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The radius of a sphere is increasing at a rate of $ 4 mm/s. $ How fast is the volume increasing when the diameter is $ 80 mm? $

Volume is increasing at the rate of 25600$\pi \approx 80424.772 \mathrm{mm}^{3} / \mathrm{s}$

04:12

Alex L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Oregon State University

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mhm. In this question we're asked that given the radius of the spheres increase, we want to find how fast the volume increases at a surge in diameter. We first need to remember what the formula for the volume of the sphere is. We know that the volume of the sphere is 4/3 times pi times r cubed. But remember that the volume and the radius are both changing as functions of time. So it's more correct to write it as as functions of time. Now we'll just differentiate both sides with respect to time. Using implicit differentiation. We know that the rate of change of the volume is the prime of T equaling four thirds times pi tons. Now this are of T cubed requires a change rule. We first need to use the power rule to bring the power down. So we're going to have three times our prime times R f T squared multiplied by the derivative of R. Of T. Which is our prime of T. Yeah. And now we can plug in. Our rates were given that the volume Is increasing if the diameter is 80 That means that the radius has to be 40. Yeah. And we're also given that the rate of increase is four mm/s. So we can plug all of this information in. And also this three and three we'll cancel. So we'll have this is equal to four times pi times 40 squared times four. And when we evaluate this, we get that. This is eagle to 25,000 600 pie and this is in millimeters cubed per second. That's how you do this question.

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