Directional derivative: angle trail is descending at

[Baron]

A hiker is going down a hill whose shape is given by z = e^(4-x^2-2y^2). Suppose that the trail the hiker is on follows the path of the steepest descent from the point P(1, 1/2, e^(2.5)). Find the angle the trail is descending at (compared to the horizontal plane).

My work:

To find the direction of the steepest descent, I found the gradient which is equal to <f_x, f_y>

gradient at P = <e^(4-x^2-2y^2)*-2x , e^(4-x^2-2y^2)*-4y>

= <-2x*e^(4-x^2-2y^2) , -4y*e^(4-x^2-2y^2) >

Plug in the points x = 1 and y = 1/2 (coordinates of p)

gradient at P = <-2e^2.5, -2e^2.5>. This is the direction of the steepest ascent. The direction of the steepest descent is in the direction of <2e^2.5, 2e^2.5 > or the unit vector <1/sqrt(2), 1/sqrt(2) >

But how do I find the angle the trail is descending at?

 

[Baron]

The angle of descent is the magnitude of the gradient

magnitude of gradient = sqrt( 4(e^2.5)^2 + 4(e^2.5)^2) = sqrt(8e^5) = 2sqrt(2)*e^(5/2)

but how is the angle of descent the magnitude of the gradient? I know the magnitude of the gradient represents the greatest rate of change of the hill but I don't see the connection between the magnitude of the gradient and the angle.

 

[DrPhil]

magnitude of gradient = sqrt( 4(e^2.5)^2 + 4(e^2.5)^2) = sqrt(8e^5) = 2sqrt(2)*e^(5/2)

but how is the angle of descent the magnitude of the gradient? I know the magnitude of the gradient represents the greatest rate of change of the hill but I don't see the connection between the magnitude of the gradient and the angle.

I should have said the magnitude of the gradient (another word for "slope") is the tangent of the angle.

 

[Baron]

I should have said the magnitude of the gradient (another word for "slope") is the tangent of the angle.

That makes sense. Because the question is asking for the angle of descent, the tangent of the angle should be equal to the negative slope or the negative magnitude of the gradient so:

tan θ = -2sqrt(2)*e^(5/2)

θ = arctan(-2sqrt(2)*e^(5/2))

Is that correct?

 

[DrPhil]

A hiker is going down a hill whose shape is given by z = e^(4-x^2-2y^2). Suppose that the trail the hiker is on follows the path of the steepest descent from the point P(1, 1/2, e^(2.5)). Find the angle the trail is descending at (compared to the horizontal plane).

My work:

To find the direction of the steepest descent, I found the gradient which is equal to <f_x, f_y>

gradient at P = <e^(4-x^2-2y^2)*-2x , e^(4-x^2-2y^2)*-4y>

= <-2x*e^(4-x^2-2y^2) , -4y*e^(4-x^2-2y^2) >

Plug in the points x = 1 and y = 1/2 (coordinates of p)

gradient at P = <-2e^2.5, -2e^2.5>. This is the direction of the steepest ascent. The direction of the steepest descent is in the direction of <2e^2.5, 2e^2.5 > or the unit vector <1/sqrt(2), 1/sqrt(2) >

But how do I find the angle the trail is descending at?

I hadn't checked your work yet. I don't like your answer - seems too large. Hmmm - maybe what I don't like is the question! It doesn't make sense that <x,y> are map coordinates and z is an elevation.

\(\displaystyle \displaystyle z = e^{4-x^2-2y^2} \)

\(\displaystyle \displaystyle \ln z = 4 - x^2 - 2\ y^2\)

\(\displaystyle \displaystyle \bigtriangledown z = < \dfrac{\partial z}{\partial x},\ \dfrac{\partial z}{\partial y} > \)

\(\displaystyle \displaystyle \dfrac{( \bigtriangledown z)}{z} = < -2x,\ -4y> \)

That is the same as you got - but what does it mean? A gradient must be a dimensionless ratio, such as feet/feet. But x and y in the function don't look like feet, and certainly don't have the same dimensions as z.

Oh well - just treat it as a math problem, not a map problem. I think you are doing ok.

 

[HallsofIvy]

The angle of descent is the magnitude of the gradient

Are you British, DrPhil? I know our British friends use the word "gradient" to mean what we in the USA call the "derivative". But, here, we are talking about the gradient vector. Its magnitude is the rate of change. The "angle of descent", the angle the vector makes with the xy-plane, is the complement the vector makes with that z-axis. And that can be calculated by looking at the unit vector in the same direction to find the direction cosines. That is, if the vector is \(\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}\), the unit vector is \(\displaystyle \frac{a}{\sqrt{a^2+ b^2+ c^2}}\vec{i}+ \frac{b}{\sqrt{a^2+ b^2+ c^2}}\vec{j}+ \frac{c}{\sqrt{a^2+ b^2+ c^2}}\vec{k}\). The angle, \(\displaystyle \theta\), is given by \(\displaystyle cos(\theta)= \frac{c}{\sqrt{a^2+ b^2+ c^2}}\)

 

[DrPhil]

Are you British, DrPhil? I know our British friends use the word "gradient" to mean what we in the USA call the "derivative". But, here, we are talking about the gradient vector. Its magnitude is the rate of change. The "angle of descent", the angle the vector makes with the xy-plane, is the complement the vector makes with that z-axis. And that can be calculated by looking at the unit vector in the same direction to find the direction cosines. That is, if the vector is \(\displaystyle a\vec{i}+ b\vec{j}+ c\vec{k}\), the unit vector is \(\displaystyle \frac{a}{\sqrt{a^2+ b^2+ c^2}}\vec{i}+ \frac{b}{\sqrt{a^2+ b^2+ c^2}}\vec{j}+ \frac{c}{\sqrt{a^2+ b^2+ c^2}}\vec{k}\). The angle, \(\displaystyle \theta\), is given by \(\displaystyle cos(\theta)= \frac{c}{\sqrt{a^2+ b^2+ c^2}}\)

SORRY - I should have gone back and corrected that post! I did "fix" my error in a subsequent post.

I treated z(x,y) as the function, so I got the gradient as a vector field in R^2. That is, at every point (x,y) the gradient has a magnitude and an azimuthal angle. At the point (1, 1/2) I agreed with the student that the azimuthal angle would be pi/4. But we wound up with a magnitude of \(\displaystyle 2\ \sqrt{2}\ z \), with \(\displaystyle z=\mathrm e ^{2.5}\), which when considered as the slope is precipitous. Do you have any comments on that result?

BTW, I did spend one year working in the UK, and was impressed with how well trained the Brits are in vectors and matrices. But I tried not to pick up the language differences...