Using the gradient operator to find the slope of a hill questions.

[knowlewj01]

Homework Statement

The shape of a hill is described by the height function:

[itex] h(x,y) = \frac{1}{\sqrt{2+x^2+y^4}}[/itex]

a) find the gradient [itex]\nabla h(x,y)[/itex]
b) find the maximum slope of the hill at the point [itex]\bf{r_0 = i+j}[/itex] [or (x,y) = (1,1)]
c) If you walk NorthEast (in the direction of the vector [itex]\bf{i+j}[/itex]) from the point [itex]\bf{r_0}[/itex], what is the slope of your path? are you going uphill or downhill?
d) find a vector in the direction of a path through the point [itex]r_0[/itex] that remains at the same height

Homework Equations

in cartesian:
[itex]\nabla = \left( \bf{i}\frac{\partial}{\partial x} + \bf{j}\frac{\partial}{\partial y} + \bf{k}\frac{\partial}{\partial z}\right)[/itex]

The Attempt at a Solution

a) [itex]\nabla h(x,y)[/itex]

[itex] = \left( \bf{i}\frac{\partial}{\partial x} + \bf{j}\frac{\partial}{\partial y} + \bf{k}\frac{\partial}{\partial z}\right) \cdot \left(2+x^2+y^4\right)^{-\frac{1}{2}}[/itex]

[itex] = -\bf{i}x\left(2+x^2+y^4\right)^{-\frac{3}{2}} -\bf{j}2y^3\left(2+x^2+y^4\right)^{-\frac{3}{2}}[/itex]

so this is the direction in which the gradient of h increases the quickest.

b) at the point (1,1) the maximum slope of the hill will be given by substituting [itex]x=1[/itex] and [itex]y=1[/itex] into the answer from part (a).

[itex]\nabla h(1,1) = -\bf{i} \frac{1}{8} -\bf{j} \frac{1}{4}[/itex]

so at the point (1,1) the maximum slope is in this direction. the magnitude of the slope is the magnitude of this vector.

[itex]\left|-\bf{i} \frac{1}{8} -\bf{j} \frac{1}{4}\right| = \sqrt{\frac{1}{64} + \frac{1}{16}}[/itex]
[itex] = \frac{\sqrt{5}}{8}[/itex]

I think I'm doing fine so far, part (c) confuses me.
Start at point (1,1) and walk in the direction [itex]\bf{i+j}[/itex].
I know the direction in which slope increases the quickest from this point, but to find the slope of the path i am taking I'm not sure what to do, would it have something to do with the angle between [itex]\nabla h(1,1)[/itex] and the vector [itex]\bf{i+j}[/itex]?

for part (d) I assume I need to find a solution to [itex]\nabla h[/itex] such that it is equal to zero. I'm not quite sure how to do this either. Can anyone point me in the right direction?

 

[Quinzio]

d) find a vector in the direction of a path through the point LaTeX Code: r_0 that remains at the same height

Image you are driving on a mountain road (the problem talks about hills, by the way).
Usually mountain roads tend to minimize the slope (as long as it is possible), so they try to remain at the same height, as your problem asks.
So, did you ever see a mountain road pointing directly towards the top of the mountain ?
Usually, while driving, you'll have the top at your ... or your ...

(Hope that helps)

 

[gabbagabbahey]

part (c) confuses me.
Start at point (1,1) and walk in the direction [itex]\bf{i+j}[/itex].
I know the direction in which slope increases the quickest from this point, but to find the slope of the path i am taking I'm not sure what to do, would it have something to do with the angle between [itex]\nabla h(1,1)[/itex] and the vector [itex]\bf{i+j}[/itex]?

The wording is a little misleading. What they mean is that the components of your velocity in the [itex]x[/itex] and [itex]y[/itex] directions are equal and positive; you aren't told anything directly about the upward component of your velocity though. Keep in mind that if you walk along a hill, your likely to stay fairly attached to it (unless you have wings, or jump a lot), that tells you that at any value of [itex]x[/itex] and [itex]y[/itex], your height is going to be [itex]h(x,y)[/itex], so your (3D) position vector will be

[tex]\textbf{r}=x\textbf{i}+y\textbf{j}+\frac{1}{\sqrt{2+x^2+y^4}}\textbf{k}[/tex]

How does position relate to velocity? What does that tell you about the upward component of your velocity?

for part (d) I assume I need to find a solution to [itex]\nabla h[/itex] such that it is equal to zero. I'm not quite sure how to do this either. Can anyone point me in the right direction?

Finding where [itex]\mathbf{\nabla}h=0[/itex] is only going to tll you where the critical points (maxima/minima/extrema/saddles) of the hill are. The gradient of the hill doesn't need to be zero at a given point for you to be able to walk in a direction that doesn't increase your height.

Instead, realize that if you walk in a path at constant height, and that path goes through the point (1,1), you must have [itex]h(1,1)=h(x,y)[/itex] at every point on the path. What does that tell you about [itex]x^2+y^4[/itex]? What is the (2D) slope of that path?

 

[knowlewj01]

ok i see what you mean about the 3D position vector, i have worked out the vector to be

[itex]\bf{i + j} + \frac{1}{2}\bf{k}[/itex]

then if i walk in the direction [itex]\bf{i+j}[/itex] the next point at which i can find the height h(x,y) is a t h(2,2).

I calculate this as:
[itex]h(2,2) = \frac{1}{\sqrt{22}}[/itex]
so my new position vector is:

[itex] 2\bf{i} + 2\bf{j} +\frac{1}{\sqrt{22}}\bf{k}[/itex]

i know that velocity is the first time derivative of position but i don't see how it applies here, the only thing i can think of at this stage is to subtract the position vector evaluated at h(1,1) from the one evaluated at h(2,2) and divide the difference by the magnitude of the position change on the x-y plane.

So in effect calculating: [itex]\frac{\Delta h}{\Delta \bf{r}}[/itex]

i realize that this is probably wrong as it is inacurate and i don't know how the height varies between h(1,1) and h(2,2).

 

[gabbagabbahey]

ok i see what you mean about the 3D position vector, i have worked out the vector to be

[itex]\bf{i + j} + \frac{1}{2}\bf{k}[/itex]

then if i walk in the direction [itex]\bf{i+j}[/itex] the next point at which i can find the height h(x,y) is a t h(2,2).

When you walk do you disappear from one point and then reappear at a point 1 meter away, or do you move through every point in between regardless of whether or not it is an integer multiple of whatever unit system you are measuring distance in?

You don't need to increment [itex]x[/itex] and [itex]y[/itex] by integers, but rather by infinitesmal displacements. When you move a small amount from the point [itex]\langle x,y,h(x,y)\rangle[/itex] while staying attached to the hill, you arrive at the point [itex]\langle x+dx,y+dy,h(x+dx,y+dy)\rangle[/itex]. If you are moving Northeast, then [itex]dx=dy[/itex] and both are positive.

i know that velocity is the first time derivative of position but i don't see how it applies here

The chain rule tells you

[tex]\frac{d}{dt}\textbf{r}=\frac{dx}{dt}\textbf{i}+\frac{dy}{dt}\textbf{j}+\frac{d}{dt}h(x,y)\textbf{k}[/tex]

According to the chain rule (work it out for yourself if you haven't seen this result before) [itex]\frac{d}{dt}h(x,y)= \textbf{v}_{2D}\cdot\mathbf{\nabla}h(x,y)[/itex]

Where [itex]\textbf{v}_{2D}\equiv \frac{dx}{dt}\textbf{i}+\frac{dy}{dt}\textbf{j}[/itex]

If you are walking Northeast, what can you say about [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex]? What does that make the upward component of your (3D) velocity ( [itex]\frac{d}{dt}h(x,y)[/itex] )?

the only thing i can think of at this stage is to subtract the position vector evaluated at h(1,1) from the one evaluated at h(2,2) and divide the difference by the magnitude of the position change on the x-y plane.

So in effect calculating: [itex]\frac{\Delta h}{\Delta \bf{r}}[/itex]

If you traveled in a straight line from <1,1,h(1,1)> to <2,2,h(2,2)>, this would be correct, but you must follow the profile of the hill. In order to calculate the slope of a curve, you need to use infinitesmals/calculus.

i realize that this is probably wrong as it is inacurate and i don't know how the height varies between h(1,1) and h(2,2).

But you do know how the height varies between those two points... you are given h(x,y).

 

[knowlewj01]

i think i have got part (c) down:

we know the slope of the hill at point (1,1)

[itex]\nabla h = \frac{1}{8}\hat{i} + \frac{1}{4}\hat{j}[/itex]

call the direction we want to travel vector s

where [/itex]\bf{s} = \hat{i} + \hat{j}[/itex]

the slope in the desired direction is just the component of [itex]\nabla h[/itex]
in the direction of s. ie, a unit vector in the direction of s.

so find the unit vector:

[itex]\hat{\bf{s}} = \frac{\bf{s}}{\left|\bf{s}\right|} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}[/itex]

now find the scalar product of [itex]\nabla h[/itex] with the unit vector:

[itex]\nabla h \cdot \hat{\bf{s}} = \frac{1}{8\sqrt{2}} +\frac{1}{4\sqrt{2}} = \frac{3\sqrt{2}}{16}[/itex]

i read this in one of my textbooks and did one of their examples which was similar to this question and the answer was right, so I am quite confident that this is ok. I'm just getting on to section d now.

 

[gabbagabbahey]

the slope in the desired direction is just the component of [itex]\nabla h[/itex]
in the direction of s. ie, a unit vector in the direction of s.

Right, but do you understand why this is true?

The slope of your path is the upward component of your velocity divided by the lateral component of your velocity.

[tex]\begin{aligned}\text{slope} &= \frac{v_z}{|\textbf{v}_{2D}|} \\ &= \frac{\frac{d}{dt}h(x,y)}{|\textbf{v}_{2D}|} \\ &= \frac{\textbf{v}_{2D}\cdot\mathbf{\nabla}h(x,y)}{|\textbf{v}_{2D}|} \\ &= \frac{\textbf{v}_{2D}}{|\textbf{v}_{2D}|} \cdot \mathbf{\nabla} h(x,y) \end{aligned}[/tex]

And [itex]\frac{\textbf{v}_{2D}}{|\textbf{v}_{2D}|}[/itex] is just the unit vector pointing in the (2D) direction of travel.

 

[knowlewj01]

Yeah, I understand that now, thanks very much.

For part (d) I also see what you mean, finding places where [itex]\nabla h = 0[/itex] won't help at all. so, what i am looking for is a path on the surface for which the function h(x,y) is always the same, regardless of what x and y are. This implies that [itex]x^2+y^4[/itex] must always be the same.

 

[knowlewj01]

so what is h(x,y) at (1,1):

[itex]h(x,y) = \frac{1}{\sqrt{2+1+1}} = \frac{1}{2}[/itex]

so, no matter what values of x, and y h(x,y) must be equal to 1/2:

this implies:

[itex]\frac{1}{\sqrt{2+x^2+y^4}} =\frac{1}{2}[/itex]

[itex]2+x^2+y^4 = 4[/itex]

[itex]x^2 + y^4 = 2[/itex]

we can see just from inspection that this is only true for [itex]x= \pm 1[/itex] and [itex]y = \pm 1[/itex]

...so height is the same at these two points, still a bit stuck here. It feels like I'm not going to solve it like this, is there a way to do it with the grad opertor?

 

[gabbagabbahey]

so what is h(x,y) at (1,1):

[itex]h(x,y) = \frac{1}{\sqrt{2+1+1}} = \frac{1}{2}[/itex]

so, no matter what values of x, and y h(x,y) must be equal to 1/2:

this implies:

[itex]\frac{1}{\sqrt{2+x^2+y^4}} =\frac{1}{2}[/itex]

[itex]2+x^2+y^4 = 4[/itex]

[itex]x^2 + y^4 = 2[/itex]

Good.

we can see just from inspection that this is only true for [itex]x= \pm 1[/itex] and [itex]y = \pm 1[/itex]

No, there is no reason to assume only integer values of [itex]x[/itex] and [itex]y[/itex]. The equation [itex]x^2+y^4=2[/itex] represents a curve...how do you find the tangent vector of a curve? Won't the tangent vector always point in the direction of travel along the curve?

 

[knowlewj01]

ok, yeah i see that, so the curve has many non-integer solutions as well such as [itex] x=\sqrt{2}, y=0[/itex] for example. So the height is going to be constant at all points on a curve which has a 2d shape with equation [itex] x^2 + y^4 = 2[/itex]. Its almost like the general equation for a circle. But surely a vector which only passes through one of these points will vary in height because there are no straight lines on the curve.
The tangent to the curve at the point (1,1) will be prependicular to the vector [itex]i + j[/itex] ie, in the direction [itex]i -j[/itex] or [itex]-i+j[/itex]. so if i choose to go around the curve counter-clockwise, the unit vector in this direction at the point (1,1) will be [itex]-\frac{1}{\sqrt{2}}\bf{i} +\frac{1}{\sqrt{2}}\bf{j}[/itex].

this vector doesn't always point in the direction of travel though. can i express this vector in terms of x and y?