anonymous
  • anonymous
The surface area of a right circular cylinder of height 5 feet and radius r feet is given by S(r)=2πrh+2πr2. Find the instantaneous rate of change of the surface area with respect to the radius, r, when r = 6.
Mathematics
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anonymous
  • anonymous
The surface area of a right circular cylinder of height 5 feet and radius r feet is given by S(r)=2πrh+2πr2. Find the instantaneous rate of change of the surface area with respect to the radius, r, when r = 6.
Mathematics
jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
\[\Large\rm S(r)=2\pi r h+2\pi r^2\]Hey Camila! :) Have you learned your derivative shortcuts at this point? Or do we have to use the limit definition for the derivative to find our instantaneous rate of change? -_-
anonymous
  • anonymous
I know how to derive.
zepdrix
  • zepdrix
We're holding the height, h, constant at 5.\[\Large\rm S(r)=2\pi r \cdot 5+2\pi r^2\]Simplifies our function a little bit, which is nice.\[\Large\rm S(r)=10\pi r+2\pi r^2\]

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zepdrix
  • zepdrix
So apply your power rule to each term, should be pretty straight forward :) Are you getting confused because of all that pi nonsense mixed in?
anonymous
  • anonymous
Yeah that was the part that confused me.
zepdrix
  • zepdrix
\[\Large\rm (10\pi x)'=10\pi(x)'=10\pi (1)\]Normal power rule, ignoring the constant coefficients :)
anonymous
  • anonymous
So I would get 10π+4πr and plug in 6 for r?
zepdrix
  • zepdrix
\[\Large\rm S'(r)=10 \pi+4\pi r\]Good good good.
anonymous
  • anonymous
So when I plug that in it would be 34π right?
zepdrix
  • zepdrix
Yayyy good job! *\c:/* Camila <*c:/*
anonymous
  • anonymous
Haha okay thank you so much!
zepdrix
  • zepdrix
np

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