<p>Radius is 2 in, Height is 12 inches. </p>
<p>R(t)= 2 in^3/min</p>
<p>What is the rate of water level drop when</p>
<p>(a) water is 6 inch deep</p>
<p>(b) cup is half full</p>
<p>I'm totally stumped, can anyone help me with this one?</p>
<p>ok, first of all, can you please type out the problems just as it says wherever you got this from?</p>
<p>what is R(t), is it the rate of change of the radius?</p>
<p>what is water level drop? Is it the change in volume of the cone? </p>
<p>just type ou the problem AS IS, and I;ll be happy to guide you through it :)</p>
<p>The Radius is 2 in and Height is 12 inches. Water is leaking out of this cone and the rate of change in volume is 2 in^3/min.</p>
<p>R(t)= 2 in^3/min(change in volume)</p>
<p>What is the rate of water level drop when</p>
<p>(a) water is 6 inch deep</p>
<p>(b) cup is half full</p>
<p>(this was from my friend and since he isn't around, I don't have the specifics, sorry) Thanks.</p>
<p>Ok, here's how to crack it (lol..quote from princeton review)</p>
<p>V = pi<em>r^2</em>h</p>
<p>ratio of radius/height = 1/6---> so radius = (height / 6)</p>
<p>so Volume = (pi/36) * h^3
dV/dt = (pi/12) * h^2 * dh/dt</p>
<p>dh/dt is the rate of change of the height; this is what your'e looking for in both parts of the problem...</p>
<p>so when water is 6-inches deep; h = 6;</p>
<p>so plug in all the variables in the above bold equation, and you get for part a.and you get your answer...</p>
<p>for part b. I guess its the same thing; because when cup is half full the height is 6 again...? Yeah so its the same answer...</p>
<p>r = 2
h = 12
dV/dt = -2</p>
<p>a) dh/dt = ? when h = 6
b) dh/dt = ? when V = 24pi</p>
<p>a)
r = h/6
h = 6</p>
<p>V = pi(r^2)h
= pi(h/6)^2<em>h
= pi/36 * h^3
dV/dt = pi/36</em>3h^2 dh/dt
-2 = pi/36<em>3</em>6^2 dh/dt
-2 = 3pi dh/dt
dh/dt = -2/(3pi)
= about -0.212 in/min</p>
<p>b)
r = h/6
V = 24pi</p>
<p>24pi = pi(h/6)^2*h
24 = 1/36 * h^3<br>
864 = h^3
h = 9.52</p>
<p>From part a)
dV/dt = pi/36*3h^2 dh/dt
-2 = pi/36 * 3(9.52^2) dh/dt
-2 = pi/36 * 271.8912 dh/dt
dh/dt = -2 * 36/pi * 1/271.8912
= -0.0843 in/min</p>
<p>I hope this is correct, but I just got up (no school today) and I'm only half awake.</p>
<p>^yeah that's right...I did my part b wrong...I guess I'm half a sleep...lol...</p>
<p>I made a mistake in my arithmetic...my TI-89 broke yesterday :(</p>
<p>so yeah, your answer is right..GOOD JOB!</p>