- #1
JFonseka
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Just when I thought I got the hang of tangent planes and surfaces there comes a question I haven't quite seen before
z = ln (x[tex]^{2}[/tex]+3y[tex]^{2}[/tex])
Find a normal vector n and the equation of the tangent plane to the surface at the point
(2, -1, ln 7)
So keeping the cartesian equation in mind:
z = z[tex]_{0}[/tex] + F[tex]_{x}[/tex](x,y)(x - x[tex]_{0}[/tex]) + F[tex]_{y}[/tex](x,y)(y - y[tex]_{0}[/tex])
Partial derivative with respect to x: [tex]\frac{2x}{x^{2} + 3y^{2}}[/tex]
Evaluate with the values and I get 4/11
Partial derivate with respect to y: [tex]\frac{6y}{x^{2} + 3y^{2}}[/tex]
Evaluate with the values and I get -6/11
Have I done this correctly?
So putting it all in the equation I get z = ln 7 + 4x/11 -6y/11 -14/11
I get the feeling I did something wrong somewhere, and of course the normal vector would be (4/11, -6/11, -1) if everything was right.
z = ln (x[tex]^{2}[/tex]+3y[tex]^{2}[/tex])
Find a normal vector n and the equation of the tangent plane to the surface at the point
(2, -1, ln 7)
So keeping the cartesian equation in mind:
z = z[tex]_{0}[/tex] + F[tex]_{x}[/tex](x,y)(x - x[tex]_{0}[/tex]) + F[tex]_{y}[/tex](x,y)(y - y[tex]_{0}[/tex])
Partial derivative with respect to x: [tex]\frac{2x}{x^{2} + 3y^{2}}[/tex]
Evaluate with the values and I get 4/11
Partial derivate with respect to y: [tex]\frac{6y}{x^{2} + 3y^{2}}[/tex]
Evaluate with the values and I get -6/11
Have I done this correctly?
So putting it all in the equation I get z = ln 7 + 4x/11 -6y/11 -14/11
I get the feeling I did something wrong somewhere, and of course the normal vector would be (4/11, -6/11, -1) if everything was right.
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