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A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s.(a) Express the radius $r$ of the balloon as a function of the time $t$ (in seconds).(b) If $V$ is the volume of the balloon as a function of the radius, find $V \circ r$ and interpret it.

a) $r(t)=2 t$b) $V \circ r=\frac{32}{3} \pi t^{3}$

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DG

David Base G.

October 26, 2020

That was not easy, glad this was able to help

KM

Kinley M.

February 2, 2021

A balloon in the shape of a sphere is being inflated at the rate of 2 cm/sec. At the time at which the radius of the balloon is 3cm, how fast is its radius increasing?

Anna Marie V.

Campbell University

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Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

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Video Transcript

we're told that the balloon is being inflated at a rate of two centimeters per second. That's the rate at Rich at which the radius is growing. So to find the radius we would find its rate of increase in centimeters per second multiplied by its time in seconds. And that would give us the centimeters. So the rate of increase is too. And we can call the time t. So we have radius equals to t. Now let's find the volume and the volume of a sphere is 4/3 pi r cubed. What we can do is we can substitute for are the two t that we just found and we will have 4/3 pi times two t cubed. And if we want to simplify that, we can cube the two and cube the tea and we get 4/3 pi times eight t cubed and then we'll multiply the eight by the 4/3 and we get 32 3rd pi times t cubed. So what that represents is the volume as a function of time, as opposed to volume as a function of radius

Oregon State University
Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp