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Need help taking partial derivatives

by paul2001
Tags: derivatives, partial
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paul2001
#1
Aug11-08, 09:55 AM
P: 2
Hello everyone,

I am trying to take partial derivatives of the following equation and I am having difficulties. The partial derivatives are of Q w.r.t D, ΔP, ρ, and w. Any help would be much appreciated. Thank you.

Paul
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Airflow Equation.JPG  
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HallsofIvy
#2
Aug11-08, 11:50 AM
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Quote Quote by paul2001 View Post
Hello everyone,

I am trying to take partial derivatives of the following equation and I am having difficulties. The partial derivatives are of Q w.r.t D, ΔP, ρ, and w. Any help would be much appreciated. Thank you.

Paul
The function is
[tex]Q= 5.9863 CD^2\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of D, that is just D2 and the derivitive of that is 2D.
[tex]Q_P= 2(5.9863) CD\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of [itex]\Delta P[/itex], that is just [itex](\Delta P)^{0.5}[/itex] and the derivative of that is [itex]0.5(\Delta P)^{-0.5}[/itex]
[tex]Q_{\Delta P}= 0.5 (5.9863) CD^2\left[\frac{1}{\Delta P\rho(1-\beta^4)}\right]^{0.5}[/tex]

As a function of [itex]\rho[/itex] it is [itex]1/\rho^{0.5}= \rho^{-0.5}[/itex] and the derivative of that is [itex]-0.5\rho^{-1.5}[/itex]
[tex]Q_{\rho}= -0.5(5.9863) CD^2\left[\frac{\Delta P}{\rho^3(1-\beta^4)}\right]^{0.5}[/tex]

I do not see any "w" in the formula so the derivative with respect to "w" would be 0!

There is a "C" that you did not mention.
[tex]Q_C= 5.9863 D^2\left[\frac{\Delta P}{\rho(1-\beta^4)}\right]^{0.5}[/tex]

There is also a "[itex]\beta[/itex] in the formula. That's a little more complicated because the function involves [itex](1- \beta^4)^{-0.5}[/itex] and the derivative of that, by the chain rule, is [itex]-0.5(1- \beta^4)^{-1.5}(-4\beta^3)= 2(1- \beta^4)^{-1.5}\beta^3[/itex]
[tex]Q_{\beta}= 2(5.9863) CD^2\beta^3\left[\frac{\Delta P}{\rho(1-\beta^{12}}\right]^{0.5}[/tex]

Since this has nothing directly to do with differential equations, I am moving it to "Calculus and Analysis".
Alex6200
#3
Aug11-08, 12:37 PM
P: 75
Just pretend that whatever variable you're not taking the partial derivative with respect to is a constant. For example if you're taking it with respect to B, just treat all of the other variables like constants, and differentiate it the way you would:

[tex]\sqrt{1 - B^4}[/tex]

using the chain rule since everything else is just a constant multiple that isn't going to be impacted by integration.

paul2001
#4
Aug11-08, 01:10 PM
P: 2
Need help taking partial derivatives

I have attached the same equation with the appropriate substitutions. The problem I am having is that the equation is iterative so I cannot simply take the partial derivatives without solving for Q. Is there a special approach I can use for this type of derivation? Thank you again.

Paul
Attached Thumbnails
Airflow Equation 2.JPG  
HallsofIvy
#5
Aug11-08, 01:24 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,549
Well, you don't have to solve for Q, you can use "implict differentiation".
Alex6200
#6
Aug11-08, 01:47 PM
P: 75
Yeah, just put dQ/dwhateva on both sides and then take it like both sides were the expression.


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