Applications of Derivatives
Related Rates
Grade 12
Question:
<p>If the surface area of a cube is increasing at a rate of 3.6 cm<sup>2</sup>/sec, retaining its shape, then the rate of change of its volume (in cm<sup>3</sup>/sec), when the length of a side of the cube is 10 cm, is:</p>
<p>(a) 18</p>
<p>(b) 10</p>
<p>(c) 9</p>
<p>(d) 20</p>
Step-by-Step Solution
Key Concept: Use related rates: differentiate both surface area and volume formulas with respect to time, and use the given rate of change of surface area to find the rate of change of volume.
<p><strong>Solution:</strong></p><p>Since the surface area of cube, $A = 6a^2$ cm<sup>2</sup></p><p>It is given, $\frac{dA}{dt} = 3.6$ cm<sup>2</sup>/sec</p><p>Differentiating $A = 6a^2$ with respect to <i>t</i>:</p><p>$\frac{dA}{dt} = 12a \cdot \frac{da}{dt}$</p><p>$3.6 = 12(10) \cdot \frac{da}{dt}$</p><p>$\frac{da}{dt} = \frac{3.6}{120} = 0.03$ cm/sec</p><p>The volume of cube, $V = a^3$</p><p>$\frac{dV}{dt} = 3a^2 \cdot \frac{da}{dt} = 3(10)^2(0.03) = 3(100)(0.03) = 9$ cm<sup>3</sup>/sec</p>
Correct Answer: C