The surface area of a spherical ball is incre | vedclass

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(B) Let $S$ be the surface area and $r$ be the radius of the spherical ball. The surface area is given by $S = 4 \pi r^2$. Given that $\frac{dS}{dt} =

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) Let $S$ be the surface area and $r$ be the radius of the spherical ball. The surface area is given by $S = 4 \pi r^2$. Given that $\frac{dS}{dt} = 4 \pi \,cm^2/s$. Differentiating $S$ with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$. When $S = 16 \pi \,cm^2$,we have $4 \pi r^2 = 16 \pi$,which implies $r^2 = 4$,so $r = 2 \,cm$. Substituting the values into the derivative equation: $4 \pi = 8 \pi (2) \frac{dr}{dt}$. $4 \pi = 16 \pi \frac{dr}{dt}$. $\frac{dr}{dt} = \frac{4 \pi}{16 \pi} = 0.25 \,cm/s$.

The surface area of a spherical ball is increasing at the rate of $4 \pi \,cm^2/s$. The rate at which the radius is increasing when the surface area is $16 \pi \,cm^2$ is: (in $\,cm/s$)

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