# Thread: slope of the tangent line to the curve of intersection of the vertical plane &surface

1. ## slope of the tangent line to the curve of intersection of the vertical plane &surface

Hello,

vert plane: x - root(3)*y + 2*root(3) - 1 = 0
surface: z = x^2 + y^2

slope of the< tangent line to the curve of intersection of the vertical plane &surface at point (1,2,5)

First, I took the gradients and plugged in (1,2,5)

del F (1,2,5) = <1, - root 3, 0> && del G (1,2,5) = <2,4,-1>

Then take cross product, gives < root 3, 1, 4+2*root(3)>. The cross product is tangent to the curve of intersection.

Now in order to get slope, I think I have to do a directional derivative, but not sure.

2. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

What, exactly, do you mean by the "slope" of a line in three dimensions?

3. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

I am not sure exactly. This is a practice problem my teacher posed. But I am pretty sure I have to do a directional derivative since it is defined as the "slope of the tangent line to the curve formed by the intersection of z = f (x; y) and the vertical plane through a point p parallel to a unit vector u". He gave us a hint, use Du F (1,2)

edit:

So the cross product gives us < root 3, 1, 4+2*root(3)> and this is the direction of the tangent line.

So, if i make a unit vector from it and then dot it with del F, that should satisfy my above definition?

4. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

Can nobody help the poor reptile?

5. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

It sounds like you have already found the direction vector for the line of intersection (I didn't check your calculations, though).

HallsofIvy is right - the idea of slope in this problem makes no sense. But I'm guessing what it means is the z component of the direction vector divided by the component in the x-y plane. If so, that's easy to calculate: $v_z=4+2\sqrt{3}$ and $v_{xy}=\sqrt{v_x^2+v_y^2}=2$, so $\text{slope}=2+\sqrt{3}$.

- Hollywood

6. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

I understand the slope oddity and your suggestion makes sense to me but he told us we would have to use D_u f (1,2), I don't know for which vector though.

unit vector u = cross product/ |cross product| = <sqrt(3), 1, 4+2*sqrt(3)> / 7.73.

okay, but what vector am I dotting u with, what would f be?

7. ## Re: slope of the tangent line to the curve of intersection of the vertical plane &sur

Since you have z as a function of x and y ( $z=x^2+y^2$), and you know the direction you're going in the x-y plane $\bold{v}=<\sqrt{3},1>$, you should be able to take the directional derivative and get the same answer.

The gradient is $<2x,2y> = <2,4>$, so the directional derivative is $<2,4>\cdot\frac{\bold{v}}{|\bold{v}|} = \frac{1}{2}<2,4>\cdot<\sqrt{3},1>=2+\sqrt{3}$.

- Hollywood