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sjmoneys96
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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?
 one year ago
 one year ago
sjmoneys96 Group Title
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?
 one year ago
 one year ago

This Question is Open

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

sjmoneys96 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.1
The second equation is with h cubed. I'm glad you understand.
 one year ago

sjmoneys96 Group TitleBest ResponseYou've already chosen the best response.0
but how do you actually get the derivative that was obtained by the professor?
 one year ago

calculusfunctions Group TitleBest ResponseYou've already chosen the best response.1
@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\[\frac{ dV }{ dt }=(3)(\frac{ 4\pi }{ 75 })(h ^{2})(\frac{ dh }{ dt })\] \[\frac{ dV }{ dt }=\frac{ 4\pi }{ 25 }h ^{2}\frac{ dh }{ dt }\]
 one year ago

Stacey Group TitleBest ResponseYou've already chosen the best response.0
When we differentiate the constants just sort of tag along.dw:1351220408198:dw
 one year ago
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