# The radius of a sphere increases at a rate

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

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krolaniaN

Step 1
Surface area of sphere is $$\displaystyle{S}={4}\pi{r}^{{{2}}}$$.
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 $$\displaystyle={d}\frac{{S}}{{\left.{d}{t}\right.}}={8}\pi{r}.{}\frac{{dr}}{{\left.{d}{t}\right.}}={8}\pi{.9}{.4}={288}\pi$$.
So options A correct.