Differentials — Question 6

Question 6

The radius $r$ of a right circular cylinder is measured to be 5 cm with a possible error of 0.1 cm. The height $h$ is 20 cm with a possible error of 0.2 cm.

• Use differentials to approximate the maximum error in calculating the volume of the cylinder.

• Find the relative and percentage errors in the volume.

Question 6 - Solution

The volume of a cylinder is: $$V = \pi r^2 h$$

We compute the differential $dV$: $$dV = \frac{\partial V}{\partial r} \, dr + \frac{\partial V}{\partial h} \, dh = 2\pi r h \, dr + \pi r^2 \, dh$$

Given: $$r = 5 \text{ cm}, \quad dr = 0.1 \text{ cm}, \quad h = 20 \text{ cm}, \quad dh = 0.2 \text{ cm}$$

Substitute into the differential: $$dV = 2\pi (5)(20)(0.1) + \pi (5)^2 (0.2) = 20\pi + 5\pi = \boxed{25\pi \approx 78.54 \text{ cm}^3}$$

(a) Maximum error: $$\boxed{dV \approx 78.54 \text{ cm}^3}$$

(b) Relative and percentage error:

Actual volume: $$V = \pi (5)^2 (20) = 500\pi \approx 1570.8 \text{ cm}^3$$

Relative error: $$\frac{dV}{V} = \frac{25\pi}{500\pi} = \frac{1}{20} = 0.05$$

Percentage error: $$0.05 \times 100 = \boxed{5\%}$$

Final Answers:

• Maximum error in volume: $\boxed{25\pi \approx 78.54 \text{ cm}^3}$

• Relative error: $\boxed{0.05}$

• Percentage error: $\boxed{5\%}$