I am lost again...
Do i just put the u value ( x^2 +y^2) back in the equation and solve for x and then do the same thing but with ∂u/∂y for y?
my x and y values were 2.0 and 2.7 respectively.
that symbol thing helps;
so
dV/du = -175*u^(-1.5)
now understanding that its a ∂u/∂x with respect to x; =2x and with respect to y; =2y
but how to i complete the equation to get an equation
-175*x^(1.5)/2x and -175*y^(1.5)/2y for the x and y directions respectively?
ok so dV/dx = dV/du * du/dx
dV = .5*350u^-.5 = 175u^-.5
du = 2x + 2y
175u^-.5/2x+2y * 2x+2y/2
=175u^-.5/2
for the x partial derivative and
175u^-.5/2 for the y partial
doesn't look right
yea the equation 350/sqrt (x^2 +y^2) is given as what voltage equals and x and y are in meters. It then asks what the strength of the electric field is at x=2 and y = 2.7. I have to partially derive the equation. Once for x and once for y. Put the values in and solve for each component. I just...
nevermind its -dv/dx which would give the -150 answer thanks...
Finished that problem and now i have to do another one involving an x and a y component. Obviously you have to break the thing down into x and y but again I'm struggling on the partial derivative part.
350/(sqrt (x^2 + y^2))
so when i solve for x=1 i get 150. I put that answer in and it still says its wrong but to check my signs.
I changed it to -150 and its correct... am i missing something?
Homework Statement
The electric potential along the x-axis is V = 50x - (100/x)V, where x is in meters. What is E_x at x = 1.0m
Homework Equations
partial derivative of dV/dx, then plug in the value for x
The Attempt at a Solution
I understand the concept of trying to solve the...
Homework Statement
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