# Tangent planes to surfaces

1. Oct 5, 2008

### JFonseka

Suppose that F(x,y) = x$$^{2}$$+y$$^{2}$$. By using vector geometry, find the Cartesian equation of the tangent plan to the surface z = F(x,y) at the point where (x,y,z) = (1,2,5). Find also a vector n that is normal to the surface at this point.

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Now the step by step working is given for this question, I however get confused at one part.
First they intersect the surface z = x$$^{2}$$+y$$^{2}$$ with the plane x = 1

Therefore z = 1 + y$$^{2}$$

The gradient F$$_{y}$$(1,2) = 4

By using the point gradient formula for a straight line, the Cartesian eq for the tangent is:

z -5 = 4(y - 2), x =1

If $$\lambda$$ = y - 2, then the equation of the tangent line in parametric vector form is

(x y z) = (1 2 5) + $$\lambda$$(0 1 4)

Now my question is, where did the (0 1 4) come from?

2. Oct 6, 2008

### tiny-tim

Hi JFonseka!

This is the tangent line in the x = 1 plane (hmm … why didn't they choose the easier z = 5 plane? ), so the x term has to be 0.

And from z -5 = 4(y - 2), the z must increase 4 times as fast as the y, so altogether it's a multiple of (0 1 4).

3. Oct 6, 2008

### HallsofIvy

I'm not sure I can answer you question because that is not quite the way I would do the problem! In fact, I am pretty sure (x,y,z)= (1, 2, 5)+ $\lambda$(0, 1, 4) is wrong. z= x2+ y2 has zx= 2x= 2 at (1, 2, 5) so that the line z= 5+ 2(x-1) Taking x= 2 in the first equation, z= 5+ 2= 7. But the equation of the plane you give has x= 1 for all y and z.

Given z= x2+ y2, think of this as a "level surface" of the function F(x,y,z)= x2+ y2- z= 0. The gradient of F is the vector function <2x, 2y, -1> and, at (x,y,z)= (1, 2, 5) that is <2, 4, -1>. The gradient vector is always perpendicular to a "level surface" so this vector, <2, 4, -1> is perpendicular to the surface and so to the tangent plane. A vector equation of a plane with normal vector <2, 4, -1> and containing point (1, 2, 5) is (x, y, z)= (1, 2, 5)+ t(2, 4, -1). That is the equation of the tangent plane and is NOT what you give.

4. Oct 6, 2008

### tiny-tim

uh-uh … that's the equation of the normal to the tangent plane.

(and (0 1 4) is perpendicular to it)

5. Oct 6, 2008

### JFonseka

I forgot to mention, there is more, I just didn't put it because I was confused at that point.

6. Oct 6, 2008

### HallsofIvy

And just to prove what an imbecile I am, I didn't even notice it was the equation of a line, not a plane! Darn- makes me wish I could just erase that post and pretend I never wrote it. I think I got confused when JFonseca asked about the tangent plane but then started talking about a tangent line.

The equation of the normal plane to z= x2+ y2 is, of course,
$$2\vec{i}+ 4\vec{j}-\vec{k} \cdot (x-1)\vec{i}+ (y-2)\vec{j}+ (z- 5)\vec{k}$$
or 2(x-1)+ 4(y-2)- (z- 5)= 0.

Yes, if we take the line tangent to the curve on z= x2+ y2 with x= 1, then z= 1+ y2 so dz/dy= 2y and dz/dy= 4 at y= 2. dy/dy= 1, of course, and, since x is the constant, 0, here, dx/dy= 0. The tangent vector is <0, 1, 4> and, as tiny-tim pointed out, that is pependicular to my <2, 4, -1> which is perpendicular to the surface itself. I'm not sure why you would want to calculate a tangent to the curve- unless you are also going to calculate the tangent vector to the curve on z= x2+ y2 with y= 2 and then take the cross product of the two tangent vectors to get the normal vector to the surface. Seems a complicated method to me.

7. Oct 6, 2008

### JFonseka

Ah...so that's how they worked it out, now I get it!

That's actually what they do in the next few steps, I just never posted that bit because it was similar and also I didn't get how they got that tangent vector <0,1,4>, but now I know. Thanks a lot tiny and hallsofivy!