Partial Derivative question! ..... tan^-1(y/x)

can anyone at all do this question please. rep will be given. :

find Gx, Gy, Gxx, Gyy, Gxy for G(x,y) = tan^-1(y/x) ... note tan^-1(u) = 1/(1+u^2).

cheers.

Reply 1

zCtpts2T

Hi, hope this helps:

G(x,y) = z = arctan(y/x) (arctan = tan^-1, just different notation)

therefore tan(z) = y/x;
differentiate both sides with respect to y:

sec^2(z) dz/dy = 1/x (sec^2(z) = tan^2(z) + 1)

and tan^2(z) = (y/x)^2;

so dz/dy = 1/(x(1 + (y/x)^2));

as for dz/dx, same procedure reveals:

dz/dx = -1/(y(1 + (x/y)^2));

differentiating again with respect to the same variables gives:

d^2z/dx^2 = 2xy/(x^2 + y^2)^2;

d^2z/dy^2 = -2xy/(x^2 + y^2)^2;

and the cross derivative gives:

d^2z/dxdy = 1/(x^2 + y^2) - 2x^2/(x^2 + y^2)^2;

I believe these to be correct, however there may be sign errors in my workings out as I rattled through these quickly. Any that aside, this is the general jist of how to do these derivatives,

Hope this helps.

Reply 2

sul

thanks!! i believe ur answers are correct!

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