Answer to Question #187475 in Differential Equations for imran

Question #187475
Determine ‘a’ and ‘b’ as to make the cylinder y2= 4ax orthogonal to the ellipsoid at the point (1, 2, 1).
Find the directional derivative of f(x,y)=2x^2+y^2 at the point (-2,3) in the direction of u ⃗=4i ̂-5j ̂.

Show that the function f(x,y)=e^x siny+e^y cosx satisfies the Laplace’s equation
1
Expert's answer
2021-05-07T10:31:51-0400

Ans:-

f(x,y)=exsiny+eycosxf(x,y)=e^x siny+e^y cosx

partial derivative with respect to x partial derivative with respect to y

fx=exsinyeysinx\frac{\partial f}{\partial x} =e^x siny-e^ysinx , fy=excosy+eycosx\frac{\partial f}{\partial y}=e^xcosy+e^ycosx


again partial derivative with respect to x again partial derivative with respect to y

2fx2=exsinyeycosx\frac{\partial^2 f}{\partial x^2}=e^xsiny-e^ycosx (i)-(i) 2fx2=exsiny+eycosx\frac{\partial^2 f}{\partial x^2}=-e^xsiny+e^ycosx (ii)-(ii)


Add these two equations

2fx2+2fy2=exsinyeycosx+(exsiny+eycosx)\Rightarrow \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}=e^xsiny-e^ycosx+(-e^xsiny+e^ycosx) =0=0


2fx2+2fy2=0\Rightarrow \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}=0

Hence Laplace's equation will be satisfied.




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