# Distance traveled winding around sphere

## Homework Statement

I have a sphere that I want to travel from the north pole to the south pole. The route I take is winding around the sphere (instead of the obvious shortest path). My path is dictated by:

##\phi(t)=wt##
##z(t)=R-v_z t## in the negative z direction per N to S

The form of the final answer is given!

##d_{tot}=\int_?^? A\sqrt{1+B^2sin^n(\theta)}d\theta##

A and B are some constants to find.

## Homework Equations

dl for spherical:##dl=dr\hat{r}+rd\theta \hat{\theta}+r sin(\theta)d\phi \hat{\phi}##

## The Attempt at a Solution

I know that R is constant since we're flying around at the same altitude. I believe the limits for ##\phi## are 0 to 2##\pi## and ##\theta## 0 to ##\pi##. Clearly according to the given integral I don't need to worry abut phi and R anyway though.

I thought that I may take ##z(t)=R-v_z t## and convert the z coordinate to spherical so that ##Rcos(\phi)=R-v_z t## but that hasn't really lead me anywhere near that integral so far.

Just not sure how to start. Anyone care to help point me in the right direction?

Simon Bridge
Homework Helper
...I believe the limits for ϕ are 0 to 2π ... convert the z coordinate to spherical so that ##R\cos\phi=R-v_zt##
If ##z=R\cos\phi##, then that makes ##\phi## the angle from the +z axis, so the limits for ##\phi## cannot be ##0## to ##2\pi##.

Remember, you want to express the equations in ##t##, that you have, in terms of equations ##\theta## and the rest. To do that, you need to understand what they are.

I'm at a complete loss as to what I should do. It seems to me that I need a way to get rid of the z component if I'm to be able to do this in spherical. And, I don't understand the final integral at all. Somehow they removed the phi dependence for the distance and reduced it to a single integral instead of a double as I would expect.

Simon Bridge
Homework Helper
OK - perhaps I was not clear: is ##\phi## the angle from the +z axis or not?
If yes - then you should adjust those limits.
If no - then it must be ##z=R\cos\theta = R-v_zt## right?

##\theta## should be the angle from the z axis. ##\phi## is the angle from the x axis.

Simon Bridge
Homework Helper
##\theta## should be the angle from the z axis. ##\phi## is the angle from the x axis.
Then ##R\cos\theta = R-v_zt## - hint: solve for t and sub into the equation for ##\phi## Or maybe the other way around - gets you ##\theta## in terms of ##\phi##.

I've done the algebra. Going to have to sleep on it to decide what I should do with it next. Not sure at all where the root or powers are coming from.

LCKurtz
Homework Helper
Gold Member
I've done the algebra. Going to have to sleep on it to decide what I should do with it next. Not sure at all where the root or powers are coming from.

Remember that the scalar arc length element ##ds## is the square root of the sum of the squares of the components of ##dl##. You can get an easy equation for ##\phi## in terms of ##\theta## as already suggested. Then express ##ds## in terms of ##d\theta## and you will see.

Okay, since ##ds=\sqrt{1+(\frac{d\phi}{d\theta})^2}##

in spherical ##dl= \hat{r}dr+rd\theta \hat{\theta}+rsin(\theta)d\phi \hat{\phi}##

I think: ##ds=rsin(\theta)\sqrt{1+(\frac{d}{d\theta}\frac{wR}{v_z}(1-cos(\theta))^2}##
##=Rsin(\theta)\sqrt{1+(\frac{wR}{v_z}sin(\theta))^2}##

##A=Rsin(θ)##
##B=(\frac{wR}{v_z})^2##
n=2

the limits would be 0 to ##\pi##

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LCKurtz
Homework Helper
Gold Member
Remember that the scalar arc length element ##ds## is the square root of the sum of the squares of the components of ##dl##. You can get an easy equation for ##\phi## in terms of ##\theta## as already suggested. Then express ##ds## in terms of ##d\theta## and you will see.

Okay, since ##ds=\sqrt{1+(\frac{d\phi}{d\theta})^2}##

in spherical ##dl= \hat{r}dr+rd\theta \hat{\theta}+rsin(\theta)\color{red}{d\phi \hat \phi}##

I think: ##ds=rsin(\theta)\sqrt{1+(\frac{d}{d\theta}\frac{wR}{v_z}(1-cos(\theta))^2}##
##=Rsin(\theta)\sqrt{1+(\frac{wR}{v_z}sin(\theta))^2}##

##A=Rsin(θ)##

But ##\theta## is not a constant.

##B=(\frac{wR}{v_z})^2##
n=2

the limits would be 0 to ##\pi##

Given that ## dr## is zero, what do you get for ##ds^2## if you take the sum of the squares of the components of ##dl##?

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I don't know what to put into dl.
dropping hats
##dl=dr+rd\theta+rsin(\theta)d\phi##
I only have 1 equation now

##\phi(t)=\frac {Rcos(\theta)-R}{v_z}##

So I have no equation for r (but dr=0 so okay) or theta. I'm left with only ##\phi(t)##

The only reason I just plugged it into the ##\frac{d\phi}{d\theta}## is because it's the only derivative I can do. What could I use for ##\frac{d\theta}{d\theta}##?

LCKurtz
Homework Helper
Gold Member
Given that ## dr## is zero, what do you get for ##ds^2## if you take the sum of the squares of the components of ##dl##?

Please answer my question before continuing. You don't just "drop the hats".

I don't know what to put into dl.
dropping hats
##dl=dr+rd\theta+rsin(\theta)d\phi##
I only have 1 equation now

##\phi(t)=\frac {Rcos(\theta)-R}{v_z}##

So I have no equation for r (but dr=0 so okay) or theta. I'm left with only ##\phi(t)##

The only reason I just plugged it into the ##\frac{d\phi}{d\theta}## is because it's the only derivative I can do. What could I use for ##\frac{d\theta}{d\theta}##?

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##ds^2=0^2\hat{r}+r^2d\theta^2\hat{\theta}+r^2sin^2(\theta)d\phi^2\hat{\phi}##

I was "dropping hats" because latex becomes cumbersome to type after a while.

LCKurtz
Homework Helper
Gold Member
##ds^2=0^2\hat{r}+r^2d\theta^2\hat{\theta}+r^2sin^2(\theta)d\phi^2\hat{\phi}##

I was "dropping hats" because latex becomes cumbersome to type after a while.

##ds^2## is a scalar. You just want the components squared. If I asked you for the length^2 of a vector ##a\vec i + b\vec j + c\vec k## would you leave the ##\vec i,\vec j, \vec k## in the answer? Fix that and lets see what you have for ##ds##.

##ds^2=r^2d\theta^2+r^2sin^2(\theta)d\phi^2##
## ds^2=r^2(d\theta^2+sin^2(\theta) d\phi^2)##

Since I have no function for ##\theta## (but it goes from 0 to ##\pi##) can I go ahead and say that:

## ds^2=r^2(\pi^2+sin^2(\theta) d\phi^2)##

LCKurtz
Homework Helper
Gold Member
##ds^2=r^2d\theta^2+r^2sin^2(\theta)d\phi^2##
## ds^2=r^2(d\theta^2+sin^2(\theta) d\phi^2)##
That's good.
Since I have no function for ##\theta## (but it goes from 0 to ##\pi##) can I go ahead and say that:

## ds^2=r^2(\pi^2+sin^2(\theta) d\phi^2)##

No. But now you do have$$ds=\sqrt{r^2(d\theta^2+sin^2(\theta) d\phi^2)}$$and thats what you want to integrate. You know ##r = R## is constant, and from the previous discussion I think you may have figured out how to get ##\phi## in terms of ##\theta## so you can get ##d\phi## in terms of ##d\theta## and you should be good to go. I will revisit this thread later today to see how you come out.

So, am I just suppose to put in

##d\phi=\frac {Rcos(\theta)-R}{v_z} d\theta## and then do algebra to bring out ##d\theta##

LCKurtz
Homework Helper
Gold Member
I can't follow stuff when you just write some equation down like that where I don't know where you got it. What is the equation giving the relation between ##\phi## and ##\theta##? Show what it is and how you get it. After you have that equation you can show how you calculate ##d\phi## from that equation.

##\phi(t)=wt##
from
##z(t)=R-v_z t## where ##z=Rcos(\theta)##
so ##Rcos(\theta)=R-v_z t##
solving for 't' and plugging back into ##\phi(t)=wt##
##\phi(t)=w\frac {Rcos(\theta)-R}{v_z}##

Kept forgetting to put 'w' in there.

LCKurtz
Homework Helper
Gold Member
##\phi(t)=wt##
from
##z(t)=R-v_z t## where ##z=Rcos(\theta)##
so ##Rcos(\theta)=R-v_z t##
solving for 't' and plugging back into ##\phi(t)=wt##
##\phi(t)=w\frac {Rcos(\theta)-R}{v_z}##

Kept forgetting to put 'w' in there.

You have a sign mistake on the last step, probably from not writing down the steps. Fix that and take differentials to get ##d\phi## in terms of ##d\theta##. Then let's see what you get for your integral.

Yeah, sorry I copied that from the wrong part of my notes. It gets a little cluttered after a while of trying different things.

##\phi(t)=w\frac {R-Rcos(\theta)}{v_z}##

1)##\frac{\partial \phi}{\partial \theta}=\frac{wR}{v_z}sin(\theta) d\theta##

2)##(\frac{\partial \phi}{\partial \theta})^2=\frac{w^2R^2}{v_z^2}sin^2(\theta)d\theta^2##??

##ds=R\sqrt{d\theta^2+sin^2(\theta)(\frac{wR}{v_z})^2sin^2(\theta) d\theta^2}##

##ds=Rd\theta\sqrt{1+(\frac{wR}{v_z})^2sin^4(\theta)}##

I'm not sure if I am suppose to square my derivative or leave it as is. For instance, should I be plugging 2 in for the integral or should I square the d##\theta in 1? The annotation isn't clear to me.

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I don't know why it had me logged in as someone else. University computer. Guess I'm not the only one here at the U of I looking for help!

LCKurtz
Homework Helper
Gold Member
##ds=Rd\theta\sqrt{1+(\frac{wR}{v_z})^2sin^4(\theta)}##

So the length is$$d_{tot}=\int_?^?R\sqrt{1+\left(\frac{wR}{v_z}\right)^2\sin^4(\theta)}~d\theta$$Looks like you are ready to answer the original question. • 1 person
Thanks for being patient with me. I really appreciate the help.