Partial Derivative Help?!
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Partial Derivative Help?!

[From: ] [author: ] [Date: 11-10-21] [Hit: ]
-dU/dy) also known as negative quadrilar of -6pi square / 6r7. Hope that helped!......
A potential energy function for a two-dimensional force is of the form U = (3x^2)y - 6x. Find the force that acts at the point (x, y). (Use x and y as appropriate.)

I know that:
F(x,y) = (-dU/dx, -dU/dy) [the partial derivatives]
I just can't remember how to do the math part of this physics problem, if someone could explain how, that would be great! Thanks!

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When taking partial derivatives, remember to differentiate only one variable at a time and hold all other variables constant.

∂U/∂x= 6xy - 6
∂U/∂y= 3x^2

F(x,y) = (-6xy +6, -3x^2)

EDIT: pay no attention to above poster he is trolling

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Ok so in this equation "X" equals Delta Phi. Based on the common principle of Derpington's first law of derivatives, You take "Delta Phi" and square it by 2. From there you will rotate it on it's axis and turn it in a tri-dimensional equation. You than take F(x,y) which as you said equals (-dU/dx, -dU/dy) also known as negative quadrilar of -6pi square / 6r7. Hope that helped!

EDIT: Pay no attention to the below poster he is trolling
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