cal 3 question, looking for horizontal tangent plane on a surface?

Question

z=(2x^2+3y^2)exp(-x^2-y^2)

i am so confused about what to do after i get dz/dx and dz/dy

these two partial derivatives are so complicated, i don't no if it is possible to solve for points where there exists a horizontal tangent plane BY HAND.
if you know how to solve this question BY HAnd, NOT BY COMPUTER, please show me how.
thx

Answer 1

Well... solve for the locations where Zx and Zy are both 0. Something that probably escaped your notice is that the exponential function is always > 0, so once you factor that out, you can simply drop it in your calculations of Zx = 0 and Zy = 0.

Zx = 4x(e^(-x^2 - y^2)) - (4x^3 + 6xy^2)(e^(-x^2 - y^2))
Zy = 6y(exp) - (4yx^2 + 6y^3)(exp)

Zx -> 4x - 4x^3 - 6xy^2 = 0
Zy -> 6y - 4yx^2 - 6y^3 = 0

Clearly (0, 0) works. First look for other points along the two axes. If x = 0, Zx = 0, and Zy = 6y - 6y^3. Zy = 0 when 1 - y^2 = 0, or y = +/- 1. In fact the same is true for when y = 0. So now we have

(0, 0), (+/- 1, 0), (0, +/- 1)

Then assume x != 0 and y != 0. Now you can simplify the two expressions to get

Zx -> 4 - 4x^2 - 6y^2 = 0
Zy -> 6 - 4x^2 - 6y^2 = 0

It seems unlikely that you will find any values of x and y that solve both equations at once.

Answer 2

Tangent Plane Calculator

Answer 3

Yes, it's possible do it by hand !!

After you take the partial derivatives , set them equal to zero, then solve for x and y. Then use the determinant test (D) to find the max or min point of the surface.

D=fxx(x,y) * fyy(x,y) - [fxy(x,y)]^2 ,
where fxx is the second partial derivative with respected to x;
where fyy is the second partial derivative with respected to y;
where fxy is the mix second partial derivative ;

Plug those critical points into the determinant test function, if the D is >0 or fxx >0 , then the critical point is a min;
if the D > 0 and fxx < 0 , then the critial point is a max.

In this case, you probably will get one of these results.

Then plug your critial point back into your oringial function to get a z value, then z= (whatever that value is) will be the
horizontal tangent plane on a surface.

Note: only the max or min point will give you horizontal tangent plane

Answer 4

Ok, after looking more closely, here we go.

Any point on the surface is (x,y,z), going along x at a given point we travel along d/dx(x,y,z)= (1,0,dz/dx) and along y (0,1,dz/dy)

These are vectors tangent to the surface at a given point in the x and y directions. Now you can take their cross product to find the normal. Knowing the normal vector at a point, you can construct the tangent plane at that point. Let me know if you need help with that.

An example:

Let z = 5x +8y +9 (yeah, it's already a plane . . . because it's supposed to be an example)

Then the two vectors are:
(1,0,5) & (0,1,8) [These happen to be coordinate independent in this case . . ]

Their cross product is:

(-5,-8,1)

Say we want the tangent plane, that's tangent to the function at (0,0,9). That means that any vector from this point to another in the plane must be perpendicular to the cross product above. So if r(point on plane) - (0,0,9) = (x,y,z-9) is always perpendicular to the above, then the dot product of h and (-5,-8,1) is zero, aka -5x+-8y +(z-9) =0 or z = 5x +8y +9

So it works.

If you are to have a HORIZONTAL tangent plane at a point, that would imply dz/dy = dz/dx = 0