Report, edit, etc...Posted by DiscipleOfAdun on 2006-10-12 at 23:39:23
Eww. I don't like. I have no clue if I did it right, so I dare not post the method unless right. But, here's the answers I have for the 10 and 24 hours, truncated to the nearest whole number:
10 - 106236
24 - 151971
Dare tell me I'm even close?
10 - 106236
24 - 151971
Dare tell me I'm even close?
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-12 at 23:57:11
Your 10 value is a little off - by about 1500 - and after 24 hours that's been 1440 minutes, so you'd want to find h(1440).
Report, edit, etc...Posted by DiscipleOfAdun on 2006-10-18 at 11:35:40
, I forgot about the constant and forgot to convert 24. Hmm...maybe tomorrow I'll get it done right. That's what I get for trying this late at night.ADDITION:
I GIVE UP. I asked around, and I don't know how to properly solve it. DTBK, I don't think this will be answered. Could you do me a favor and PM me the solution?
Report, edit, etc...Posted by Mune'R0x on 2006-10-18 at 14:44:56
TeH Algebra Thread
Time to review Jr. High!
Jr. High, WTF are you talking about?
Time to review Jr. High!
Jr. High, WTF are you talking about?
Report, edit, etc...Posted by WoAHorde2 on 2006-10-18 at 15:15:50
Most people will do/do/did The ebginnings of Algebra in 7th or 8th grade.
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-18 at 17:40:39
I think he's referring to the fact that my second-semester Calculus isn't quite junior-high level.
Since I doubt anyone will solve the heat problem, here's the solution:
We are given four facts: dI/dt, the rate of input of heat, is 15000 J/s; dO/dt, the rate of output of heat, is proportional to the amount of heat, and is thus equal to kh, where k is the constant of proportionality; h(0), the initial heat, is zero, and h(2), the heat after 2 minutes, is 28000J.
dh/dt, the total propagation of heat through the water is equal to the input minus the output.
Thus, dh/dt = 15000 - kh
By algebra,
dh/dt = 15000 - kh
1/(15000-kh) dh = dt
∫1/(15000-kh) dh = ∫dt
By rules of integration:
∫1/(15000-kh) dh = ∫dt
(-1/k)ln|15000-kh| + C[sub]1[/sub] = t + C[sub]2[/sub]
By algebra*:
(-1/k)ln|15000-kh| + C[sub]1[/sub] = t + C[sub]2[/sub]
(-1/k)ln|15000-kh| = t + C
C = (-1/k)ln|15000-kh| - t
*C[sub]2[/sub] - C[sub]1[/sub] = C, just another constant.
Plugging in that h(0) = 0:
C = (-1/k)ln|15000-kh| - t
C = (-1/k)ln|15000-k(0)| - 0
C = (-1/k)ln(15000)
Plugging back in:
(-1/k)ln|15000-kh| = t + C
(-1/k)ln|15000-kh| = t + (-1/k)ln(15000)
By algebra:
(-1/k)ln|15000-kh| = t + (-1/k)ln(15000)
ln|15000-kh| = -kt + ln(15000) {multiply through by -k}
e[sup]ln|15000-kh|[/sup] = e[sup]-kt + ln(15000)[/sup]
15000 - kh = e[sup]-kt[/sup]e[sup]ln(15000)[/sup] {a[sup]m+n[/sup] = a[sup]m[/sup]a[sup]n[/sup]}
15000 - kh = 15000e[sup]-kt[/sup]
-kh = 15000e[sup]-kt[/sup] - 15000
-kh = 15000 (e[sup]-kt[/sup] - 1)
h = -15000/k (e[sup]-kt[/sup] - 1)
h = 15000/k (1 - e[sup]-kt[/sup])
Plugging in that h(2) = 28000:
28000 = 15000/k (1 - e[sup]-2k[/sup])
15000/k (1 - e[sup]-2k[/sup]) - 28000 = 0
By calculator-assisted solving of the above equation (graph and find the zero), k ~= 0.06980.
Therefore:
h(t) ~= 15000/0.0698 (1 - e[sup]-0.0698t[/sup])
h(t) ~= 214899.7 (1 - e[sup]-0.0698t[/sup])
h(10) ~= 107,920.02J
h(1440) ~= 214,899.7J
Try this one: Prove that the volume of a sphere radius 10 centered about the origin (x[sup]3[/sup] + y[sup]3[/sup] + z[sup]3[/sup] = 1000) is equal to 4000π/3 by dividing the sphere into circles parallel to the xy-plane.
Since I doubt anyone will solve the heat problem, here's the solution:
We are given four facts: dI/dt, the rate of input of heat, is 15000 J/s; dO/dt, the rate of output of heat, is proportional to the amount of heat, and is thus equal to kh, where k is the constant of proportionality; h(0), the initial heat, is zero, and h(2), the heat after 2 minutes, is 28000J.
dh/dt, the total propagation of heat through the water is equal to the input minus the output.
Thus, dh/dt = 15000 - kh
By algebra,
dh/dt = 15000 - kh
1/(15000-kh) dh = dt
∫1/(15000-kh) dh = ∫dt
By rules of integration:
∫1/(15000-kh) dh = ∫dt
(-1/k)ln|15000-kh| + C[sub]1[/sub] = t + C[sub]2[/sub]
By algebra*:
(-1/k)ln|15000-kh| + C[sub]1[/sub] = t + C[sub]2[/sub]
(-1/k)ln|15000-kh| = t + C
C = (-1/k)ln|15000-kh| - t
*C[sub]2[/sub] - C[sub]1[/sub] = C, just another constant.
Plugging in that h(0) = 0:
C = (-1/k)ln|15000-kh| - t
C = (-1/k)ln|15000-k(0)| - 0
C = (-1/k)ln(15000)
Plugging back in:
(-1/k)ln|15000-kh| = t + C
(-1/k)ln|15000-kh| = t + (-1/k)ln(15000)
By algebra:
(-1/k)ln|15000-kh| = t + (-1/k)ln(15000)
ln|15000-kh| = -kt + ln(15000) {multiply through by -k}
e[sup]ln|15000-kh|[/sup] = e[sup]-kt + ln(15000)[/sup]
15000 - kh = e[sup]-kt[/sup]e[sup]ln(15000)[/sup] {a[sup]m+n[/sup] = a[sup]m[/sup]a[sup]n[/sup]}
15000 - kh = 15000e[sup]-kt[/sup]
-kh = 15000e[sup]-kt[/sup] - 15000
-kh = 15000 (e[sup]-kt[/sup] - 1)
h = -15000/k (e[sup]-kt[/sup] - 1)
h = 15000/k (1 - e[sup]-kt[/sup])
Plugging in that h(2) = 28000:
28000 = 15000/k (1 - e[sup]-2k[/sup])
15000/k (1 - e[sup]-2k[/sup]) - 28000 = 0
By calculator-assisted solving of the above equation (graph and find the zero), k ~= 0.06980.
Therefore:
h(t) ~= 15000/0.0698 (1 - e[sup]-0.0698t[/sup])
h(t) ~= 214899.7 (1 - e[sup]-0.0698t[/sup])
h(10) ~= 107,920.02J
h(1440) ~= 214,899.7J
Try this one: Prove that the volume of a sphere radius 10 centered about the origin (x[sup]3[/sup] + y[sup]3[/sup] + z[sup]3[/sup] = 1000) is equal to 4000π/3 by dividing the sphere into circles parallel to the xy-plane.
Report, edit, etc...Posted by Centreri on 2006-10-18 at 17:47:01
All I can say is that you scare me. That is not junior high. I was 2 years ahead of the rest of my class in math, and those little letters and numbers scare me 
.
I'll do this problem later; this seems more solvable.
.I'll do this problem later; this seems more solvable.
Report, edit, etc...Posted by Deathawk on 2006-10-18 at 18:11:29
YOU CHEATER!!! THIS WAS AN ALGEBRA THREAD!!!
Report, edit, etc...Posted by DiscipleOfAdun on 2006-10-18 at 22:22:39
DTBK, you crazy person you. Anyway, I'd like to see a non-calculator solving of 28000 = 15000/k (1 - e[sup]-2k[/sup]) for k. It's the only part I wasn't able to do.
Dunno if this is exactly what you wanted...
The equation of y = √(100-z[sup]2[/sup]) has y as the radius of every circle parallel to the xy-plane for each z from (-10, 10). The volume of the sphere is given by adding the volumes of cylnders(π*r[sup]2[/sup]h, where the height tends to 0(given by dz)), making the integral of ∫π * (100 - z[sup]2[/sup])dz evaluated from -10 to 10.
By Integration, it is
π(100)z - πz[sup]3[/sup] / 3 evaluted from 10 to -10.
Evaluation gives
(1000π - (1000π)/3) - (1000π + (1000π/3))
Simplifying gives
3000π / 3 - 1000π / 3 + 3000π / 3 - 1000π / 3 =
4000π / 3
Hmm...another question for you all. Given that D[sub]x[/sub] tanh[sup]-1[/sup] = 1 / 1-x[sup]2[/sup], prove that tanh[sup]-1[/sup] = 1/2 ln(1+x/1-x).
Edited to fix sign.
Dunno if this is exactly what you wanted...
The equation of y = √(100-z[sup]2[/sup]) has y as the radius of every circle parallel to the xy-plane for each z from (-10, 10). The volume of the sphere is given by adding the volumes of cylnders(π*r[sup]2[/sup]h, where the height tends to 0(given by dz)), making the integral of ∫π * (100 - z[sup]2[/sup])dz evaluated from -10 to 10.
By Integration, it is
π(100)z - πz[sup]3[/sup] / 3 evaluted from 10 to -10.
Evaluation gives
(1000π - (1000π)/3) - (1000π + (1000π/3))
Simplifying gives
3000π / 3 - 1000π / 3 + 3000π / 3 - 1000π / 3 =
4000π / 3
Hmm...another question for you all. Given that D[sub]x[/sub] tanh[sup]-1[/sup] = 1 / 1-x[sup]2[/sup], prove that tanh[sup]-1[/sup] = 1/2 ln(1+x/1-x).
Edited to fix sign.