October 30th, 2016, 01:03 PM  #1 
Newbie Joined: Oct 2016 From: Toronto Posts: 3 Thanks: 0  Question about related rate
A downward cone is melting. Assume that the radius r and the height h always satisfy h = 6r. The radius r, height h, and slant height L. 1. Express l as a function of r. L(r)= sqrt((6pi^2)+r^2) is this correct? 2 Find a formula for the lateral surface area A as a function of r. La(r)= pi*r*L is this correct? 3. Find a formula for the volume V as a function of the slant height L. V(L)= La + pi*r^2 is this correct ? 4. Suppose that the lateral surface area is decreasing at a constant rate of 30pi mm2/hr. Find an equation which gives dr/dt in terms of r. How do I do this question? 5. At what rate is the radius changing when r = 300mm? 
October 30th, 2016, 01:13 PM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1474  Quote:
$L = \sqrt{r^2 + h^2} = \sqrt{r^2 + (6r)^2} = \sqrt{37r^2} = r\sqrt{37}$ $L(r) = \sqrt{37}r$ Quote:
Use the fact that $h=6r$ to express $L$ as a function of $r$ (i.e. problem 1) Quote:
find the formula for the volume of a cone in $r$ and $h$ and then use the results of (1) and $h=6r$ to convert that formula into one solely in terms of $L$ Quote:
Last edited by romsek; October 30th, 2016 at 01:15 PM.  
October 30th, 2016, 05:40 PM  #3  
Newbie Joined: Oct 2016 From: Toronto Posts: 3 Thanks: 0  Quote:
La(r)= pi*r*L > pi*r*r*sqrt(37) > pi*r^2*sqrt(37) Is this correct ? 3. Find a formula for the volume V as a function of the slant height L. The volume of the cone is 1/3*pi*r^2*h. I don't understand how to do next.  

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