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Anyone any good at basic applied mathematics? (Closest Approach) - Final update

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ArmadX

I have a (small) test on it tomorrow, and although I know most of the topics pretty well, I've been trying some of the questions from the book, and this one has really got me stumped.

My working is pretty messy, and probably incomprehensible, but maybe it'll help someone spot where I've gone wrong.

The answer is 1.92km (probably rounded)

Thank you to the many talented people of SG with better minds than me ;)

UPDATE: Here's some much neater (and probably better) working

UPDATE 2: Next piece of working. I gave up after this, since we haven't learnt how to find minimum values, but I'll post an update after asking my professor tomorrow.

FINAL UPDATE: I asked my professor about the question, which he glanced at, glanced at my working, and then said there's a much easier way to do it. When I asked him what it was he said he'll tell me after the holidays. Nooooooooooooooooooo!

PS I also received my test results, I got 99%, so looks like it didn't matter too much that I couldn't solve this question :D

11 years ago*

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33 Comments
XXvincent

i believe i just heard my brain shutting down after watching those problems

11 years ago
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ArmadX

May want to look at the sauce :)

11 years ago
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loco2turco

All i have figured out is that when cyclist A has driven 1.5km, cyclist B has reached the crosspoint which means that their distance is 2km...

11 years ago
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ArmadX

Yeh, I got that too :/ And then I got that the velocity of A relative to B is 25i-(15sqrt3)j, but I'm stuck on what to do next

11 years ago
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loco2turco

omg why did you draw a right-angled triangle? you already went wrong with that :l

11 years ago
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ArmadX

I drew several right-angled triangles, and don't see why any of them would be wrong.

Which one do you mean?

11 years ago
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minanu

c=(a^2+b^2-2abcos60)^(1/2)

Use this, angle in the center will be always 60 degrees. Put the length of a(distance between A and junction) and b(distance between B and the junction) as a function of time. Take derivative, find the place where it is minimum. Then use approximation

11 years ago
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ArmadX

Ok, I've done that rule :)

Using it, I get that c=(sqrt37)/2

However, I haven't done calculus yet (I believe derivations are a part of calculus), so there must be another way to solve it

11 years ago
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Nanabanana

So this requires some combining of functions.
Point A travels along vector A. Point B travels along vector B.

To calculate the distance between point A and point B, determine a right triangle, with the hypothenuse connecting those points:

  1. Use sine 60 degrees = X / ( distance A to intersection point) -> X = sin 60 * ( distance A to intersection point). This will be one side of the right triangle we're looking for.
  2. Use cosine 60 degrees = Z / ( distance A to intersection point) -> Z = cos 60 * ( distance A to intersection point).
  3. Say Y = Z - (distance B to intersection point). This is the other side of the right triangle.
  4. Pythagoras then states that (Distance between point A and B) ^ 2 = X ^ 2 + Y ^ 2.

That's the easy part.
Now, in the above, replace the distances to intersection point with the linear functions relative to time.

dA = starting distance A - velocity A time -> dA = 3.5 - 30 t
dB = starting distance B - velocity B time -> dB = 2 - 40 t.

So distance between A and B as a function of time becomes:
(Distance between point A and B) ^ 2 = ( sin 60 ( 3.5 - 30 t ) ) ^ 2 + ( (2 - 40 t) - ( cos 60 ( 3.5 - 30 * t ))) ^ 2.

Simplify that and you should end up with a fairly basic equasion to find the minimum value for.

11 years ago
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ArmadX

I'm assuming for part 3 you meant to say Y=(distance to B)-Z, since Z<(distance to B)
That gives this

I'm fine up until you say (Distance between point A and B) ^ 2 = ( sin 60 ( 3.5 - 30 t ) ) ^ 2 + ( (2 - 40 t) - ( cos 60 ( 3.5 - 30 t ))) ^ 2.*
Here, aren't you using Pythagoras' rule (hypotenuse^2=(one side of the triangle)^2+(other side of the triangle)^2?
If the triangle you're doing this for is AOB, then how come you can do that, seeing as you can't assume it's a right angled triangle?

EDIT: After some thinking, I decided to use the cosine rule, using the distances to O relative to time as the lengths
This is the result. I understand that I now need to find what t is for the minimum value of d. Not sure how to do that though. Oh well, I guess I'll just hope the questions aren't this hard in the test :)

I'll ask my professor about it and see if he had a better way, using methods we've learnt, and will post an update.

11 years ago
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Nanabanana

Well I was doing this from the top of my head, but the idea is that you can effectively split the OAB triangle into two straight ones by drawing a line from point A on a 90 degrees angle to the line OB is on.

For the sake of the rest of the math, it doesn't really matter if Y = Z - (distance B to intersection point) turns negative, as it doesn't matter if both triangles overlap or not. It'd be easier to sketch it out, but I don't have the tools at hand for now :(

11 years ago
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ArmadX

Yeh no worries :)

I know it would all work out in the end, but it's still simpler if Y > 0

11 years ago
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Nanabanana

Eh, you'll start disagreeing on that once you start calculating vectors. Drop a sign there and you end up wondering why your arrows are pointing the wrong way all of a sudden.

11 years ago
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ArmadX

I've done vectors already (There are vectors in my first bit of working) :) I meant for that particular example, I know it doesn't apply for everything xD

11 years ago
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Toca1FTW

Well I managed to get the correct answer!

Using the cosine rule, feeding in known data and simlifying, you get distance squared is: (x is hours)
9.25 - 170x + 1300x^2

Differentiate that to get
2600x - 170 = 0

so the closest approach is after 0.06538 hours, which feeding back into the first equation is 1.92km

11 years ago
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ArmadX

Haven't done differentiating yet, so can't use that :/

11 years ago
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patetico

Given f(x) = ax^2 + bx + c, the max/min of f(x) is -(b^2 - 4ac)/4a

You can thank me later =)

11 years ago
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ArmadX

I'll make a note of it for the future :)

11 years ago
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Toca1FTW

What patetico is talking about is quadratic function. If you've covered that you might be expected to know one of the formulas.

We got to
1300x^2 - 170x + 9.25
and I just used calculus there, but it's a quadratic, so at the mimimum x is
-b/2a
= - (-170)/(2*1300) = 0.065 hours

or you might be expected to remember the equation patetico posted, which gives the distance squared directly.

BTW: just saw your edit 2, I think you have an error on your second line, you are dividing everything at the end by 2, but you forgot the 2 that was there to start with, so they cancel out (2 x cos(60) = 1).

11 years ago
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patetico

For the record: I would have used your first method too.
I didn't remember the formula itself, just remembered that it existed =p

11 years ago
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ArmadX

Yes, I've done f(x) = ax^2 + bx + c :)

If I've done finding the minimum though, I don't remember it, I'm used to just entering the function into my graphic calculator and finding the minimum from it.

Ah yes, there is :) Even then, it still doesn't help me with solving it, and the professor would do it a completely different way

11 years ago
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hulaballoon

I'm not even sure what you're supposed to find LOL.

11 years ago
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ArmadX

"Calculate the shortest distance between the cyclists in subsequent motion."
There you go :D

11 years ago
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jatan11t
(Super Moderator)

Applied mathematics? And here I was thinking you were going to ask a complex analysis question :(

Maybe I'll be really evil and make a math quiz haha

11 years ago
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ArmadX

You're welcome to include this question ;) Teehee I already know the answer :D

11 years ago
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jatan11t
(Super Moderator)

Your question is not complex analysis. I'm talking about something like this

11 years ago
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ArmadX

I was only replying to "Maybe I'll be really evil and make a math quiz haha" :P

11 years ago
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Solnaga

Oh wow... I didn't have to take higher levels of math for my profession but I really like this problem :o

11 years ago
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ArmadX

Well, there are plenty more in my book like it ;) I'll take photos of more questions if you like :)

11 years ago
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AterVulpes

Try this: Law of sines It'll make your life easier.

11 years ago
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ArmadX

I know that one :) Except we haven't done circumcircles yet

11 years ago
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MaximumBunny

Are you doing any math tutoring for us dummies? (T^T)

11 years ago
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ArmadX

Haha I wish I had the time, would be a nice way to make a little cash :)

11 years ago
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Closed 11 years ago by ArmadX.