# Finding Angles Between Horizontal Planes and Multivariable Functions

1. Oct 15, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Suppose a mountain is described by the function z = 10x^2 * y − 5x^2 − 4y^2 − x^4 − 2y^4 and that you are standing at the point (1,1,−2). The positive x-axis points east and the positive y- axis points north. If you walk in the northeast direction what angle above the horizontal does your path make?

2. Relevant equations
cos(θ) = (v • w)/(|v||w|) (But I'm not sure this is even relevant.)
z - z0 = fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0)

3. The attempt at a solution
I am not convinced that this will work, but I intend to find a tangent plane at (1,1,-2). Then, using the normal vectors I have between the tangent and the xy-plane, I will calculate the angle.

fx = 20xy - 10x - 4x^3
fy = 10x^2 - 8y - 8y^3

Evaluated at (1,1) gives,

fx = 6
fy = -6

so,

z + 2 = 6(x-1) - 6(y-1), which I rearrange to give,

0 = 6(x-1) - 6(y-1) - 1(z + 2)

So, I think I have normal vector <6,-6,-1>, and a normal vector on the xy-plane, <0,0,1>.

So,

cos(θ) = (-1)/sqrt(73). So θ = arccos((-1)/sqrt(73)), which is obtuse, so, replacing <0,0,1> with its negative gives θ = arccos((1)/sqrt(73))

Last edited: Oct 15, 2011
2. Oct 16, 2011

### HallsofIvy

Staff Emeritus
What you need is the directional derivative: if $D_\theta f(x, y)$ is the derivative of f(x,y) in the direction that makes angle $\theta$ with the positive x-axis (so that $\partial f/\partial x= D_0 f$ and $\partial f/\partial y= D_{\theta/2} f$, then
$$D_\theta f= \frac{\partial f}{\partial x} cos(\theta)+ \frac{\partial f}{\partial y} sin(\theta)$$

Similarly, if $\vec{v}$ is a unit vector the derivative in the direction of that vector is
$$\nabla f \cdot \vec{v}$$.
"Northeast", given that the positive x-axis is east and the positive y-axis north, is at angle $\theta= \pi/4$ which is the same as the direction of unit vector $\vec{v}= (\vec{i}+ \vec{j})/\sqrt{2}$

Once you know the "rate of change", "rise over run", in that direction, just take the inverse tangent to get the angle.

No, $\theta$ is NOT "$arccos(1)/\sqrt{73}$".

"fx = 6 fy = -6" should give you the answer immediately.

3. Oct 16, 2011

### TranscendArcu

So, by the definition of the directional derivative: ∇f ⋅ v⃗. Then my directional derivative should be <6,-6> • <1/sqrt(2),1/sqrt(2)>. This gives a dot product of 0.

Arctan(0) = 0, which does not seem correct to me.