anonymous
  • anonymous
the radius of a copne is increasing at a rate of 3in/sec, and the height of the cone is 3 times the radius. Find the rate of change for the volume of that cone when the radius is 7 inches
Mathematics
chestercat
  • chestercat
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nowhereman
  • nowhereman
So you know the volume of a cone is \(\frac{1}{3}πr^2h\) so if as given \(h = 3r\) you get \(V = πr^3\). You also know that \(\dot{r} = 3in/s\) and thus \(\dot{V} = 3πr^2\dot r\) which for \(r = 7 in\) gives you \(\dot V = 3^2\cdot 7^2πin^3/s\).
anonymous
  • anonymous
Some how I think that I am missing a step. [V=1\div3\times \Pi \times r^2\ times\h\] knowing that dr/dt = 3 in/sec h = 3r V =(3) * (1/3) * pi * r^2 * (dr/dt ) dv/dt =(3) * (1/3) * pi * r^2 * (3 )
anonymous
  • anonymous
thank you nowhereman very helpfull

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