RD Sharma Class 12 Solutions Chapter 14 - Differentials, Errors and Approximations (Ex 14.1) Exercise 14.1 - Free PDF
FAQs on RD Sharma Class 12 Solutions Chapter 14 - Differentials, Errors and Approximations (Ex 14.1) Exercise 14.1
1. How do you find the approximate value of (82)^(1/4) using the method from RD Sharma Chapter 14?
To find the approximate value of (82)^(1/4), we define the function as y = f(x) = x^(1/4). We choose a value of x whose fourth root is known and is close to 82. So, we let x = 81 and the change Δx = 1. The approximation formula is f(x + Δx) ≈ f(x) + dy. First, calculate f(x) = (81)^(1/4) = 3. Next, find the derivative f'(x) = (1/4)x^(-3/4). Then, calculate the differential dy = f'(x)Δx, which is [(1/4)(81)^(-3/4)] * 1. This simplifies to (1/4) * (1/27) = 1/108 ≈ 0.0092. Finally, the approximate value is f(82) ≈ f(81) + dy = 3 + 0.0092 = 3.0092.
2. What is the step-by-step process to calculate the approximate error in the volume of a sphere, as per the problems in Exercise 14.1?
To calculate the approximate error in the volume of a sphere when there is an error in measuring its radius, you should follow these steps as per the CBSE 2025-26 syllabus guidelines:
First, state the formula for the volume of a sphere: V = (4/3)πr³.
Differentiate the volume V with respect to the radius r to find the rate of change: dV/dr = 4πr².
The approximate error in volume, ΔV, is represented by the differential dV.
Use the formula for the differential of volume: dV = (dV/dr) * Δr, where Δr is the given error in the radius.
Substitute the known values of r and Δr into the equation dV = (4πr²)Δr to find the final approximate error in the volume.
3. What is the key difference between the actual change Δy and the approximate change dy when solving problems in this chapter?
The key difference is that Δy represents the exact change in a function's value, calculated as Δy = f(x + Δx) - f(x). In contrast, dy represents the approximate change derived from the tangent line at point x, calculated as dy = f'(x)Δx. While Δy measures the change along the actual curve of the function, dy measures the corresponding change along the tangent line. For very small values of Δx, the tangent line is a very close approximation of the curve, making dy a reliable estimate for Δy.
4. How are relative error and percentage error calculated for a quantity using differentials?
To find the relative and percentage error using the methods in this chapter, you follow this correct procedure:
Let y = f(x) be the given quantity. The absolute error is Δy, which is approximated by the differential dy = f'(x)Δx.
The relative error is the ratio of the absolute error to the original value of the quantity. It is calculated as Δy/y, which is approximated by dy/y.
The percentage error is the relative error expressed as a percentage. The correct formula is (Δy/y) × 100, which is approximated by (dy/y) × 100.
5. Why is the formula f(x + Δx) ≈ f(x) + f'(x)Δx so effective for approximations?
The formula f(x + Δx) ≈ f(x) + f'(x)Δx is effective because it essentially uses the tangent line to the function's curve at point 'x' to estimate the value at a nearby point 'x + Δx'. The term f'(x) is the slope of this tangent. For a very small change (Δx), the curve of the function and its tangent line are extremely close to each other. Therefore, the change in 'y' along the tangent line, which is dy = f'(x)Δx, serves as a very good approximation for the actual change in 'y' along the curve, which is Δy.
6. In what scenarios might the approximation using differentials become less accurate?
The approximation using differentials, dy ≈ Δy, becomes less accurate and should be used with caution in the following scenarios:
When the value of Δx is large. As the distance from the point of tangency increases, the function's curve and its tangent line diverge, leading to a significant error.
When the function has a high degree of curvature (a large second derivative, f''(x)) near the point of approximation. A rapidly changing slope means the tangent line is only a good representation for a very small interval.
At points where the function is not differentiable, such as at a sharp corner or cusp, as a unique tangent does not exist.
7. How do the RD Sharma solutions for Chapter 14 guide you to solve problems involving percentage change, like in y = x³ if x increases by 2%?
The solutions demonstrate the correct method to handle percentage changes. If y = x³ and x increases by 2%, it implies the relative change in x is Δx/x = 0.02. To find the approximate percentage change in y, first find the differential dy = (dy/dx)Δx. Since dy/dx = 3x², we have dy = (3x²)Δx. Next, to find the relative change in y, we calculate dy/y = (3x²Δx) / x³ = 3(Δx/x). By substituting the given relative change in x, we get dy/y = 3 × 0.02 = 0.06. This means the approximate percentage change in y is 6%.
