Gradients and Directional Derivatives

[Alw]

Homework Statement

(a) Find two linear equations in the variables x,y,z such that their solution set is the tangent line to the graph of f(x,y) = x^{2} - y^{2} at (2,1,3) which is parallel to the x,z plane. (b) Represent the line of (a) by parametric equations.

Homework Equations

\nablaf(x,y,z) = f(f_{x}, f_{y}, f_{z})

The tangent plane to the surface at a point:
0 = f_{x}(a,b,c)(x-a) + f_{y}(a,b,c)(y-b) + f_{z}(a,b,c)(z-c)

The Attempt at a Solution

I know how to solve for the plane (line in a function defined in two variables) tangent to the graph, and how to represent it parametrically, so I started with part b.

\nablaf = (2x, -2y)

plugging into the tangent equation the point (2,1,3) for (a,b,c) and the components of the gradient gives

0 = 4(x-2) - 2(y-1)

and represented parametrically:

x = 2 + 4t, y = 1 - 2t (solution to part b)

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Looking again at part (a), he wants two linear equations such that their solution set is 0 = 4(x-2) - 2(y-1). I've been through the sections a few times and I haven't found any examples asking for this, and the other students in my class were struggling with this part too. Any ideas?

 

[HallsofIvy]

Homework Statement

(a) Find two linear equations in the variables x,y,z such that their solution set is the tangent line to the graph of f(x,y) = x^{2} - y^{2} at (2,1,3) which is parallel to the x,z plane. (b) Represent the line of (a) by parametric equations.

Homework Equations

\nablaf(x,y,z) = f(f_{x}, f_{y}, f_{z})

I'm not sure why you have the "f" in there.
\nablaf(x,y,z)= f_x \vec{i}+ f_y\vec{j}+ f_z\vec{k}
I presume that was what you meant.

The tangent plane to the surface at a point:
0 = f_{x}(a,b,c)(x-a) + f_{y}(a,b,c)(y-b) + f_{z}(a,b,c)(z-c)

The Attempt at a Solution

I know how to solve for the plane (line in a function defined in two variables) tangent to the graph, and how to represent it parametrically, so I started with part b.

\nablaf = (2x, -2y)

plugging into the tangent equation the point (2,1,3) for (a,b,c) and the components of the gradient gives

0 = 4(x-2) - 2(y-1)

and represented parametrically:

x = 2 + 4t, y = 1 - 2t (solution to part b)

Remember that you are working in 3 dimensions, not 2. The gradient of
f(x,y,z)= x^2- y^2 is 2x\vec{i}- 2y\vec{j}+ 0\vec{k}
At (2, 1, 3) that is 4\vec{i}- 2\vec{k}. Since the gradient is perpendicular to the tangent plane, any tangent vector must be perpendicular to that. Such vector would be 4\vec{i}+ 2\vec{j}+ z\vec{k} where z can be any value (obviously, the dot product of that with 4\vec{i}- 2\vec{3} is 0 for any z. Because the problem says also "parallel to the x,z plane", z= 0. The tangent vector for the line referred to is 4\vec{i}+ 2\vec{j} and so the line is given by the parametric equations x= 4t+ 2, y= 2t+ 1, z= 3. Of course,that line is the intersectio of two planes. The equations asked for are the equations of those two planes (which will involve x, y, and z. There are, of course, many such planes and many solutions to this problem.

------------------------------------

Looking again at part (a), he wants two linear equations such that their solution set is 0 = 4(x-2) - 2(y-1). I've been through the sections a few times and I haven't found any examples asking for this, and the other students in my class were struggling with this part too. Any ideas?