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How would I do this problem I am not sure if I did it correctly.

The volume $V$ of a spherical balloon is increasing at a constant rate of 32 cubic feet per minute.

A. How fast is the radius r increasing when the radius is exactly 3 feet.

I know $\dfrac{dv}{dt}=32$ $,r=3$

the formula for volume is

$$v=\frac{4}{3}\pi r^3$$

$$32=\frac{4}{3}\pi3r^2$$

$$32=4\pi 9\frac{dr}{dt}$$

$$\frac{dr}{dt}=\frac{8\pi}{9}$$ I am not sure would this be correct?

My second question is

B.How fast is the surface area a increasing when the radius is 4 feet

I know

$$A=4\pi r^2$$

but then what would I do

$$\frac{dA}{dt}=4\pi2r \frac{dr}{dt}$$

but what would I plug in for $\dfrac{dr}{dt}$?

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2 Answers 2

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Your $\dfrac{dr}{dt}$ should have $\pi$ in the denominator: $$\frac{dr}{dt}=\frac{8}{9\pi}\tag{1}$$

Use $\dfrac{dr}{dt}$ from computing $dr/dt$ as you did for probem $1$ but using $r = 4$

$$32=\frac{4}{3}\pi3r^2 \implies 32 = \frac {3\cdot4}3 (4^2)\pi\frac{dr}{dt} \iff \frac{dr}{dt} = \frac {1\cdot 32}{64 \pi} = \frac 1{2\pi}$$

And simplify. Then substitute into the equation you found:

$$\frac{dA}{dt}=4\pi2r \frac{dr}{dt}$$

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  • $\begingroup$ yes that make sense. $\endgroup$ Commented Apr 6, 2013 at 22:45
  • $\begingroup$ one quick question I have is can I use the dr/dt from my first equation when my second one has a radius of 4 pi because in my first question it has only 3 pi the radius would I need to find dr/dt again? $\endgroup$ Commented Apr 6, 2013 at 22:48
  • $\begingroup$ Yup, you did all the work! Just putting pieces together. Then solve for $dA/dt$ at r = 4. $\endgroup$
    – amWhy
    Commented Apr 6, 2013 at 22:50
  • $\begingroup$ Yes, Fernando you can use your computation from above to evaluate dr/dt when r = 4. $\endgroup$
    – amWhy
    Commented Apr 6, 2013 at 22:53
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    $\begingroup$ See my edit: you can calculate $\frac{dr}{dt}$ from the first part, substituting $4$ into the equation, as you did for $3$. $\endgroup$
    – amWhy
    Commented Apr 6, 2013 at 22:59
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You have $\dfrac{8\pi}{9}$ where you need $\dfrac{8}{9\pi}$.

If you happen to know that the surface area is $4\pi r^2$, then you can say the rate at which the volume is increasing is the surface area times the rate at which the radius is increasing.

Your reasoning concerning the surface area is alright. You can find the value of $\dfrac{dr}{dt}$ at that time by the same method you used to find $\dfrac{dr}{dt}$ in the first question.

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