What is the electric potential derivative when x=2 and y=1?

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SUMMARY

The electric potential derivative for the equation V = 200/(x² + y²)^(1/2) at the point (x=2, y=1) is calculated to be dV = -53.7 V/m. The solution involves applying the chain rule and exponent rules to derive the partial derivatives with respect to x and y. The final expression for dV incorporates both variables and confirms the correctness of the calculations performed by the user.

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Homework Statement

The electrical potential can be described by the following equation:
V= 200/(x2 + y2 )1/2 find dV when x=2 and y=1


Homework Equations


n/a


The Attempt at a Solution


dv/d(x,y) = ∂/∂x + ∂/∂y
=200/(x2 + y2)(1/2) +200/(x2 + y2)(1/2)
replace variables with C where necessary and rearrange for easy of deriving =200(x2 + C)(1/2) +200(C + y2)(1/2)
Chain rule and exponent rule
=200(-1/2)(2x)(x2 + C)(-3/2) +200(-1/2)(2y)(C + y2)(-3/2)
=-200x(x2 + C)(-3/2) + -200y)(C + y2)(-3/2)
=-200x(x2 + y2) )(-3/2) + -200y)(x2+ y2)(-3/2)
Then I plug in the values x=2 and y=1 and dV=-53.7 V/m, I'm not very confident with multivariable calculus and just wanted to check to see if this is right.
 
Last edited:
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The result looks okay.
 

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