As a spherical balloon is being inflated, its radius (in centimeters) after $t$ minutes is given by $r(t)=3\left[(t+8)^{1/3}\right]$ where $0<t<10$. What is the rate of change with respect to $t$ of each of the following at $t=8$?
- $r(t)$
- the surface area
I solved 1 by taking the derivative and plugging in $t$, so I approached 2 similarly. However, plugging in $8$ into the $r(t)$ equation and plugging that into the surface area equation doesn't seem to be working.
Thank you so much!
You cannot plug in the derivative of $r(t)$ into the expression for the surface area. You want the derivative of $A(t)$, where $A$ is the surface area.
The equation of the surface area of a sphere is $$A = 4\pi r^2$$ So, plugging in the expression for $r$, $$A = 4\pi (3(t+8)^{\frac13})^2$$$$A = 4\pi (9(t+8)^{\frac23})$$$$A = 36\pi (t+8)^{\frac23}$$
Now take the derivative of that and plug in $t=8.$