Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 7

The volume (V) of the rod is given by:

$V = \pi r^2 \cdot 5x = 5\pi r^2 x$

Differentiating both sides with respect to time, we use the product rule:

$\frac{dV}{dt} = 5 \pi (r^2 \frac{dx}{dt} + x \frac{d(r^2)}{dt})$

We know that:

• When (x = 2), we can substitute (x = 2) into the equation, and we already found (\frac{dr}{dt}) above.
• Since (r = 2) cm at this instant, we can substitute:

$\frac{dV}{dt} = 5\pi \left(2^2 \cdot \frac{dx}{dt} + 2 \cdot 5(2) \frac{dr}{dt}\right)$

We substitute (\frac{dr}{dt} = 0.0025 \text{ cm/s} ) and (\frac{dx}{dt} = 0) (since it is not specified, we often assume a constant for the x-direction):

$\frac{dV}{dt} = 5\pi \left(0 + 2 \cdot 5(2) imes 0.0025\right) \approx 0.48 \text{ cm}^3/s$

Thus, the rate of increase of the volume of the rod when (x = 2) is approximately (0.48 \text{ cm}^3/s).