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Partial derivatives of four different variables
PostPosted: Sun Oct 15, 2006 8:29 pm
by Ingemar
Hello lovely CAA members. I have a calculus problem (well, an error propagation problem) but I don't exactly have the time to dig through my OOOOLD calc notes.
I need the derivative of tau with respect to each of the four variables--t1, V1, t2 and V2.
The problem is right on the attatchment.
PostPosted: Sun Oct 15, 2006 11:11 pm
by Slater
in before Technomancer
(btw, calculus is magic, and I ain't no magician, sorry ^^; )
PostPosted: Sun Oct 15, 2006 11:17 pm
by Ingemar
The t variables seem easy enough but the V variables look like a real pain.
PostPosted: Mon Oct 16, 2006 5:45 am
by Warrior4Christ
Multiplying by 'e' does not remove the ln. You'd have to have e^ln(x) to cancel them.
So the answers are:
d tau/dt_1 = -1/ln(V_1/V_2)
d tau/dt_2 = 1/ln(V_1/V_2)
d tau/dV_1 = (t_1 - t_2)/(ln(V_1/V_2))^2 * (1/V_1)
d tau/dV_2 = (t_2 - t_1)/(ln(V_1/V_2))^2 * (V_1/(V_2)^2)
I hope that's understandable...
PostPosted: Mon Oct 16, 2006 12:25 pm
by Ingemar
How did you come to that conclucsion?
And for the first two, why did the other t variable disappear?
PostPosted: Wed Oct 18, 2006 5:32 am
by Warrior4Christ
Ingemar wrote:How did you come to that conclucsion?
And for the first two, why did the other t variable disappear?
Take all other variables but the you are interested in as constants. Recall that the derivative of a constant term is zero.
Use the derviative rules:
(f(x)^n)` -> n*f(x)^(n-1) * f`(x)
eg. d(5 * t_1 + k* t_2)/dt_2 = k, as the t_1 is treated as a constant term.
(ln(f(x))` -> f`(x)/f(x)
eg. d(ln(V_1/V_2))/dV_1 = (1/V_2)/(V_1/V_2) = 1/V_1
PostPosted: Wed Oct 18, 2006 9:32 pm
by Icarus
First off, let's rename the variable for simplicity.
tau = T
t_1 =t
t_2 = s
v_1 = v
v_2 = y
Also, from the rules of logarithms, ln(v/y) = ln v - ln y. Thus,
dy/dT = (s-t) / [(ln v - ln y)^2 ] * y^-1
PostPosted: Thu Oct 19, 2006 5:32 am
by Warrior4Christ
Warrior4Christ wrote:Multiplying by 'e' does not remove the ln. You'd have to have e^ln(x) to cancel them.
So the answers are:
d tau/dt_1 = -1/ln(V_1/V_2)
d tau/dt_2 = 1/ln(V_1/V_2)
d tau/dV_1 = (t_1 - t_2)/(ln(V_1/V_2))^2 * (1/V_1)
d tau/dV_2 = (t_2 - t_1)/(ln(V_1/V_2))^2 * (V_1/(V_2)^2)
I hope that's understandable...
Err, yeah.. sorry, that last one should be:
d tau/dV_2 = (t_2 - t_1)/(ln(V_1/V_2))^2 * (V_1/(V_2)^2) / (V_1/V_2)
-> d tau/dV_2 = (t_2 - t_1)/(ln(V_1/V_2))^2 * (1/V_2)