \): \( (0.999)^{1/10} \approx 1 + (-0.0001) = 0.9999 \). Therefore, the approximate value of \( (0.999)^{1/10} \) is \( 0.9999 \).In simple words: We want to find the 10th root of 0.999. We start with 1, which is easy to find the 10th root of. Then we figure out how much the answer changes when we subtract a very small amount (0.001) from 1, using calculus. Adding this change to our starting answer gives us the approximate value.

🎯 Exam Tip: Remember to handle negative \( \Delta x \) values carefully. The formula \( y + \Delta y \) still applies, meaning you add the (potentially negative) change to the original function value.

 

Question 3. Approximate the value of \( \sqrt{0.0037} \).
Answer: Let the function be \( y = \sqrt{x} \). We choose \( x = 0.0036 \) because it is a perfect square near 0.0037. First, find \( y \) for \( x = 0.0036 \): \( y = \sqrt{0.0036} = 0.06 \). So, \( \Delta x = 0.0037 - 0.0036 = 0.0001 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x}) = \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 0.0036 \) is \( \frac{1}{2\sqrt{0.0036}} = \frac{1}{2 \times 0.06} = \frac{1}{0.12} \). So, \( \Delta y = \frac{1}{0.12} \times 0.0001 = \frac{0.0001}{0.12} \approx 0.000833 \). The approximate value of \( \sqrt{0.0037} \) is \( y + \Delta y \): \( \sqrt{0.0037} \approx 0.06 + 0.000833 = 0.060833 \). Therefore, the approximate value of \( \sqrt{0.0037} \) is \( 0.0608 \).In simple words: To find the square root of 0.0037, we pick a nearby number, 0.0036, which has an easy square root. Then, we use a formula involving calculus to see how much the square root changes for the tiny difference between 0.0036 and 0.0037. Adding this small change to the square root of 0.0036 gives us our estimated answer.

🎯 Exam Tip: When approximating square roots, always look for the nearest perfect square to simplify the initial `y` calculation and derivative evaluation.

 

Question 4. Approximate the value of \( \frac{1}{(2.002)^2} \).
Answer: Let the function be \( y = \frac{1}{x^2} = x^{-2} \). We choose \( x = 2 \) because it is a simple integer near 2.002. So, \( \Delta x = 2.002 - 2 = 0.002 \). First, find \( y \) for \( x = 2 \): \( y = \frac{1}{2^2} = \frac{1}{4} = 0.25 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{-2}) = -2x^{-2-1} = -2x^{-3} = \frac{-2}{x^3} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 2 \) is \( \frac{-2}{(2)^3} = \frac{-2}{8} = -\frac{1}{4} = -0.25 \). So, \( \Delta y = (-0.25) \times 0.002 = -0.0005 \). The approximate value of \( \frac{1}{(2.002)^2} \) is \( y + \Delta y \): \( \frac{1}{(2.002)^2} \approx 0.25 + (-0.0005) = 0.2495 \). Therefore, the approximate value of \( \frac{1}{(2.002)^2} \) is \( 0.2495 \).In simple words: To find the value of 1 divided by 2.002 squared, we use a nearby whole number, 2. We calculate 1 divided by 2 squared, which is 0.25. Then, we use calculus to estimate how much this answer changes because of the small extra bit (0.002) in 2.002. We subtract this small change from 0.25 to get our final estimate.

🎯 Exam Tip: When working with reciprocal functions like \( \frac{1}{x^n} \), remember to use the power rule for differentiation: \( \frac{d}{dx}(x^{-n}) = -nx^{-n-1} \). This avoids errors in the derivative calculation.

 

Question 5. Approximate the value of \( (15)^{1/4} \).
Answer: Let the function be \( y = x^{1/4} \). We choose \( x = 16 \) because it is a perfect fourth power near 15. First, find \( y \) for \( x = 16 \): \( y = (16)^{1/4} = (2^4)^{1/4} = 2 \). So, \( \Delta x = 15 - 16 = -1 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{1/4}) = \frac{1}{4} x^{(1/4) - 1} = \frac{1}{4} x^{-3/4} = \frac{1}{4x^{3/4}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 16 \) is \( \frac{1}{4(16)^{3/4}} = \frac{1}{4( (2^4)^{1/4} )^3} = \frac{1}{4(2)^3} = \frac{1}{4 \times 8} = \frac{1}{32} \). So, \( \Delta y = \frac{1}{32} \times (-1) = -\frac{1}{32} = -0.03125 \). The approximate value of \( (15)^{1/4} \) is \( y + \Delta y \): \( (15)^{1/4} \approx 2 + (-0.03125) = 1.96875 \). Therefore, the approximate value of \( (15)^{1/4} \) is \( 1.9688 \).In simple words: To find the fourth root of 15, we pick 16, which is close to 15 and whose fourth root is easy to find (it's 2). Then we calculate how much the fourth root changes because 15 is slightly less than 16. We subtract this small change from 2 to get our approximate answer.

🎯 Exam Tip: Be careful with the calculations involving fractional exponents, especially when raising a number to a fractional power like \( x^{3/4} \). It often helps to break it down into \( (x^{1/4})^3 \) or \( (x^3)^{1/4} \).

 

Question 6. Approximate the value of \( \sqrt{401} \).
Answer: Let the function be \( y = \sqrt{x} \). We choose \( x = 400 \) because it is a perfect square near 401. First, find \( y \) for \( x = 400 \): \( y = \sqrt{400} = 20 \). So, \( \Delta x = 401 - 400 = 1 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 400 \) is \( \frac{1}{2\sqrt{400}} = \frac{1}{2 \times 20} = \frac{1}{40} \). So, \( \Delta y = \frac{1}{40} \times 1 = \frac{1}{40} = 0.025 \). The approximate value of \( \sqrt{401} \) is \( y + \Delta y \): \( \sqrt{401} \approx 20 + 0.025 = 20.025 \). Therefore, the approximate value of \( \sqrt{401} \) is \( 20.025 \).In simple words: To find the square root of 401, we use 400, which is a perfect square and very close. We find its square root, which is 20. Then we use a small change formula to estimate how much the square root increases when we go from 400 to 401. Adding this small increase to 20 gives us our estimated answer.

🎯 Exam Tip: For square root approximations, always identify the closest perfect square. This makes the initial `y` calculation and the derivative evaluation straightforward, reducing calculation errors.

 

Question 7. Approximate the value of \( (3.968)^{3/2} \).
Answer: Let the function be \( y = x^{3/2} \). We choose \( x = 4 \) because it is a simple integer and easy to work with for fractional powers near 3.968. First, find \( y \) for \( x = 4 \): \( y = (4)^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \). So, \( \Delta x = 3.968 - 4 = -0.032 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{3/2}) = \frac{3}{2} x^{(3/2) - 1} = \frac{3}{2} x^{1/2} = \frac{3}{2} \sqrt{x} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 4 \) is \( \frac{3}{2} \sqrt{4} = \frac{3}{2} \times 2 = 3 \). So, \( \Delta y = 3 \times (-0.032) = -0.096 \). The approximate value of \( (3.968)^{3/2} \) is \( y + \Delta y \): \( (3.968)^{3/2} \approx 8 + (-0.096) = 7.904 \). Therefore, the approximate value of \( (3.968)^{3/2} \) is \( 7.904 \).In simple words: To find 3.968 raised to the power of 3/2, we use 4 as a simple starting point, which is easy to calculate for. We find that 4 to the power of 3/2 is 8. Since 3.968 is a bit less than 4, we use calculus to estimate how much the answer will decrease. Subtracting this small decrease from 8 gives us our approximate answer.

🎯 Exam Tip: When dealing with fractional powers like \( x^{m/n} \), calculate \( (x^{1/n})^m \) or \( (x^m)^{1/n} \). For example, \( 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \), which simplifies the process.

 

Question 8. Approximate the value of \( (32.15)^{1/5} \).
Answer: Let the function be \( y = x^{1/5} \). We choose \( x = 32 \) because it is a perfect fifth power near 32.15. First, find \( y \) for \( x = 32 \): \( y = (32)^{1/5} = (2^5)^{1/5} = 2 \). So, \( \Delta x = 32.15 - 32 = 0.15 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^{1/5}) = \frac{1}{5} x^{(1/5) - 1} = \frac{1}{5} x^{-4/5} = \frac{1}{5x^{4/5}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 32 \) is \( \frac{1}{5(32)^{4/5}} = \frac{1}{5( (2^5)^{1/5} )^4} = \frac{1}{5(2)^4} = \frac{1}{5 \times 16} = \frac{1}{80} \). So, \( \Delta y = \frac{1}{80} \times 0.15 = \frac{0.15}{80} = 0.001875 \). The approximate value of \( (32.15)^{1/5} \) is \( y + \Delta y \): \( (32.15)^{1/5} \approx 2 + 0.001875 = 2.001875 \). Therefore, the approximate value of \( (32.15)^{1/5} \) is \( 2.001875 \).In simple words: To find the fifth root of 32.15, we use 32, which is a number whose fifth root is easy to find (it's 2). Then we calculate how much the fifth root changes because 32.15 is slightly more than 32. We add this small increase to 2 to get our approximate answer.

🎯 Exam Tip: Identifying the correct 'x' (a perfect nth power if you're finding the nth root) is crucial. A good choice simplifies the derivative calculation and minimizes errors.

 

Question 9. Approximate the value of \( \sqrt{0.6} \).
Answer: Let the function be \( y = \sqrt{x} \). We choose \( x = 0.64 \) because it is a perfect square near 0.6. First, find \( y \) for \( x = 0.64 \): \( y = \sqrt{0.64} = 0.8 \). So, \( \Delta x = 0.6 - 0.64 = -0.04 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 0.64 \) is \( \frac{1}{2\sqrt{0.64}} = \frac{1}{2 \times 0.8} = \frac{1}{1.6} \). So, \( \Delta y = \frac{1}{1.6} \times (-0.04) = \frac{-0.04}{1.6} = -0.025 \). The approximate value of \( \sqrt{0.6} \) is \( y + \Delta y \): \( \sqrt{0.6} \approx 0.8 + (-0.025) = 0.775 \). Therefore, the approximate value of \( \sqrt{0.6} \) is \( 0.775 \).In simple words: To find the square root of 0.6, we use 0.64, which is a close number with an easy square root (0.8). Since 0.6 is less than 0.64, we calculate how much the square root decreases due to this small difference. We subtract this small change from 0.8 to get our approximate answer.

🎯 Exam Tip: When dealing with decimals, quickly identify the nearest perfect square that makes calculations straightforward. Remember that \( \sqrt{0.ab} = 0.\sqrt{ab} \) is not true; instead, consider it as \( \sqrt{\frac{ab}{100}} = \frac{\sqrt{ab}}{10} \).

 

Question 10. Approximate the value of \( \log_{10} (10.1) \), when \( \log_{10} e = 0.4343 \).
Answer: Let the function be \( y = \log_{10} x \). We know that \( \log_{10} x = \log_{10} e \cdot \log_e x \). So, \( y = (\log_{10} e) \cdot \log_e x \). We choose \( x = 10 \) because it is a simple integer near 10.1. First, find \( y \) for \( x = 10 \): \( y = \log_{10} 10 = 1 \). So, \( \Delta x = 10.1 - 10 = 0.1 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (\log_{10} x) = \frac{d}{dx} ((\log_{10} e) \cdot \log_e x) = (\log_{10} e) \cdot \frac{1}{x} \) Given \( \log_{10} e = 0.4343 \). So, \( \frac{dy}{dx} = \frac{0.4343}{x} \). Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 10 \) is \( \frac{0.4343}{10} = 0.04343 \). So, \( \Delta y = 0.04343 \times 0.1 = 0.004343 \). The approximate value of \( \log_{10} (10.1) \) is \( y + \Delta y \): \( \log_{10} (10.1) \approx 1 + 0.004343 = 1.004343 \). Therefore, the approximate value of \( \log_{10} (10.1) \) is \( 1.004343 \). This shows how logarithms change in small steps.In simple words: To find the log base 10 of 10.1, we start with the log base 10 of 10, which is simply 1. Then we use a special calculus method to figure out how much the log value increases when we go from 10 to 10.1. Adding this small increase to 1 gives us our estimated answer.

🎯 Exam Tip: Remember the change of base formula for logarithms, \( \log_b a = \frac{\log_c a}{\log_c b} \), or specifically, \( \log_{10} x = \log_{10} e \cdot \log_e x \) when differentiating `log_10 x`.

 

Question 11. Approximate the value of \( \log_e (10.02) \), when \( \log_e 10 = 2.3026 \).
Answer: Let the function be \( y = \log_e x \). We choose \( x = 10 \) because it is a simple integer near 10.02. First, find \( y \) for \( x = 10 \): \( y = \log_e 10 = 2.3026 \) (given). So, \( \Delta x = 10.02 - 10 = 0.02 \). Next, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (\log_e x) = \frac{1}{x} \) Now, calculate \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): \( \frac{dy}{dx} \) at \( x = 10 \) is \( \frac{1}{10} = 0.1 \). So, \( \Delta y = 0.1 \times 0.02 = 0.002 \). The approximate value of \( \log_e (10.02) \) is \( y + \Delta y \): \( \log_e (10.02) \approx 2.3026 + 0.002 = 2.3046 \). Therefore, the approximate value of \( \log_e (10.02) \) is \( 2.3046 \).In simple words: To find the natural logarithm (log base e) of 10.02, we start with the natural log of 10, which is given as 2.3026. Then we use calculus to calculate how much this log value increases because 10.02 is slightly more than 10. Adding this small increase to our starting value gives us the approximate answer.

🎯 Exam Tip: For natural logarithms (`log_e x`), the derivative is simply `1/x`, which simplifies calculations. Always ensure you are using the correct base for the given logarithm function.

 

Question 12. If \( y = x^2 + 4 \) and \( x \) changes from 3 to 3.1, then use differential to approximate change in \( y \).
Answer: Given the function \( y = x^2 + 4 \). The initial value of \( x \) is \( x = 3 \). The change in \( x \) is \( \Delta x = 3.1 - 3 = 0.1 \). First, find the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx} (x^2 + 4) = 2x \) Now, calculate the approximate change in \( y \), which is \( \Delta y \approx \frac{dy}{dx} \cdot \Delta x \): Evaluate \( \frac{dy}{dx} \) at \( x = 3 \): \( 2 \times 3 = 6 \). So, \( \Delta y = 6 \times 0.1 = 0.6 \). Therefore, the approximate change in \( y \) is \( 0.6 \). This method gives a good estimate for small changes.In simple words: When \( x \) changes a little bit, we want to know how much \( y \) changes in the function \( y = x^2 + 4 \). We use a special formula that involves the derivative of the function, which tells us the rate of change. We multiply this rate by the small change in \( x \) to find the approximate change in \( y \).

🎯 Exam Tip: Remember that differential \( dy \) approximates the actual change \( \Delta y \). This approximation is more accurate for smaller values of \( \Delta x \). Always calculate \( \frac{dy}{dx} \) at the initial \( x \) value.

 

Question 13. Show that the relative error in computing the volume of a cubical box, due to an error in measuring the edge, is approximately equal to the three times the relative error in the edge.
Answer: Let the length of the edge of the cubical box be \( x \). The volume of the cubical box is given by \( V = x^3 \). First, find the derivative of \( V \) with respect to \( x \): \( \frac{dV}{dx} = \frac{d}{dx} (x^3) = 3x^2 \) The approximate change in volume, \( \Delta V \), due to a small error in measuring the edge, \( \Delta x \), is given by \( \Delta V \approx \frac{dV}{dx} \cdot \Delta x \). So, \( \Delta V = 3x^2 \cdot \Delta x \). Now, let's find the relative error in volume, which is \( \frac{\Delta V}{V} \): \( \frac{\Delta V}{V} = \frac{3x^2 \cdot \Delta x}{x^3} \) We can simplify this expression: \( \frac{\Delta V}{V} = \frac{3 \cdot \Delta x}{x} \) The term \( \frac{\Delta x}{x} \) represents the relative error in the measurement of the edge. Thus, \( \frac{\Delta V}{V} = 3 \left( \frac{\Delta x}{x} \right) \). This shows that the relative error in computing the volume is approximately three times the relative error in measuring the edge. This principle is widely used in physics and engineering to understand measurement uncertainties.In simple words: If you make a small mistake when measuring the side of a cube, that mistake will cause a much bigger mistake when you calculate the cube's volume. Specifically, the percentage error in the volume will be about three times the percentage error in the side measurement.

🎯 Exam Tip: Relative error is defined as \( \frac{\Delta \text{Quantity}}{\text{Original Quantity}} \). This question requires differentiating the volume formula and then forming the ratio to show the relationship between relative errors.

 

Question 14. The radius of a sphere shrinks from 10 cm to 9.8 cm. Find the approximate decrease in its volume.
Answer: Let the radius of the sphere be \( r \). The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). The original radius is \( r = 10 \text{ cm} \). The change in radius is \( \Delta r = 9.8 - 10 = -0.2 \text{ cm} \). First, find the derivative of \( V \) 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 approximate change in volume, \( \Delta V \), due to a small change in radius, \( \Delta r \), is given by \( \Delta V \approx \frac{dV}{dr} \cdot \Delta r \). Evaluate \( \frac{dV}{dr} \) at \( r = 10 \text{ cm} \): \( 4 \pi (10)^2 = 4 \pi \times 100 = 400 \pi \text{ cm}^2 \). So, \( \Delta V = (400 \pi) \times (-0.2) = -80 \pi \text{ cm}^3 \). The negative sign indicates a decrease in volume. Therefore, the approximate decrease in its volume is \( 80 \pi \text{ cm}^3 \). This calculation helps predict volume changes from small radius adjustments.In simple words: When the radius of a ball gets a bit smaller, its volume also decreases. We can calculate this approximate decrease in volume by using the derivative of the volume formula and multiplying it by the small change in the radius.

🎯 Exam Tip: Pay close attention to the units (cm, cm\(^3\)) and the sign of \( \Delta r \). A negative \( \Delta r \) indicates a decrease, leading to a negative \( \Delta V \) which signifies a decrease in volume.

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