<p>The total surface area of a spherical ice-ball is decreasing at a rate of $2$ cm$^{2}$/s. Find the rate of change of the volume when its radius is $0.5$ cm.</p>

The total surface area of a spherical ice-ball is decreasing at a rate of $2$ cm$^{2}$/s. Find the rate of change of the volume when its radius is $0.5$ cm.

Applications of Differentiation

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

$\displaystyle\frac{\mathrm{d}A}{\mathrm{d}t} = -2$ cm²/s

$\displaystyle\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}A} \times \frac{\mathrm{d}A}{\mathrm{d}t}$

$= \displaystyle\frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}A} \times \frac{\mathrm{d}A}{\mathrm{d}t}$

where $V = \displaystyle\frac{4}{3}\pi r^3$, $A = 4\pi r^2$

$\displaystyle\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2$, $\displaystyle\frac{\mathrm{d}A}{\mathrm{d}r} = 8\pi r$

$= 4\pi r^2 \times \displaystyle\frac{1}{8\pi r} \times (-2)$

$\displaystyle\frac{\mathrm{d}V}{\mathrm{d}t}\bigg|_{r=0.5} = \frac{4\pi(0.5)^2}{8\pi(0.5)} \times (-2) = -\frac{1}{2}$ cm³/s

Applications of Differentiation

Question a72098af

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