Question

Melting Ice A spherical iron ball is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 8 \(\mathrm{mL} / \mathrm{min}\) , how fast is the outer surface area of ice decreasing when the outer diameter (ball plus ice) is 20 \(\mathrm{cm} ? \)

Short answer

The outer surface area of the ice is decreasing at the rate of 0.005027 \(\mathrm{m}^2 / \mathrm{min}\).

Step-by-step solution

Identify known values

The rate of ice melting (\(dm/dt\)) is given as -8 \(\mathrm{mL} / \mathrm{min}\) (negative because it is decreasing). The outer diameter of the ball plus ice is given as 20 \(\mathrm{cm}\), which means the radius (\(r\)) is 10 \(\mathrm{cm}\) = 0.1 \(\mathrm{m}\). As 1 mL of water = 1 \(\mathrm{cm}^3\), we convert the rate into \(m^3 / min\) to make it consistent with \(r\). Hence, \(dm/dt = -8 * 10^{-6} m^3/min\).

Set up the volume formula

Recall the formula for the volume of a sphere, \(V = (4/3)πr^3\). The ice melts at a rate of \(dm/dt\), which also equals \(dV/dt\), as ice volume is decreasing.

Differentiate the volume formula with respect to time

Differentiating both sides of \(V = (4/3)πr^3\) with respect to \(t\), we get \(dV/dt = 4πr^2*(dr/dt)\). Substitute \(dV/dt = dm/dt = -8 * 10^{-6}\) into the equation, we get \(-8 * 10^{-6} = 4π * (0.1)^2 * (dr/dt)\). Solving, we get \((dr/dt) = -2 * 10^{-3} m/min\).

Set up the surface area formula & Differentiate

The formula for the surface area of a sphere is \(A = 4πr^2\). Differentiating both sides with respect to \(t\), we get \(dA/dt = 8πr*(dr/dt)\).

Substitute the known values

Substituting \(r = 0.1 m\), and \((dr/dt) = -2 * 10^{-3} m/min\) from step 3 into the equation, we get \(dA/dt = 8π * 0.1 * (-2 * 10^{-3})\). Solving, we get \(dA/dt = -5.027*10^{-3} m^{2}/min\). Due to the negative sign, the surface area is decreasing at a rate of 0.005027 \(\mathrm{m}^2 / \mathrm{min}\) .