Partial derivatives of four different variables

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Partial derivatives of four different variables

Postby Ingemar » Sun Oct 15, 2006 8:29 pm

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.
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Postby Slater » Sun Oct 15, 2006 11:11 pm

in before Technomancer

(btw, calculus is magic, and I ain't no magician, sorry ^^; )
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Postby Ingemar » Sun Oct 15, 2006 11:17 pm

The t variables seem easy enough but the V variables look like a real pain.
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Postby Warrior4Christ » Mon Oct 16, 2006 5:45 am

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...
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Postby Ingemar » Mon Oct 16, 2006 12:25 pm

How did you come to that conclucsion?

And for the first two, why did the other t variable disappear?
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Postby Warrior4Christ » Wed Oct 18, 2006 5:32 am

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
Everywhere like such as, and MOES.

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Postby Icarus » Wed Oct 18, 2006 9:32 pm

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
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Postby Warrior4Christ » Thu Oct 19, 2006 5:32 am

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)
Everywhere like such as, and MOES.

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