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# !!!HELP???

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nnnnnn

Guest Nov 21, 2018
edited by Guest  Nov 21, 2018

#1
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$$V=\dfrac 4 3 \pi r^3\\ \dfrac{dV}{dr} = 4 \pi r^2\\ \text{ok a bit of magic here}\\ dV = 4\pi r^2 dr\\ \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} \\ 2 = 4 \pi (3)^2 \dfrac{dr}{dt}\\ \dfrac{dr}{dt} = \dfrac{1}{18\pi}~in/hr$$

Rom  Nov 21, 2018
#1
+3190
+2

$$V=\dfrac 4 3 \pi r^3\\ \dfrac{dV}{dr} = 4 \pi r^2\\ \text{ok a bit of magic here}\\ dV = 4\pi r^2 dr\\ \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} \\ 2 = 4 \pi (3)^2 \dfrac{dr}{dt}\\ \dfrac{dr}{dt} = \dfrac{1}{18\pi}~in/hr$$

Rom  Nov 21, 2018
#2
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Hi Guest,

You should not delete your question, because other people may have that question as well.

- Daisy

dierdurst  Nov 21, 2018
#3
+656
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nnnnnn? Why did you delete your question? Other people may have the same or similar questions.

PartialMathematician  Nov 22, 2018
#4
+3190
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The question was:

A balloon is losing 2 cubic inches of air per hour.

At what rate is it's radius decreasing when the balloon is 3 inches in radius.

Rom  Nov 23, 2018