Differentials — Question 8

Question 8 - Solution

We are given: $$r = 10 \text{ cm}, \quad dr = 0.05 \text{ cm}$$

The volume of a sphere is: $$V = \frac{4}{3} \pi r^3$$

Differentiate using differentials: $$dV = \frac{d}{dr}\left( \frac{4}{3} \pi r^3 \right) \cdot dr = 4\pi r^2 \cdot dr$$

Substitute known values: $$dV = 4\pi (10)^2 \cdot 0.05 = 4\pi \cdot 100 \cdot 0.05 = 20\pi \approx \boxed{62.83 \, \text{cm}^3}$$

(a) Maximum error in volume: $$\boxed{dV \approx 62.83 \, \text{cm}^3}$$

(b) Relative and percentage error:

Actual volume: $$V = \frac{4}{3} \pi (10)^3 = \frac{4000}{3} \pi \approx 4188.79 \, \text{cm}^3$$ $$\text{Relative error} = \frac{dV}{V} \approx \frac{62.83}{4188.79} \approx 0.015$$ $$\text{Percentage error} = 0.015 \times 100 = \boxed{1.5\%}$$

Final Answers:

• Maximum error in volume: $\boxed{62.83 \, \text{cm}^3}$

• Relative error: $\boxed{0.015}$

• Percentage error: $\boxed{1.5\%}$