Differentials — Question 9

Question 9 - Solution

We are given: $$s = 5 \text{ cm}, \quad ds = 0.1 \text{ cm}$$

The volume of a cube is: $$V = s^3$$

Differentiate using differentials: $$dV = \frac{d}{ds}(s^3)\, ds = 3s^2\, ds$$

Substitute values: $$dV = 3(5)^2(0.1) = 3 \cdot 25 \cdot 0.1 = \boxed{7.5 \, \text{cm}^3}$$

(a) Maximum error in volume: $$\boxed{7.5 \, \text{cm}^3}$$

(b) Relative and percentage error:

$$V = 5^3 = 125 \, \text{cm}^3$$ $$\text{Relative error} = \frac{dV}{V} = \frac{7.5}{125} = 0.06$$ $$\text{Percentage error} = 0.06 \times 100 = \boxed{6\%}$$

Final Answers:

• Maximum error in volume: $\boxed{7.5 \, \text{cm}^3}$

• Relative error: $\boxed{0.06}$

• Percentage error: $\boxed{6\%}$