Melting a CubeDate: 12/20/2001 at 13:51:46 From: Asma Subject: Differentiation (calculus) Question: How long does it take for an ice cube to melt? Assume it retains its cubical shape as it melts: V = s^3, s is edge length. Assume the cube volume decreases at a rate proportional to its surface area: dv/dt = -k(6s^2) K>0 a constant. We need one more piece of information: how long will it take for a specific percentage of the ice cube to melt? This depends on many factors that are determined from the situation in which the ice cube sits. Let us set the conditions here that the cube loses 1/4 of its volume in the first hour. Thus given V = s^3 and dV/dt = -k(6s^2) V = Vo when t=0 (initially) V = (3/4).Vo when t = 1 hour Find the volume of t when V = 0 Note/hint: side length s also changes; so S(initial) is side length at t = 0rs. Date: 12/21/2001 at 03:32:26 From: Doctor Jeremiah Subject: Re: Differentiation (calculus) Hi Asma, We need to know s in terms of t (how fast s is changing). If we differentiate V = s^3, we get: V = s^3 d/dt( V ) = d/dt( s^3 ) dV/dt = 3s^2 ds/dt But dV/dt = -k(6s^2) so: -k(6s^2) = dV/dt = 3s^2 ds/dt -k(6s^2) = 3s^2 ds/dt -2k = ds/dt So ds/dt is a constant value, a straight line. The advantage of a straight line is that it doesn't matter how far apart two points are, the slope is always accurate (not true of dV/dt = -k(6s^2) because it's a parabola). The slope of a straight line is: m = ds/dt = (s1 - s0)/(t1 - t0) And that means: m = ds/dt = (s1 - s0)/(t1 - t0) -2k = (s1 - s0)/(1 - 0) -2k = (s1 - s0) We have two points on this graph: s1 is the side length at t = 1 and s0 is the side length at t = 0 V0 = s0^3 V0^(1/3) = s0 V1 = s1^3 3V0/4 = s1^3 (3V0/4)^(1/3) = s1 So then: m = ds/dt = (s1 - s0)/(t1 - t0) -2k = (s1 - s0)/(1 - 0) -2k = (s1 - s0) -2k = ((3V0/4)^(1/3) - V0^(1/3)) Now we know what k is. Since it is a straight line slope, the straight line would have a form of y = m x + b or, with our variables, s = -2k x + C. We can also get that (even though we don't need to) if we integrate the slope like this: ds/dt = -2k ds = -2k dt / / | ds = | -2k dt / / s = -2k t + C Which, as I said, is a straight line where -2k is the slope (m) and C is the y-intercept. So now we have a straight line and two points on that line, and the slope: s = -2k t + C m = ds/dt = (s1 - s0)/(t1 - t0) point at t = 0: s0 = V0^(1/3) point at t = 1: s1 = (3V0/4)^(1/3) At this point it's easy to solve because we just have to substitute in these two points and solve for the answers. So now we have a straight line and we know the slope is -2k = (3V0/4)^(1/3) - V0^(1/3), so we can substitute in a known value to get C: s = -2k t + C s0 = -2k t0 + C V0^(1/3) = -2k 0 + C V0^(1/3) = C Now we can put these back into our line equation: s = -2k t + C s = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3) And solve for when s = 0 (we don't need to solve for when the volume gets to 0 because the side length will get to 0 at the same time) s = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3) 0 = ((3V0/4)^(1/3) - V0^(1/3)) t + V0^(1/3) -V0^(1/3) = ((3V0/4)^(1/3) - V0^(1/3)) t Which gives us: t = -V0^(1/3) / ((3V0/4)^(1/3) - V0^(1/3)) So if we know the original volume V0, we can calculate how big t is when s = 0. What saved us is that the slope ds/dt (how fast s is changing) was a straight line. I don't see any other way to solve this. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 12/21/2001 at 07:37:00 From: Asma Subject: Differentiation (calculus) Thank you Dr. Jeremiah - you have been a lot of help. I appreciate your time and thinking spend on my problem. Once again thank you. |
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