## sjmoneys96 2 years ago In the third video on related rates the Professor differentiates 1/3pi(2h/5)^2h and gets pi/3(2/5)^2(3h^2)(dh/dt). How did he get this? I know you're trying to find rate of change of h with respect to t. So I get the dh/dt at the end, I guess I am just getting confused on some usage of the chain rule. Can anyone help?

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1. calculusfunctions

If$V =\frac{ 1 }{ 3 }\pi(\frac{ 2 }{ 5 }h)^{2}h$If we simplify this, we get$V =\frac{ 4\pi }{ 75 }h ^{3}$I haven't differentiated yet because first I wanted to know if you understand what I have done here, first. Do you understand?

2. sjmoneys96

Is that an h^3 at the end of the equation or an h^2? If it is an h^3 then yes I do understand. I thought it was an h^2, but then when I went to type a reply i realized that the equation editor on this site makes the h^2 and h^3's look exactly the same.

3. calculusfunctions

The second equation is with h cubed. I'm glad you understand.

4. sjmoneys96

but how do you actually get the derivative that was obtained by the professor?

5. calculusfunctions

@sjmoneys96 In related rate problems we differentiate all unknown quantifiable variables with respect to time. Thus if$V =\frac{ 4\pi }{ 75 }h ^{3}$then$\frac{ dV }{ dt }=(3)(\frac{ 4\pi }{ 75 })(h ^{2})(\frac{ dh }{ dt })$ $\frac{ dV }{ dt }=\frac{ 4\pi }{ 25 }h ^{2}\frac{ dh }{ dt }$

6. Stacey

When we differentiate the constants just sort of tag along.|dw:1351220408198:dw|