# 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
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.
$$288\pi m^{2}/\sec$$
$$36\pi m^{2}/\sec$$
$$18\pi m^{2}/\sec$$
$$72\pi m^{2}/\sec$$
$$144\pi m^{2}/\sec$$

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.