Question

The radius of a sphere increases at a rate of 4m/\sec. Find the rate at which the surface area increases when the radius is 9m. 288\pi m^{2}/\sec 36\pi m^{2}/\sec 18\pi m^{2}/\sec 72\pi m^{2}/\sec 144\pi m^{2}/\sec

Derivatives
ANSWERED
asked 2021-05-04
The radius of a sphere increases at a rate of \(4m/\sec\). Find the rate at which the surface area increases when the radius is 9m.
\(288\pi m^{2}/\sec\)
\(36\pi m^{2}/\sec\)
\(18\pi m^{2}/\sec\)
\(72\pi m^{2}/\sec\)
\(144\pi m^{2}/\sec\)

Answers (1)

2021-05-05
Step 1
Surface area of sphere is \(S= 4\pi r^{2}\).
And rate of change means derivatives.
Step 2
Now let radius of sphere is =r.
So dr/dt=4.
So rate of change of surface area \(= dS/dt= 8\pi r.dr/dt =8\pi .9.4=288\pi\).
So options A correct.
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