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...
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omg why did you draw a right-angled triangle? you already went wrong with that :l
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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
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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:
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.
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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.
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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 :(
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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.
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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
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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).
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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
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I'm not even sure what you're supposed to find LOL.
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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
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Try this: Law of sines It'll make your life easier.
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Are you doing any math tutoring for us dummies? (T^T)
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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
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