# Tangent planes

Ok, this is probably easy...but i'm stuck
f(x,y) = y^2 + xy - x^2 +2
find the gradient of the normal to the level curve at the point (3,-2)
my answer is -1/root65 , but it's supposed to be 1/8.

I did it by finding Fx(x,y) = y-2x and Fy(x,y) = 2y+x then finding the absolute value of the gradient..

You got the gradient vector correct, you must have made a mistake in translating it to the plane equation. This is a great page for help on this:

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HallsofIvy
Homework Helper
It was probably the use of the word "gradient" that was confusing you- it confused me! One use of the word "gradient" is: gradient f(x,y) is the vector with components fx and fy. Of course, the gradient vector is perpendicular to a level curve. You correctly found that, for this function, that is (y-2x, 2y+x) and, at x= 3, y= -2, (-8,-1). The length of that (I wouldn't say "absolute value" for a vector) is $$sqrt{65}$$.

However, "gradient" is also often used for "slope": rise over run, of a line or vector. Here the "rise", the y component of the vector, is -1 and the "run", the x component of the vector, is -8. The "gradient" in this sense is -1/(-8)= 1/8.

Personally, I think its a bad choice of words.

OlderDan
Homework Helper
HallsofIvy said:
It was probably the use of the word "gradient" that was confusing you- it confused me! One use of the word "gradient" is: gradient f(x,y) is the vector with components fx and fy. Of course, the gradient vector is perpendicular to a level curve. You correctly found that, for this function, that is (y-2x, 2y+x) and, at x= 3, y= -2, (-8,-1). The length of that (I wouldn't say "absolute value" for a vector) is $$sqrt{65}$$.

However, "gradient" is also often used for "slope": rise over run, of a line or vector. Here the "rise", the y component of the vector, is -1 and the "run", the x component of the vector, is -8. The "gradient" in this sense is -1/(-8)= 1/8.

Personally, I think its a bad choice of words.

I agree that the words are confusing. Gradient of the normal suggests an operation on a vector that is supposed to be an operation on a scalar. Your interpretation that it means the slope of something seems to be the only reasonable interpretation, but I found myself asking what it was the slope of, and finally arrived at the conclusion that was probably obvious to you that it is the slope of the normal to the curve y vs x along a line of constant z through the point (3,-2,f(3,-2)) = (3,-2,-9). The slope of that curve can also be found by implicit differentiation of f(x,y) = constant to find the slope at (3,-2), where y' = -8, so of course the slope of the normal through that point is 1/8.

But since the title of this thread is "tangent plane" I thought it would be nice to connect this result to finding the plane tangent to the surface defined by f(x,y). The slope of the normal to the curve y vs x for constant z gives the slope of the projection of the vector normal to the surface onto the plane of constant z, but what does the tangent plane and the full normal vector look like? The procedure for finding the complete normal vector and the equation of the tangent plane is nicely outlined in the link posted by whozum (except for repeated references to the "tangent line" that I think should say "tangent plane"; I dropped a note to the author about that). The result is that the normal vector has components (-8,-1,-1), and the equation of the plane is

$$z = - 8x - y + 13$$

The figure shows the original function and the tangent plane in a 15x15x15 cube surrounding the point (3,-2,-9). To the left is a zoomed in view of the surface cut by the surface of constant z = -9.

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