Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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 \[z-ln(5)=z_x(0,2)(x-0)+z_y(0,2)(y-2)\]
so \[z=\frac{4}{5}x+\ln(5)\]
I don't see where I am going wrong....

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

nevermind. I see where I went wrong haha.
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.
er, I mean didn't turn it into a 3-variable function...
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 \[z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)\]
I just accidentally put f_y where f_x should have been...
ah, gotchya thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question