Question

A solid sphere of gold, of radius 5 cm, is being melted in a furnace such that the radius is decreasing uniformly.

When its radius is 1 cm, show that the rate at which its surface area is decreasing with respect to time is twice the rate at which its volume is decreasing with respect to time.

Answer 1

The rate at which the surface area is decreasing with respect to time is twice the rate at which its volume is decreasing with respect to time, as demonstrated in the explanation.

Answer 2

Let $r$ be the radius of the gold sphere.
Let $A$ be the surface area of the sphere.
Let $V$ be the volume of the sphere.

The formulas for the surface area and volume of a sphere are: Surface Area: $A = 4\pi r^2$ Volume: $V = \frac{4}{3}\pi r^3$

We are given that the radius is decreasing uniformly, which means the rate of change of radius with respect to time, $\frac{dr}{dt}$, is a constant. Since the radius is decreasing, $\frac{dr}{dt}$ will be a negative constant. Let $\frac{dr}{dt} = -k$, where $k$ is a positive constant.

Now, we differentiate $A$ and $V$ with respect to time $t$ using the chain rule:

Rate of change of surface area: $\frac{dA}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \cdot (2r) \frac{dr}{dt} = 8\pi r \frac{dr}{dt}$

Rate of change of volume: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi \cdot (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$

We need to evaluate these rates when the radius $r = 1$ cm.

Substitute $r = 1$ cm into the expressions for $\frac{dA}{dt}$ and $\frac{dV}{dt}$:

At $r = 1$ cm: $\frac{dA}{dt} = 8\pi (1) \frac{dr}{dt} = 8\pi \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi (1)^2 \frac{dr}{dt} = 4\pi \frac{dr}{dt}$

Since the radius is decreasing, $\frac{dr}{dt}$ is negative. Let $\frac{dr}{dt} = -k$, where $k > 0$.

So, at $r = 1$ cm: $\frac{dA}{dt} = 8\pi (-k) = -8\pi k$ $\frac{dV}{dt} = 4\pi (-k) = -4\pi k$

The rate at which the surface area is decreasing is $-\frac{dA}{dt}$. The rate at which the volume is decreasing is $-\frac{dV}{dt}$.

Rate of decrease of surface area: $-\frac{dA}{dt} = -(-8\pi k) = 8\pi k$

Rate of decrease of volume: $-\frac{dV}{dt} = -(-4\pi k) = 4\pi k$

Now, we compare these two rates: We observe that $8\pi k = 2 \times (4\pi k)$. Therefore, the rate at which its surface area is decreasing with respect to time ($-\frac{dA}{dt}$) is twice the rate at which its volume is decreasing with respect to time ($-\frac{dV}{dt}$).