Find the point on the graph of z=−x^2+y^2 at which vector n= is normal to the tangent plane.

Question

Find the point on the graph of z=−x^2+y^2  at which vector n=<4,−6,1>  is normal to the tangent plane.

anonymous · Answer

How do I approach this? I'm completely lost right now!

amistre64 · Answer

hmm

i beleive the derivative of a surface IS the normal vector equations

amistre64 · Answer

$$D[z=-x^2+y^2]\to <\frac{dz}{dx},\frac{dz}{dy},\frac{dz}{dz}>$$ $$<-2x, 2y, 1>$$

amistre64 · Answer

so when x=-2 and y = 3, we should have it

anonymous · Answer

Nope. WeBWorK tells me that is not the correct answer.

amistre64 · Answer

then ill have to review how to find a gradient of a surface ...

amistre64 · Answer

http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx last problem i believe is the same structure is this one

amistre64 · Answer

y = -3 might be better since we are trying to get a -6 from 2y

amistre64 · Answer

plug in for x=-2, and y=-3 to find the value for z then your point is:

(-2,-3,z)

anonymous · Answer

I tried that with z=1, and it told me that -3 is correct for the y-coordinate, but the other two were still wrong. Changing the sign on the x-coordinate gave me two correct coordinates, but I still don't know how to get the z-coordinate.

amistre64 · Answer

you are given that:  z = -x^2 + y^2

if you know x and y, plug them in to get "z"  :)

anonymous · Answer

Oh...duh!

amistre64 · Answer

-2x = 4;  when x=-2   so either your problem is entered wrong on here or ther eis a misprint in your material

anonymous · Answer

I copied it directly, so I'm guessing it's wrong on the server side of WeBWorK. Thank you so much though!

amistre64 · Answer

youre welcome :)  and good luck