Volume of a sphere:
<em>V</em> = 4/3 <em>π</em> <em>r</em> ³
Differentiate <em>V</em> with respect to time <em>t</em> :
d<em>V</em>/d<em>t</em> = 4<em>π</em> <em>r</em> ² d<em>r</em>/d<em>t</em>
Surface area of a sphere:
<em>A</em> = 4<em>π</em> <em>r</em> ²
Differentiate <em>A</em> with respect to <em>t</em> :
d<em>A</em>/d<em>t</em> = 8<em>π</em> <em>r</em> d<em>r</em>/d<em>t</em>
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You're given that the volume is increasing at a constant rate, d<em>V</em>/d<em>t</em> = 10 cm³/s, so
10 cm³/s = 4<em>π</em> <em>r</em> ² d<em>r</em>/d<em>t</em>
Solve for d<em>r</em>/d<em>t</em> (the rate of change of the sphere's radius):
d<em>r</em>/d<em>t</em> = 5/(2<em>π</em> <em>r</em> ²) cm³/s
Substitute this into the equation for d<em>A</em>/d<em>t</em> :
d<em>A</em>/d<em>t</em> = 8<em>π</em> <em>r</em> (5/(2<em>π</em> <em>r</em> ²) cm³/s)
d<em>A</em>/d<em>t</em> = 20/<em>r</em> cm³/s
At the moment the radius is <em>r</em> = 5 cm, the surface area is increasing at a rate of
d<em>A</em>/d<em>t</em> = 20/(5 cm) cm³/s = 4 cm²/s
When the surface area is increasing at a rate of d<em>A</em>/d<em>t</em> = 2 cm²/s, the radius <em>r</em> is such that
2 cm²/s = 20/<em>r</em> cm³/s
which happens when the radius is
<em>r</em> = 20/2 (cm³/s) / (cm²/s) = 10 cm
