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 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
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 Originally Posted by Apple11 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?
where would $\pi$ come in?

$L = \sqrt{r^2 + h^2} = \sqrt{r^2 + (6r)^2} = \sqrt{37r^2} = r\sqrt{37}$

$L(r) = \sqrt{37}r$

Quote:
 2 Find a formula for the lateral surface area A as a function of r. La(r)= pi*r*L is this correct?
They want you to express this solely as a function of $r$.

Use the fact that $h=6r$ to express $L$ as a function of $r$ (i.e. problem 1)

Quote:
 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 ?
you've got $r$'s in your formula. That's not allowed.

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:
 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?
get 1-3 correct first and we'll go on to 4 and 5

Last edited by romsek; October 30th, 2016 at 01:15 PM. October 30th, 2016, 05:40 PM   #3
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 Originally Posted by romsek where would $\pi$ come in? $L = \sqrt{r^2 + h^2} = \sqrt{r^2 + (6r)^2} = \sqrt{37r^2} = r\sqrt{37}$ $L(r) = \sqrt{37}r$ They want you to express this solely as a function of $r$. Use the fact that $h=6r$ to express $L$ as a function of $r$ (i.e. problem 1) you've got $r$'s in your formula. That's not allowed. 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 $L$ get 1-3 correct first and we'll go on to 4 and 5
2 Find a formula for the lateral surface area A as a function of r.
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. Tags question, rate, related Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post latzi Calculus 1 July 25th, 2009 06:07 AM titans4ever0927 Algebra 2 February 25th, 2009 06:15 PM titans4ever0927 Calculus 1 February 25th, 2009 11:48 AM squeeze101 Algebra 2 October 31st, 2008 04:10 PM mangox Calculus 9 October 25th, 2008 02:50 PM

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