Question

The volume, V cm3 , of a cone height h is
(pi)(h^3) / 12

If h increases at a constant rate of 0.2 cm/sec and the initial height is 2 cm, express V in terms of t and find the rate of change of V at time t.

Answer 1

Answer on Question #50591 – Math – Calculus

Question

The volume, $V$ (cm³), of a cone height $h$ is

$$V = \pi \frac{h^3}{12}.$$

If $h$ increases at a constant rate of 0.2 cm/sec and the initial height is 2 cm, express $V$ in terms of $t$ and find the rate of change of $V$ at time $t$.

Solution:

We know that $h$ increases at a constant rate of 0.2 cm/sec, it means that

$$\frac{dh}{dt} = 0.2 \text{ (cm/sec)}.$$

Integrate both sides of this equality with respect to $t$ in order to find

$$h(t) = \int 0.2 dt = 0.2t + c.$$

Because the initial height is

$$h(0) = 2 \text{ (cm)},$$

then

$$h(0) = 0.2 \cdot 0 + c = 2,$$ $$c = 2.$$

So

$$h(t) = 0.2t + 2 = 0.2(t + 10) \text{ (cm)}.$$

Thus, the volume of cone expressed in terms of $t$ is

$$V(t) = \pi \frac{h^3}{12} = \pi \frac{(0.2(t + 10))^3}{12} = \frac{\pi}{1500}(t + 10)^3 \text{ (cm}^3\text{)}.$$

And finally the rate of change of $V$ at time $t$ is the derivative of $V(t)$ with respect to $t$:

$$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi}{1500}(t + 10)^3 \right) = \frac{\pi}{1500} \cdot \frac{d}{dt} \left( (t + 10)^3 \right) = \frac{\pi}{1500} \cdot 3(t + 10)^2 = \frac{\pi}{500}(t + 10)^2 \text{ (cm}^3\text{/sec)}.$$

Answer: $V(t) = \frac{\pi}{1500}(t + 10)^3$ (cm³), $\frac{dV}{dt} = \frac{\pi}{500}(t + 10)^2$ (cm³/sec)

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