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Harman [31]
3 years ago
15

The spherical balloon is inflated at the rate of 10 m³/sec. Find the rate at which the surface area is increasing when the radiu

s of the sphere is 3m?
Mathematics
1 answer:
rjkz [21]3 years ago
4 0

The balloon has a volume V dependent on its radius r:

V(r)=\dfrac43\pi r^3

Differentiating with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}

If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of

10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}

The surface area of the balloon is

S(r)=4\pi r^2

and differentiating gives

\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}

so that at the moment the radius is 3 m, its area is increasing at a rate of

\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}

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