4.5a Solving Related Rates Problems, Derivative, Cylinder, Volume Change - AP Calculus BC

This video provides a worked example on solving related rates problems.
This video is part of Unit 4 of AP Calculus BC on Contextual Applications of Differentiation.
You can contact me through the comments and community section, or email me at [e-mail] for questions.

[@BraydenWolframe]

Buddy explained this better than my college prof in 30 seconds compared to 6 hours of class.

[@drofeng]

Thanks for watching, I am glad it’s useful.

[@davidthomas5614]

I get that dr/dt is 0 so it doesn’t matter but i think it would be better for people that don’t know that when differentiating pi r^2 it should be 2pi r. Maybe also show product rule because i did get confused for a second. Otherwise good video

[@drofeng]

Thanks for watching. Sure, I will consider that in future videos on this topic.

[@momuali6652]

what happens if you use the power rule for r^2. then you get dv/dt=2(pi)(r)(dh/dt)

[@drofeng]

If r was also a function of time, then by the product rule: dv/dt = pi*r^2 dh/dt + pi * d(r^2)/dt * h Then apply the chain rule to the second term ( d/dt = d/dr * dr/dt ) dv/dt = pi*r^2 dh/dt + 2*pi*r*h*dr/dt Given that r is constant in this problem, dr/dt evaluates to zero.

[@Komo411]

Hello! Might be a little late on this question, but is there a trick for figuring out a related rates equations? The algebra isn’t too hard, but finding out the equation is very difficult for me.

[@drofeng]

It's all in the wording of the problem and doing a lot of practice problems until you get better at it. You can break down the information given in a few logical steps, noting that when we talk about rates, the derivatives are evaluated w.r.t. time: 1) The volume of the cylinder is stated, so you need to know the equation for the volume of a cylinder 2) The volume is increasing, so you need to evaluate the rate of change of volume with respect to time dV/dt (positive because it's increasing), and dV/dt is given as 0.1m^3/s 3) The volume is increasing w.r.t. height (while the radius is 3cm). Hence, when evaluating dV/dt, you differentiate the height w.r.t. time (dh/dt) while keeping the radius constant) Then substitute the known values to determine the unknown. Try to find this type of pattern when evaluating different problems and let me know how you go e.g.: kzbin.infolCGeq9N9ATM kzbin.infoMj4BhaKlYvw

[@Komo411]

@@drofeng I didn’t expect such a quick reply, thank you so much! My exam is in a few days so I appreciate the detailed response!

[@drofeng]

You’re welcome. All the best with your exam.

[@philasandenkosi8568]

Why not use product rule in pir²h ?

[@drofeng]

The question says the volume is increasing wrt the height, so the radius is constant. Hence, even if you use the product rule, dr/dt is zero and you get the same result.

[@maccyote7650]

where did the radius=0.03 come from?

[@drofeng]

Radius of 3cm is given in the problem which is converted to 0.03m to make the units consistent.

[@amierbahan3905]

Yeah you need to convert it to meters to continue solving the problem