Question

The slant height of a cone is fixed at 7 cm. The rate of increase in the volume of the cone corresponding to the rate of increase of 0.3 cm/s in the height when h = 4 cm is

• A. $\frac{33\pi }{10}cc/s$
• B. $\frac{3\pi }{10}cc/s$
• C. $\frac{\pi }{5}cc/s$
• D. $\frac{7\pi }{10}cc/s$

Answer 1

The correct option is A $\frac{33\pi }{10}cc/s$

Using Pythagoras Theorem on the triangle formed, we have,

$r^2+4^2=7^2{r}^2=49-16r^2=33$

Volume of the cone $=\frac{1}{3}\pi r^{2}h$

Rate of increase in volume $=\frac{1}{3}\pi r^2\times \left(\text{rate of}h\right)$

$=\frac{1}{3}\pi \times 33\times 0.3\text{cm/s}$

$=\pi \times 3.3{\text{cm}}^3/\text{sec}$

$=\frac{33\pi }{10}{\text{cm}}^3/\text{sec}$