Monday, May 15, 2006

Episode 6: Solutions

a) dT/dh = k(T-5050)
Separate the variables.
dT(1/(T-5050)) = k dh
Antidifferentiate both sides of the equation.
ln(T-5050) = kh +C
T -5050 = e^(kh+C)
T = e^(kh+C) + 5050

b) Using the general solution, and the points (0,5000) and (150,3610.5405), solve for the exact equation.
T = e^(kh+C) + 5050
First use the point (0,5000) to solve for one of the variables.
5000 = (e^(k(0)))(e^(C)) +5050
-50 = e^(C)
T = -50e^(kh) + 5050
Now using the other point (150,3610.5405), solve for the last unknown.
3610.5405 = -50(k(150)) +5050
-1439.4595 = -50e^(150k)
Solve for k.
-1439.4595/-50 = e^(150k)
28.7892 = e^(150k)
ln(28.7892) = 150k
k = (1/150)ln(28.7892)
k = 0.0224
T = -50e^(0.0224h) +5050

c) Using the fact that the element can only be used at 4444 Kelvin, you have to insert that into the equation and solve for the height, h.
T = -50e^(0.0224h) +5050
4444 = -50e^(0.0224h) +5050
-606 = -50e^(0.0224h)
-606/-50 = e^(0.0224h)
12.12 = e^(0.0224h)
ln(12.12) = 0.0224h
h = (1/0.0224)ln(12.12)
h = 111.3775 cm
The height that the element has to be held at in order to reforge the sword is 111.3775 cm.

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