<p>My mid-term is tomorrow, and I cannot get the answer to this problem for the life of me. If you would explain it to me, I would be forever thankful hahaha.</p>
<p>A cone shaped cup is 12 inches across the top, and 15 inches deep (vertex down). It is being filled at 3 cubic inches per second. At what rate is the radius changing when the height is 10 inches. V=1/3pi(r)^2(h) for a cone.</p>
<p>THANK YOU!!!!</p>
<p>I don't know, I probably messed up, but I think I got 15/(8 pi) in/sec</p>
<p>that's not what my teacher's key says... it says the answer is 3/40pi in/sec.... and i don't really know how to do the problem</p>
<p>The teacher is probably right, I suck at related rates.</p>
<p>What I did was first establish a relationship between r and h using the given 15 and 6. I got r = 6/15 * h. Then I filled it in the volume equation for r and got V = 12<em>pi</em>h^3/225.
Then I took the derviative of that, filled in 10 for h and 3 for dV/dT and solved for dH/dT.</p>
<p>I probably messed up somewhere.</p>
<p>Oh crap! I solved for the rate the height was changing, my fault!</p>
<p>yeah lol, i know how to do that.... i just don't really know how to solve for the rate of the radius changing while still using the 10 for h...</p>
<p>Yeah, I got dR/dT = 3/ (40 pi)</p>
<p>Oh, you have to write it interms of r.</p>
<p>When you establish that r = 6/15 * h as a relationship, you fill in 10 for h. solving then you get r= 4 when h = 10.</p>
<p>Go back to the volume equation and fill in h=15/6*r for the h.</p>
<p>You get V= (5 pi) r^3/6.</p>
<p>After the derivative, you fill in 4 for r and 3 for dV/dT, and solve for dR/dT. you get 3/ 40 pi</p>
<p>oh ok that makes sense!!! thank you so much</p>
<p>You're Welcome :)!</p>