# Thread: Rate of Change 2

1. ## Rate of Change 2

A metel sphere is dissolving in acid. It remains spherical and the rate at which its volume decreases is proportional to its surface area. Show that its radius is decreasing at a constant rate.

What information and chain rules that can get from this question?

2. Originally Posted by sanikui
A metel sphere is dissolving in acid. It remains spherical and the rate at which its volume decreases is proportional to its surface area. Show that its radius is decreasing at a constant rate.

What information and chain rules that can get from this question?
$\frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt}$.

Given: $\frac{dV}{dt} = k A = 4 k \pi r^2$.

Geometry: $V = \frac{4}{3} \pi r^3 \Rightarrow \frac{dV}{dr} = \, ....$

3. Originally Posted by mr fantastic
$\frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt}$.

Given: $\frac{dV}{dt} = k A = 4 k \pi r^2$.

Geometry: $V = \frac{4}{3} \pi r^3 \Rightarrow \frac{dV}{dr} = \, ....$
Area of sphere: A =4 pi r^2.
Let $\frac{dV}{dt} = k A = 4 k \pi r^2$. ( So, that: $\frac{dV}{dt} = k A = 4 k \pi r^2$. need to diff?)

4. Originally Posted by sanikui
Area of sphere: A =4 pi r^2.
Let $\frac{dV}{dt} = k A = 4 k \pi r^2$. ( So, that: $\frac{dV}{dt} = k A = 4 k \pi r^2$. need to diff?)
The problem was to "Show that its radius is decreasing at a constant rate." In order to show that, you have to look at dr/dt.