High School Calculus help on word problem.

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dungas

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If two men are running away from each other at 45 degrees and man1 is running at 12 km/h and man2 is running at 10 km/h; How fast are they moving apart at 2.5 hours?

Grade 12.

Do I use special triangles for this question?

How do I even begin with this question?

Please help.
 
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I think that you need to begin by drawing a picture.

Next, introduce a suitable coordinate system, and then start assigning symbols to represent things. (Hint: put the 45-degree angle's vertex at the origin with one ray on the positive x-axis and the other ray in QuadI.)

Pick a representative point in time, and figure out how you would calculate the distance between them at that time.

That is, if you were to come up with a function for this distance in terms of time, then you could use its derivative to answer the question.
 
mmm4444bot said:
I think that you need to begin by drawing a picture.

Next, introduce a suitable coordinate system, and then start assigning symbols to represent things. (Hint: put the 45-degree angle's vertex at the origin with one ray on the positive x-axis and the other ray in QuadI.)

Pick a representative point in time, and figure out how you would calculate the distance between them at that time.

That is, if you were to come up with a function for this distance in terms of time, then you could use its derivative to answer the question.
Click to expand...


Do I make man 1 represent as 12t and man 2 as 10t? I had the picture drawn beforehand I just need help with making an equation.


edit: ​< link to objectionable page removed >
 
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dungas said:
I had the picture drawn beforehand I just need help with making an equation.
Click to expand...

That's not what you said, in either of your first two posts!

Your sketch is not correct; the men are not running on the same line.
 
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mmm4444bot said:
That's not what you said, in either of your first two posts!

Your sketch is not correct; the men are not running on the same line.
Click to expand...

​< link to objectionable page removed >

Like this then?


and yeah I should've uploaded the pic first.(so I can at least get that right first)
 
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Hello, dungas!

Your diagram is still wrong.
There are no right angles in the problem.

Two men are running away from each other at 45o.
Man-1 is running at 12 km/hr and Man-2 is running at 10 km/hr.
How fast are they moving apart at 2.5 hours?
Click to expand...

I assume the two men start from the same point
. . and start running at the same time.

Code: 
                    B
                    *
                  *  *
            10t *     *  x
              *        *
            *           *
          * 45[SUP]o[/SUP]          *
      O *  *  *  *  *  *  * A
                12t
In \(\displaystyle t\) hours, Man-1 runs from \(\displaystyle O\) to \(\displaystyle A,\)
. . a distance of \(\displaystyle 12t\) km.

In the same \(\displaystyle t\) hours, Man-2 runs from \(\displaystyle O\) to \(\displaystyle B,\)
. . a distance of \(\displaystyle 10t\) km.

Their distance is: \(\displaystyle x \,=\,AB.\)


Law of Cosines: .\(\displaystyle x^2 \:=\: (10t)^2 + (12t)^2 - 2(10t)(12t)\cos45^o\)

This simplifies to: .\(\displaystyle x^2 \:=\: \left(244-120\sqrt{2}\right)t^2 \;=\;4\left(61-30\sqrt{2}\right)t^2\)

And we have: .\(\displaystyle x \:=\: \left(2\sqrt{61-30\sqrt{2}}\right)t\)

Hence: .\(\displaystyle \dfrac{dx}{dt} \;=\;2\sqrt{61-30\sqrt{2}} \)


Therefore, at any time \(\displaystyle t\), their distance is changing
. . at the constant rate of about 8.6 km/hr.
 
dungas said:
<link removed>

Like this then?


and yeah I should've uploaded the pic first.(so I can at least get that right first)
Click to expand...

Now suppose:

The person running along the x-axis is A and the person running along the on 45° line is B

the time spent after they started together from the origin = t

Then at time 't'

location of A = [10 * t, 0]

location of B = [ 12*cos(45°) * t, 12*sin(45°) * t]

What is the distance(s) between A and B at time 't'?

What is the rate of change of the distance(\(\displaystyle \frac{ds}{dt}\)) between A and B at time 't'?
 
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soroban said:
Hello, dungas!

Your diagram is still wrong.
There are no right angles in the problem.


I assume the two men start from the same point
. . and start running at the same time.

Code: 
                    B
                    *
                  *  *
            10t *     *  x
              *        *
            *           *
          * 45[SUP]o[/SUP]          *
      O *  *  *  *  *  *  * A
                12t
Click to expand...

How can you tell there is not a right angle?

Is it because they are not running at same speed


thanks for helping!
 
Last edited:
dungas said:
How can you tell there is not a right angle?

Is it because they are not running at same speed
Click to expand...

Basically, yes, but you may quickly confirm that the triangles are not right triangles, using right-triangle trigonometry.

Your second sketch shows a right triangle with base 10 (I'm letting t = 1 hr for purposes of discussion here), and the 45-degree angle's vertex is located at the left end of the base.

The hypotenuse of this right triangle needs to be 12 at t = 1, agree?

Do you remember: cos(theta) = adjacent/hypotenuse

Substitute 45 degrees for theta and 10 for the adjacent side, and then calculate the hypotenuse. You will not get 12!

Cheers :cool:
 
It's not so much a matter of "knowing it is not a right triangle" as "not knowing that it is a right triangle". The cosine law applies to any triangle, whether it is a right triangle or not.
 
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