RELATED RATES SPHERE Volume Problem

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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?
As with all of the other related rates problems I've worked through, we are going to be following the same four step process.
1. Draw a sketch
2. Come up with your equation
3. Implicit differentiation
4. Solve for the desired rate of change
For a list of other related rates practice problems and solutions, visit jakesmathlesso... (opens in a new tab)
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[@jeebo.m]

Thank you for the explanation

[@JakesMathLessons]

You’re welcome!

[@guyguyungtin7660]

Thank you so much greatly appreciated

[@JakesMathLessons]

You’re welcome!

[@ivannaot]

Thank you so so much!

[@JakesMathLessons]

You’re welcome!

[@RanBlakePiano]

Very ,very helpful!

[@JakesMathLessons]

Thank you, I’m glad!

[@Youwishyouknewwhohuh_]

did you forget to times by 4? cause 4pi(40^2) = 25,600pi but 4pi(40^2)*4= 102,400pi

[@johnnolen8338]

4 π ·(40)^2 = 6 400 π , my friend. If you multiply 6 400 π by 4, you get 25 600 π.

[@johnnolen8338]

Good job. My only criticism is that your explanation took way too long. All related rates problems are applications of the chain rule; a chain rule calculation that only needs to be performed once by the way. You don't need to keep repeating that "We don't have an explicit formula for volume (or radius) in terms of time." Because we don't have an explicit formula is why we use implicit differentiation.
The general form of the solution to this problem is given by dV/dt = dV/dr · dr/dt (the Chain Rule).
V = 4/3 π r^3 implies dV/dr = 4 π r^2.
Then dV/dr = 4 π (80/2)^2 = 6 400 π.
The derivative of the radius with respect to time, dr/dt = 4 mm/s, is given to you. Plug in these values, turn the crank, and you're done.
dV/dt = 6 400 π × 4 = 25 600 π mm^3/sec. ◼

[@RoyTagliaferro-k2r]

My calculation was right but i thought that the answer was wrong because it was too large of a number, i guess im only overthinking again

[@coltonhalter3836]

Thank you so much!!!

[@JakesMathLessons]

You’re welcome!