Applications of Derivatives
Rate of Change and Approximation
Grade 12
Question:
<p>The approximate change in the volume of a cube of side <i>x</i> m caused by increasing the side by 3% is:</p>
<p>(a) 0.06<i>x</i><sup>3</sup> m<sup>3</sup></p>
<p>(b) 0.6<i>x</i><sup>3</sup> m<sup>3</sup></p>
<p>(c) 0.09<i>x</i><sup>3</sup> m<sup>3</sup></p>
<p>(d) 0.9<i>x</i><sup>3</sup> m<sup>3</sup></p>
Step-by-Step Solution
Key Concept: Use differential approximation: ΔV ≈ (dV/dx)·Δx to find the change in volume when the side increases by a small percentage.
<p><strong>Solution:</strong></p><p>We know that the volume <i>V</i> of a cube of side <i>x</i> is given by: $V = x^3$</p><p>Differentiating with respect to <i>x</i>: $\frac{dV}{dx} = 3x^2$</p><p>Let $\Delta x$ be change in side = 3% of <i>x</i> = 0.03<i>x</i></p><p>Now, change in volume: $\Delta V = \frac{dV}{dx} \cdot \Delta x = (3x^2)(0.03x) = 0.09x^3$ m<sup>3</sup></p><p>Hence, the approximate change in the volume of the cube is 0.09<i>x</i><sup>3</sup> m<sup>3</sup>.</p>
Correct Answer: C