Differentials — Question 2

Question 2

Suppose the radius $r$ of a sphere is measured to be 5 cm with a possible error of 0.1 cm.

• Use differentials to approximate the possible error in the calculated volume of the sphere.

• Estimate the relative error and percentage error in the volume.

Question 2 - Solution

We use the volume formula of a sphere: $$V = \frac{4}{3} \pi r^3$$

(a) Approximate error using differentials:

$$dV = \frac{dV}{dr} \, dr = 4\pi r^2 \, dr$$

$$dV = 4\pi (5)^2 (0.1) = 4\pi (25)(0.1) = 10\pi \approx 31.42 \text{ cm}^3$$

(b) Relative error: $$\frac{dV}{V} = \frac{10\pi}{\frac{4}{3}\pi(5)^3} = \frac{10\pi}{\frac{500}{3}\pi} = \frac{10 \cdot 3}{500} = \frac{30}{500} = 0.06$$

Percentage error: $$0.06 \times 100 = \boxed{6\%}$$

Final Answer:

• Approximate error in volume: $\boxed{31.42 \, \text{cm}^3}$

• Relative error: $\boxed{0.06}$

• Percentage error: $\boxed{6\%}$