# Related Rates, Optimization, Integrals

1. Dec 12, 2005

### scorpa

Hi everyone,
Well its that great time of year....the time of feverishly studying for finals and I have been doing some practice questions and there are a few that I'm stuck on. My first question I am embarassed to ask, it should be so simple, yet I cannot get the right answer.
1) A highway patrol airplane flies 3 km above a level, straight road at a speed of 120 km/hr. The pilot sees an oncoming car and with the radar determines that the line-of-sight distance from the airplane to the car is 5 km and decreasing at the rate of 160 km/hr. Find the car’s speed along the highway
So for this I know the answer is 80 km/h because we have the answers provided but still knowing how to actually get the answer would be nice to
Anyway, I started off with drawing my diagram , and labelling it with all of the givens. Next I went with pythagoreans theorem x^3 +y^2-=z^2, and differentiated it. At this point I ended up with 4x+(3)(120) =(5)(160) which gives an answer of 110km/h which of course is not right. I know I am just doing something really stupid here but I dont know what .
2) $$/int$$ abs(3x-2)dx where b=2 and a=0.
I thought this could just be done normally but when I did it I didn't get the answer of 10/3. What I did was I took the function and integrated it to get 3x^2/2 -2x, but try as I might I always ended up with 2 as my answer when I put the numbers in.
3) Find the equation of the line through the point (2,7) such that the area in the first quadrant bound by the line, the x-axis and the y-axis is a minimum
For this one I quite honestly have no idea of where to even start. The answer is supposed to be y = (-7/2)x +14, I dont really know where to start so if there is anything that could just get me on the right track I would appreciate that.

Last edited: Dec 12, 2005
2. Dec 12, 2005

### Fermat

1)
Let h be the height of the plane above the road.
Let s be the line-of-sight distance.
Let L be the horizontal distance along the road, from directly underneath the plane, to the oncoming car.

s = 5 km, h = 3 km, so by pythagoras, L = 4 km.

Also,

s² = L² + h²

differentiate both sides wrt time,

2s.ds/dt = 2L.dL/dt + 2h.dh/dt
s.ds/dt = L.dL/dt + h.dh/dt

But the horizontal height is constant, therefore dh/dt = 0, giving

s.ds/dt = L.dL/dt
=============

You are given s and ds/dt and have worked out L, from which you can now get dL/dt.
From dL/dt. you can work out the speed of the oncoming car.

2)
You are working with the abs function, therefore you have to ensure that any value it takes actually is positive when you do the integration.

(3x - 2) is negative for x e [0,2/3], so to make it positive, write it as (2 - 3x).
You integral should now look like this,

$$\int _0 ^{2/3} (2 - 3x)\ dx + \int _{2/3} ^2 (3x - 2)\ dx$$

3)
The line passes through the point (2,7), so the eqn of a straight line passing through that point is,

y - 7 = m(x - 2)

Use this eqn to get the intercepts on the y-axis and the x-axis. Now get the area of the triangle that you have to minimise the area of. You will get the area in terms of m, the gradient of the straight line.

3. Dec 12, 2005

### scorpa

Ok thanks a lot, I have been doing exactly what you said for the airplane one but wasn't getting the right answer so I'll go back and see if I'm making a stupid error somewhere.

I forgot for the second one that it would be necessary to break it up into parts so thats where I was going wrong with that, thank you very much.

I see what you are saying for the third one, I never thought of doing that, it's so simple and easy yet I never even thought of that. Gotta love how that happens.

Thanks a lot I really appreciate it!!!

4. Dec 13, 2005

### scorpa

Ok for the first question I am getting dL/dt to be 200, I did just what you said and solved for dx/dt from (4)(dx/dt)=(5)(160) I'm not quite sure where I'm going wrong.

5. Dec 13, 2005

### Fermat

dL/dt = 200 km/hr is correct.

Imagine the plane has a shadow vertically underneath it on the road. As the plane flies along, 3km above the road, at 120 km/hr, so also does the shadow move along the road at 120 km/hr.

dL/dt is the rate at which the distance L is decreasing. In other words it is the speed of approach between the oncoming car and the plane's shadow.

Can you get it now ?

6. Dec 13, 2005

### scorpa

Oh you just subtract 200km/h from 120km/h to get the speed of the car because they are coming towards each other?

7. Dec 13, 2005

### Fermat

That's it

8. Dec 13, 2005

### scorpa

Awesome, thank you very much!!!

9. Dec 13, 2005

### scorpa

OK I thought I had the last one but apparently not. I know this is a stupid question, but how do you figure out the x and y intercepts when you have more than one variable in the equation....meaning you dont know the slope of the line?

10. Dec 13, 2005

### scorpa

******bump*****

11. Dec 14, 2005

### Fermat

The slope of the line is your variable in the expression for the area of the triangle.

Get the axis-intercepts in terms of m.
Get the area of the triangle in terms of m, A = f(m), say.
As you vary the slope-value, the area will alter.
Get a turning point of f(m) to find the minimum area.

12. Dec 14, 2005

### scorpa

Hmmmm ok, not quite sure if I'm following that, but I'll go back and see what I can come up with. Thanks

13. Dec 15, 2005

### Fermat

Let the line, y - 7 = m(x - 2), intersect the x-axis and y-axis at the points A and B. A has the coords (xm, 0) and B has the coords (0, ym).
The triangle OAB (where O is the origin) is the triangle that you have to minimise the area of, where the area is, A = ½OA.OB, or,
A = ½.xm.ym

All you have to do now is find xm and ym in terms of m using the eqn of the line given above.

14. Dec 15, 2005

### scorpa

mmmmm now I think I see, sorry but I'm a little slow when it comes to math sometimes...haha. Thanks