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valentinak56 [21]
3 years ago
8

The volume of sphere is increaseing

Mathematics
1 answer:
Artemon [7]3 years ago
4 0

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>

<em />

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

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