💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Air is being pumped into a spherical weather balloon. At any time $t,$ the volume of the balloon is $V(t)$ and its radius is $r(t).$(a) What do the derivatives $dV/dr$ and $dV/dt$ represent? (b) Express $dV/dt$ in terms of $dr/dt.$

## a. $d V / d r$ represents how quickly the volume is changing in response to unitchange in radius, or how fast the volume changes in comparison to theradius, while $d V / d t$ represents the change in volume per unit time, or howquickly it changes with respect to time.b. $\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$

Derivatives

Differentiation

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

We're talking about a spherical balloon and so we have the the volume are the radius and tea time, and we want to look at the different expressions of rates. So if you see DVD are, think about what V means and what are means. V means volume are means radius, so this would be the rate of change of the volume with respect to the radius. However, if you have DVD t think about what V means and what team means, that would be the rate of change of the volume with respect to time. Okay. Now, for the second part of the problem, the rate of change of the volume with respect to time could be expressed as the rate of change of the Radius DVD R times, the rate of change of the radius with respect to time DRD t. Now let's find ER DVD are remember, the volume of a sphere is 4/3 pi r cubed. So if we want to find the derivative of that with respect to our, we can leave the constant 4/3 pi and use the power rule, bring down the three and raised the art of the second and when we simplify that we have four pi r squared So we substitute that in for DVD are and we have four pi r squared times drd t and that's DBT.

Oregon State University

Derivatives

Differentiation

Lectures

Join Bootcamp