Question #281682

Let the electric equipotential lines (curves of constant potential) between two concentric cylinders with the z-axis in space be given by u(x,y) = x^2:+:y^2 = c (these are circular cylinders in xyz-space). Using the method in the text, find their orthogonal trajectories (the curves of electric force).


1
Expert's answer
2021-12-21T17:56:41-0500

Solution:


I) Differentiate the equation x2+y2=cx^2+y^2=c with respect to x:



2x+2yy=0    y=xy2x+2yy^{'}=0\iff y^{'}=-\frac{x}{y}

II) Since the equation of orthogonal trajectories is:



y=1f(x,y)y^{'}=-\frac{1}{f(x,y)}

and in this case f(x,y)=xy,f(x,y)=-\frac{x}{y}, we obtain:


y=1xy=yxy^{'}=-\frac{1}{\frac{x}{y}}=\frac{y}{x}

III) Now, let`s solve the differential equation:


y=yx    dydx=yx    dyy=dxxy^{'}=\frac{y}{x}\iff \frac{dy}{dx}=\frac{y}{x}\iff\frac{dy}{y}=\frac{dx}{x}



IV) Integrate the left side in relation to y, and the right side in relation to x:



dyy=dxx    lny=lnx+c\int \frac{dy}{y}= \int \frac{dx}{x} \iff ln|y|=ln|x|+c

V) By taking exponents, we obtain:



y=Cxy=Cx


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