Rate of change help please

Macy · 2011-09-10 19:34:12

1. A train is travelling at 15km/hr along a straight track. A fixed camera, 500m away from the track, is focused on the train.
a. How fast is the distance from the camera to the train changing when the train is 1km from the camera?
b. How fast is the camera rotating (in rad/hr) at the moment when the train is 1km from the camera?

2. Two poles, one 1m tall and one 2m tall are 3m apart. A length of wire is attached to the top of each pole and it is also
staked to the ground somewhere between the two poles. Where should the wire be stacked so that the minimum amount
of wire is used?

Thank you all.
M

Bob · 2011-09-10 20:20:11

hi Macy

I've been persuaded to re-think this answer. I now think all of question 1 is WRONG so please see post #8 for my corrected version. Question 2 is ok as stated here.

I've drawn a diagram, see below.

(1) a.
TC = √(500x500 + 1000x1000)

TX is changing at 15Km/hr
so TC will change at 15 x TC/TX Km/hr.

(note: It doesn't matter that I've got the distances in metres and the speed in Km / hr as I'm only using the distances for a ratio. Use the Km equivalents if it makes you feel happier.)

b.

rads = velocity / radius = 15 / TC rads / hr. (but here the 15 and TC must be in the same units so change one of these)

I'M JUST READING THIS THROUGH AND I CAN SEE I NEED TO ADJUST THIS TO GET THE 'PERPENDICULAR' VELOCITY AS THE 'LINE OF SIGHT' VELOCITY WILL CONTRIBUTE ZERO ANGULAR VELOCITY. HANG ON A MO WHILE I ALTER THE CALCULATION.

OK got it.

angle CTX = arctan(500/1000)

Split the 15 Km/hr into two components

15 cos CTX towards C

15 sin CTX at right angles to CT.

The first component has zero angular velocity as it's directed straight towards the camera. The value you want is

= 15 sin CTX / CT (in Km)

(2) Again I've made a diagram and divided the ground lengths into x and 3 - x as shown.

Use Pythagoras to get the hypotenuse for each part of the wire.

Then add these together.

Then differentiate and set this equal to zero. This will give the minimum length. (Strictly, you should show that it is a minimum and not a maximum!)

Post again if you need more details.

Bob

Last edited by Bob (2011-09-12 03:56:48)

Macy · 2011-09-10 20:46:38

Hi Bob, You are such an enthusiastic person that I have not seen in my life..
I love u, dear and learn from you too.
Okie I understand question num 2 now but I still not get my head around the conversion
to rad from Velocity. Let me think a bit more tonite

Bob · 2011-09-10 21:35:19

hi Macy

Happy to help.

The formula I used was:

there's an article at

http://en.wikipedia.org/wiki/Angular_velocity

It gets complicated fairly quickly, but the first diagram shows why you need the sine of the angle. You don't need to look at the rest. (unless you want to, of course! )

Bob

Last edited by Bob (2011-09-10 21:39:11)

Macy · 2011-09-11 12:46:37

Hi Bob, for question 1a I do not understand why you do TX/TC. Which rate is it.
As fas as i know if we use dx/dt = dx/dv . dv/dt then we can find it. We have dv/dt = 15 km/h but
how do we work out dx/dv

Bob · 2011-09-11 19:42:38

hi Macy,

dx/dt = dx/dv . dv/dt

This might arise when the velocity is changing with time. Your train has a constant velocity, so v = dx/dt = 15 for all 't'.

I thought as follows:

As T moves towards X the distances TX and CX get smaller.

CX is longer so it will lose relatively more length.

For me, that word 'relatively' is the key. I thought the rate of change of CX should be in proportion to the distances of these two measurements; hence the ratio CX/TX, which would be over 1 and therefore would make the rate of change of CX bigger than 15.

Now you've questioned that I've started to have doubts so I need to think about it and either explain it thoroughly or put a new idea in its place. I'll come back to this post later and edit it as needed so watch this place.

First edit: Just re-read the question properly. It says the train is 1 Km from the camera so my original diagram is not even correct. I am so sorry. I should have put CX = 1000 m not TX. That will alter the angle (arhh, it's exactly 30 now) so I'll re-do my first post with the revised figures. Back when I've resolved the ratio issue.

Bob

Last edited by Bob (2011-09-11 19:48:15)

anonimnystefy · 2011-09-11 20:11:00

hi guys

i have got a geometrical answer for two,and seems a little easier to do,and shorter to write,but maybe a little bit more complicated to understand:

in geometry class we learned that the minimum length in this case we could get by turning one pole around with symmetry,then connect the top of the inverted pole with the top of the other pole.where the poles intersect you will get the minima.

if you want to get the length of the wire you will see that it is a hypotenuse of a right triangle whose one side is the sum of heights of the two poles,and the other side is the distance between the two poles.

now to get a picture of what i mean check Figure 1:http://www.springerlink.com/content/n51 … 1&locus=66

if bob understood this maybe he could make a clear picture?

Bob · 2011-09-11 20:12:37

hi Macy,

Revised question 1.

Firstly my apologies for a poor attempt at your question. I think this version is correct so I'm going to edit a note on the first version pointing people to this one, in case someone else reads it and thinks it is correct.

Also, thanks to you for being persistent about understanding my answer. If you had just accepted what I wrote, we would both have gone away thinking the question was done properly. That's the thing about maths; it has got to be convincing. Let's hope this version is.

See the new diagram. With the measurements like that the angle CTX is 30 degrees (arcsin 500/1000) so I've used that value in the diagram. It seems to fit better because the question setter wanted to test the 'mechanics' of the situation, not your ability with trig.

I've resolved the vector 15 into two components, 15cos30 along the line TC and 15sin30 at right angles to this line.

So for part a. we have TC is changing at 15cos30 Km / hr.

For part b. the angular velocity is still given by velocity / radius. The component along the line TC has no angular velocity as it's along the line of sight so the angular velocity is

15sin30/1

(I've put the distance back into Km for this so that the units are consistent.)

Now, is that convincing?

Bob

Last edited by Bob (2011-09-11 20:15:14)

anonimnystefy · 2011-09-11 20:17:40

hi bob

it looks like we simu-posted,so could you check the post above.i kinda need your help.

Bob · 2011-09-11 20:26:14

hi Stefy,

Yes, that's a good method. I'll have a go at a diagram.

I've put on a "not shortest" and the "shortest" line.

It will make me feel better about making a hash of question 1. (embarrassed smile).

Bob

Last edited by Bob (2011-09-11 21:11:21)

anonimnystefy · 2011-09-11 20:29:18

hi bob

thanks very much.

you shouldn't worry about being wrong.everyone is from time to time,but no one seems to care.

Bob · 2011-09-11 20:35:54

hi Stefy,

Thanks for the words of encouragement. I can live with being wrong but I'd hate someone to get their work wrong because of me.

Got to go for now. Back later.

Bob

Last edited by Bob (2011-09-11 20:36:11)

anonimnystefy · 2011-09-11 20:39:38

hi bob

yes it would be a pity but you cannot worry about it so much.you have done your share in helping our world.everything else is just good will.

see you.

Bob · 2011-09-11 21:21:19

hi Macy,

A post script.

If a formula / method works, it should word at other points on the train's travel.

With 15 x TC/TX, the answer gets bigger as T gets nearer to X, which is clearly wrong. At the extreme when T is at X the values goes off to infinity ( 15 x 0.5 / 0 ) so I should have realised it was rubbish.

With my revised version, 15 cos(angle), as T approaches X the angle increases and the cos (angle) decreases. That's more like it! In the extreme, when T is at X, 15 cos 90 = zero, which makes sense as the line TC isn't getting bigger or smaller at this moment.

And the angular velocity, 15 sin(angle), gets larger as the train crosses our field of view more rapidly (good) and when angle = 90 it becomes 15/1 which is correct as the direction of travel is now perpendicular to the line of sight.

Hurrah!

Bob

reconsideryouranswer · 2011-09-12 03:14:14

Regarding the color used in post #2:

Would you be able to use a color other than the one you used
to reply to Macy, because the contrast is relatively harsh against the
normal color background?

If not, thank you anyway.

Last edited by reconsideryouranswer (2011-09-12 03:17:40)

Bob · 2011-09-12 03:59:34

Thanks for the suggestion. I wanted the error to be clear. I've made it bold green. In theory yellow should have worked but it looked dreadful.

Bob

Macy · 2011-09-12 13:36:25

Bob do not feel embrassed here. We learn from each other ..IF I think it does not make sense
then I will post my question again. However I am so busy today so I will think about your logic
tonite.
I still have my love for u

Math Is Fun Forum

#1 2011-09-10 19:34:12

Rate of change help please

#2 2011-09-10 20:20:11

Re: Rate of change help please

#3 2011-09-10 20:46:38

Re: Rate of change help please

#4 2011-09-10 21:35:19

Re: Rate of change help please

#5 2011-09-11 12:46:37

Re: Rate of change help please

#6 2011-09-11 19:42:38

Re: Rate of change help please

#7 2011-09-11 20:11:00

Re: Rate of change help please

#8 2011-09-11 20:12:37

Re: Rate of change help please

#9 2011-09-11 20:17:40

Re: Rate of change help please

#10 2011-09-11 20:26:14

Re: Rate of change help please

#11 2011-09-11 20:29:18

Re: Rate of change help please

#12 2011-09-11 20:35:54

Re: Rate of change help please

#13 2011-09-11 20:39:38

Re: Rate of change help please

#14 2011-09-11 21:21:19

Re: Rate of change help please

#15 2011-09-12 03:14:14

Re: Rate of change help please

#16 2011-09-12 03:59:34

Re: Rate of change help please

#17 2011-09-12 13:36:25

Re: Rate of change help please

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