Rate of Change 2

**sanikui** · March 30th 2009, 04:50 AM

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?

**mr fantastic** · March 30th 2009, 04:54 AM

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

**sanikui** · March 30th 2009, 05:43 AM

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?)

**HallsofIvy** · March 30th 2009, 06:35 AM

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.

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