The sun is shining and a spherical snowball of volume 340 ft3 is melting at a rate of 17 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 7 hours?

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The sun is shining and a spherical snowball of volume 340 ft3 is melting at a rate of 17 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 7 hours?

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Answer 1 · 2016-11-14T14:47:45

$V = \frac{4}{3} r^{3} π$ $V = 4/3r^3pi$

$\frac{d V}{d t} = \frac{4}{3} (3 r^{2}) \frac{d r}{d t} π$ $(dV)/(dt) = 4/3(3r^2)(dr)/dtpi$

$\frac{d V}{d t} = (4 r^{2}) \frac{d r}{d t} π$ $(dV)/(dt) = (4r^2)(dr)/(dt) pi$

Now we look at our quantities to see what we need and what we have.

So, we know the rate at which the volume is changing. We also know the initial volume, which will allow us to solve for the radius. We want to know the rate at which the radius is changing after $7$ $7$ hours.

$340 = \frac{4}{3} r^{3} π$ $340 = 4/3r^3pi$

$255 = r^{3} π$ $255 = r^3pi$

$\frac{255}{π} = r^{3}$ $255/pi = r^3$

$\sqrt[3]{\frac{255}{π}} = r$ $root(3)(255/pi) = r$

We plug this value in for "r" inside the derivative:

$\frac{d V}{d t} = 4 {(\sqrt[3]{\frac{255}{π}})}^{2} \frac{d r}{d t} π$ $(dV)/(dt) = 4(root(3)(255/pi))^2(dr)/(dt)pi$

We know that $\frac{d V}{d t} = - 17$ $(dV)/(dt) =-17$ , so after $7$ $7$ hours, it will have melted $- 119 {ft}^{3}$ $-119" ft"^3$ .

$- 119 = 4 {(\sqrt[3]{\frac{255}{π}})}^{2} \frac{d r}{d t} π$ $-119 = 4(root(3)(255/pi))^2(dr)/(dt)pi$

Solving for $\frac{d r}{d t}$ $(dr)/(dt)$ , we get:

$\frac{d r}{d t} = - 0.505 \frac{ft}{hour}$ $(dr)/(dt) = -0.505" ft"/"hour"$

Hopefully this helps!

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The sun is shining and a spherical snowball of volume 340 ft3 is melting at a rate of 17 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 7 hours?

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