Some vector calculus questions

[sa1988]

Homework Statement

2. The shape of a hill is described by the height function h(x,y) = (2 + x² + y⁴)^-1/2

(a) Find the gradient ∇h(x,y)

(b) What is the maximum slope of the hill at the point r₀ = i+j [or (x,y) = (1,1)]?

(c) If you walk north-east (in the direction of the vector i+j) from the point r₀, 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 r₀ that remains at the
same height.

Homework Equations

The Attempt at a Solution

For parts a), b) and c) I think I'm okay.

a) ∇h(x,y) = ([-x(2 + x² + y⁴)^-3/2], [-y³(2 + x² + y⁴)^-3/2])

b) At point (1,1) just plug x and y into ∇h(x,y) to get (-1/8 , -1/4)

c) Along path of vector (i+j) you will be going in a downhill direction since it is in a direction 'away' from the maximum slope found in the previous question. But is there a way I can mathematically prove this?

d) This is the part I'm unsure of. It's been a while since I did work with vectors. I'm guessing I need to find a vector such that infinitesimal change in the x-y direction gives no change in height. So that would mean d(h(x,y)) = 0...? Since this would signify no change in the value of h(x,y). Overall I'm a bit lost.

Thanks.

 

[SteamKing]

Homework Statement

2. The shape of a hill is described by the height function h(x,y) = (2 + x² + y⁴)^-1/2

(a) Find the gradient ∇h(x,y)

(b) What is the maximum slope of the hill at the point r₀ = i+j [or (x,y) = (1,1)]?

(c) If you walk north-east (in the direction of the vector i+j) from the point r₀, 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 r₀ that remains at the
same height.

Homework Equations

The Attempt at a Solution

For parts a), b) and c) I think I'm okay.

a) ∇h(x,y) = ([-x(2 + x² + y⁴)^-3/2], [-y³(2 + x² + y⁴)^-3/2])

You might want to recheck your calculation of ∂h / ∂y.

 

[sa1988]

You might want to recheck your calculation of ∂h / ∂y.

Sorry, typo... should be -2y³ there.

 

[SteamKing]

Sorry, typo... should be -2y³ there.

Good. Recalculate your results from that point on.

 

[sa1988]

Good. Recalculate your results from that point on.

No need. I'll get the same result. I've already done all of this on paper.The mistake you saw was purely a typing error.

The only bit I'm stumped on is part d)

 

[Ray Vickson]

Homework Statement

2. The shape of a hill is described by the height function h(x,y) = (2 + x² + y⁴)^-1/2

(a) Find the gradient ∇h(x,y)

(b) What is the maximum slope of the hill at the point r₀ = i+j [or (x,y) = (1,1)]?

(c) If you walk north-east (in the direction of the vector i+j) from the point r₀, 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 r₀ that remains at the
same height.

Homework Equations

The Attempt at a Solution

For parts a), b) and c) I think I'm okay.

a) ∇h(x,y) = ([-x(2 + x² + y⁴)^-3/2], [-y³(2 + x² + y⁴)^-3/2])

b) At point (1,1) just plug x and y into ∇h(x,y) to get (-1/8 , -1/4)

c) Along path of vector (i+j) you will be going in a downhill direction since it is in a direction 'away' from the maximum slope found in the previous question. But is there a way I can mathematically prove this?

d) This is the part I'm unsure of. It's been a while since I did work with vectors. I'm guessing I need to find a vector such that infinitesimal change in the x-y direction gives no change in height. So that would mean d(h(x,y)) = 0...? Since this would signify no change in the value of h(x,y). Overall I'm a bit lost.

Thanks.

If you go along the surface $z = h(x,y)$ a short distance from a point with $x\,y$ coordinates $(x_0,y_0)$ to a neighboring point with $x\,y$ coordinates $(x_0+ \Delta x , y_0 + \Delta y)$, the change in "height" $z$ is given as
$$\Delta z = h_x(x_0,y_0) \Delta x + h_y(x_0,y_0) \Delta y$$
to first order in the small step sizes $\Delta x, \Delta y$. Here, $h_x = \partial h/\partial x$ and $h_y = \partial h/ \partial y$. The equation above is standard Calculus II material.

 

[sa1988]

If you go along the surface $z = h(x,y)$ a short distance from a point with $x\,y$ coordinates $(x_0,y_0)$ to a neighboring point with $x\,y$ coordinates $(x_0+ \Delta x , y_0 + \Delta y)$, the change in "height" $z$ is given as
$$\Delta z = h_x(x_0,y_0) \Delta x + h_y(x_0,y_0) \Delta y$$
to first order in the small step sizes $\Delta x, \Delta y$. Here, $h_x = \partial h/\partial x$ and $h_y = \partial h/ \partial y$. The equation above is standard Calculus II material.

Sure, I get that, but how does it apply to part d) of the question?

Thanks.

 

[Ray Vickson]

Sure, I get that, but how does it apply to part d) of the question?

Thanks.

PF Rules forbid us from telling you that.

 

[sa1988]

PF Rules forbid us from telling you that.

If it's not possible to get even a single nudge in the right direction, then it must mean that single nudge would give the whole answer, so I assume the whole thing must be far easier than I've been anticipating!

For what it's worth, I've actually gone beyond even Calc IV at this point. I'm just rusty with vectors as I took a year out of university.

I'll mull things over and try to see the glaringly obvious thing that I've missed.

Thanks!

 

[Ray Vickson]

If it's not possible to get even a single nudge in the right direction, then it must mean that single nudge would give the whole answer, so I assume the whole thing must be far easier than I've been anticipating!

For what it's worth, I've actually gone beyond even Calc IV at this point. I'm just rusty with vectors as I took a year out of university.

I'll mull things over and try to see the glaringly obvious thing that I've missed.

Thanks!

Hint: what is $\Delta z$ when you stay at the same level?

 

[sa1988]

Hint: what is $\Delta z$ when you stay at the same level?

Well it's 0, however I have no idea how that relates to finding a vector pointing in a direction such that its change in the z-direction is 0

Intuitively it seems to be that a vector that is orthogonal to ∇h(x,y) will be at that point moving in a direction such that it has no change in z-position.

... which in the case of my given answer would be (1/4 i - 1/8 j ) ...?

However I've done no calculus to get to that, which seems to go against the advice you're giving regarding $\Delta z$ etc. However I did work out ∇h(x,y) and the direction of steepest slope in the previous parts of the question, if that's what you were referring to.

 

[Ray Vickson]

Well it's 0, however I have no idea how that relates to finding a vector pointing in a direction such that its change in the z-direction is 0

Intuitively it seems to be that a vector that is orthogonal to ∇h(x,y) will be at that point moving in a direction such that it has no change in z-position.

... which in the case of my given answer would be (1/4 i - 1/8 j ) ...?

However I've done no calculus to get to that, which seems to go against the advice you're giving regarding $\Delta z$ etc. However I did work out ∇h(x,y) and the direction of steepest slope in the previous parts of the question, if that's what you were referring to.

I don't get why you are having a problem. Saying that $\Delta z = 0$ is saying that $z$ (the height) does not change. And, since we already have a formula for $\Delta z$ in terms of $\Delta x, \Delta y$ and the partial derivatives of $h(x,y)$, we have everything we need. Of course, that only determines the direction (the ratio of $\Delta y$ to $\Delta x$) and not the absolute magnitudes of $\Delta x, \Delta y$, so we are free to "scale" the results however we like.

 

[sa1988]

. Of course, that only determines the direction (the ratio of $\Delta y$ to $\Delta x$) and not the absolute magnitudes of $\Delta x, \Delta y$, so we are free to "scale" the results however we like.

Sure but does ∇h(x,y) not determine the direction of x and y such that the resultant vector is pointing in the direction of the steepest slope at that point of the function? Whereas my challenge is to find the vector which has no steepness at that point. Hence me intuitively suggesting it must be orthogonal to the gradient vector. Was I wrong?

Thanks

 

[Ray Vickson]

Sure but does ∇h(x,y) not determine the direction of x and y such that the resultant vector is pointing in the direction of the steepest slope at that point of the function? Whereas my challenge is to find the vector which has no steepness at that point. Hence me intuitively suggesting it must be orthogonal to the gradient vector. Was I wrong?

Thanks

Why are you arguing with me? I said exactly the same as what you have said, except that I gave reasons instead of just writing down a prescription without explanation.

 

[sa1988]

Why are you arguing with me? I said exactly the same as what you have said, except that I gave reasons instead of just writing down a prescription without explanation.

Oh wow no I wasn't arguing, sorry if it came across that way. It was just the first line of your previous post seemed to infer that I was still wrong overall, so I remained confused.

If all is good then that's fine, thanks.

 

[morez]

simple calculate here and answer it.
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