# directional derivative problem

#### Em Yeu Anh

At what point on the paraboloid $$\displaystyle y=x^2+z^2$$ is the tangent plane parallel to the plane $$\displaystyle x+2y+7z=2$$?

Really stuck on this one.

Edit: not $$\displaystyle y=x^2+y^2$$

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#### matheagle

MHF Hall of Honor
At what point on the paraboloid $$\displaystyle y=x^2+y^2$$ is the tangent plane parallel to the plane $$\displaystyle x+2y+7z=2$$?

Really stuck on this one.

I assume you meant $$\displaystyle z=x^2+y^2$$

#### Em Yeu Anh

I assume you meant $$\displaystyle z=x^2+y^2$$
Sorry, $$\displaystyle y=x^2+z^2$$

#### HallsofIvy

MHF Helper
At what point on the paraboloid $$\displaystyle y=x^2+z^2$$ is the tangent plane parallel to the plane $$\displaystyle x+2y+7z=2$$?

Really stuck on this one.

Edit: not $$\displaystyle y=x^2+y^2$$
Well, that's pretty straight forward isn't it? Rewrite this as $$\displaystyle f(x,y,z)= x^2+ z^2- y= 0$$ so you can think of this paraboloid as being a "level surface" for f(x,y,z). What is the gradient of f at any point (x,y,z)? What is a normal vector to the plane? where are those two vectors parallel?

• Em Yeu Anh

#### Em Yeu Anh

Well, that's pretty straight forward isn't it? Rewrite this as $$\displaystyle f(x,y,z)= x^2+ z^2- y= 0$$ so you can think of this paraboloid as being a "level surface" for f(x,y,z). What is the gradient of f at any point (x,y,z)? What is a normal vector to the plane? where are those two vectors parallel?
Gradient vector of f is $$\displaystyle <2x,-1,2z>$$
Normal to the plane is $$\displaystyle <1,2,7>$$

Tangent plane is parallel to that plane if the corresponding normal vectors are parallel, so
$$\displaystyle <2x_o,-1,2z_o> = c<1,2,7>$$
or equivalently $$\displaystyle <x_o,\frac{-1}{2},z_o> = k<1,2,7>$$

$$\displaystyle x_o = k$$
$$\displaystyle \frac{-1}{2} = 2k$$
$$\displaystyle z_o = 7k$$

Solving for k yields $$\displaystyle k=\frac{-1}{4}, x_o=\frac{-1}{4}, z_o=\frac{-7}{4}, y_o=\frac{50}{16}$$

Did I do that correctly?

#### HallsofIvy

MHF Helper
Yes, good! (Clapping) (Although I would have written $$\displaystyle y_0$$ as $$\displaystyle \frac{25}{8}$$.)