<p>I'm having a lot of trouble getting these problems right. So can someone give me a general idea on how to solve related rate problems? </p>
<p>So here's an example from Barron's:
The diameter and height of a paper cup in the shape of a cone are both 4 inches, and water is leaking out at the rate of 1/2 cubic inch per second. Find the rate at which the water level is dropping when the diameter of the surface is 2 inches.</p>
<p>First figure out what you’re given. Identity unknowns and link them to some variable. Here, h = 4 inches, r = 4/2 inches, and dV/dt = .5 cubic inches /sec. dV/dh is unknown because water level refers to height I think. </p>
<p>Do not substitute values until the very end. And draw a sketch of the problem.</p>
<p>Now you need a formula to relate the height and volume of the cone. The formula is V=1/3<em>pi</em>r^2<em>h What you need to do before you differentiate is find r in terms of h. To do this you should use the fact that the cone’s diameter and height is 4 inches so d = h and because radius is 2 inches then r = 1/2h. Now V = 1/3</em>pi<em>(h^3/4) or 1/12pi</em>h^3</p>
<p>Now you can differentiate and substitute to find your answer. </p>
<p>Anyway, do you have a calculus textbook? Calc textbooks will have guidelines of problem solving and are applied to relate rates / optimization problems</p>
<p>with related rates you use implicit differentiation 90% of the time</p>
<p>We know that V = (πr^2*h)/3. Note that r = h/2 (since diameter = height), so we can find V in terms of h:</p>
<p>V = (π/12)h^3</p>
<p>Here we find dV/dt. Since h is also a function of t, we must use the chain rule.</p>
<p>dV/dt = (π/12)(3h^2)(dh/dt) = (π/4)(h^2)dh/dt.</p>
<p>You know what dV/dt is, it’s 1/2 in^3 per second. Also, when d = 2 inches, h = 2 inches (since the diameter and height are always equal). Substitute everything in and solve for dh/dt.</p>
<p>thanks guys!</p>