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richyw
Group Title
Find an equation for the tangent plane to the surface given by\[z=\ln{\left(1+x^2+y^2\right)}\]
 one year ago
 one year ago
richyw Group Title
Find an equation for the tangent plane to the surface given by\[z=\ln{\left(1+x^2+y^2\right)}\]
 one year ago
 one year ago

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richyw Group TitleBest ResponseYou've already chosen the best response.0
Sorry, at the point (0,2,ln5). why can't I just say\[z_x=\frac{2y}{\left(1+x^2+y^2\right)}\]\[z_y=\frac{2x}{\left(1+x^2+y^2\right)}\]And then that \[zln(5)=z_x(0,2)(x0)+z_y(0,2)(y2)\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
so \[z=\frac{4}{5}x+\ln(5)\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I don't see where I am going wrong....
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
nevermind. I see where I went wrong haha.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
is it that you don't have the coefficient for z, since you didn't turn it into a function? I want to know because your way is different than mine.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
er, I mean didn't turn it into a 3variable function...
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
my notation was sloppy. All I do to find the plane tangent to \(z=f(x,y)\) at the point \(P\left(a,b,f(a,b)\right)\) is use the formula \[zf(a,b)=f_x(a,b)(xa)+f_y(a,b)(yb)\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I just accidentally put f_y where f_x should have been...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
ah, gotchya thanks
 one year ago
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