What, exactly, do you mean by the "slope" of a line in three dimensions?
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.
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)
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?
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: and , so .
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?
Since you have z as a function of x and y ( ), and you know the direction you're going in the x-y plane , you should be able to take the directional derivative and get the same answer.
The gradient is , so the directional derivative is .