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?

Answer 1
#V = 4/3r^3pi#
#(dV)/(dt) = 4/3(3r^2)(dr)/dtpi#
#(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# hours.
#340 = 4/3r^3pi#
#255 = r^3pi#
#255/pi = r^3#
#root(3)(255/pi) = r#

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

#(dV)/(dt) = 4(root(3)(255/pi))^2(dr)/(dt)pi#
We know that #(dV)/(dt) =-17 #, so after #7# hours, it will have melted #-119" ft"^3#.
#-119 = 4(root(3)(255/pi))^2(dr)/(dt)pi#
Solving for #(dr)/(dt)#, we get:
#(dr)/(dt) = -0.505" ft"/"hour"#

Hopefully this helps!

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Answer 2

The rate at which the radius is changing after 7 hours is approximately ( \frac{1}{20} ) feet per hour.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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