Related Rates, Optimization, Integrals

In summary: The oncoming car is travelling at 80 km/h when it is 5 km away from the plane and 160 km/h when it is 3 km away from the plane.
  • #1
scorpa
367
1
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 :rolleyes:
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 don't know what .
2) [tex]/int[/tex] 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 don't really know where to start so if there is anything that could just get me on the right track I would appreciate that.
Thanks in advance for any advice, I appreciate it.
 
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  • #2
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,

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

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
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 that's 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
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
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
Oh you just subtract 200km/h from 120km/h to get the speed of the car because they are coming towards each other?
 
  • #7
That's it :smile:
 
  • #8
Awesome, thank you very much!
 
  • #9
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 don't know the slope of the line?
 
  • #10
******bump*****
 
  • #11
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
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
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
mmmmm now I think I see, sorry but I'm a little slow when it comes to math sometimes...haha. Thanks
 

FAQ: Related Rates, Optimization, Integrals

1. What are related rates?

Related rates are a mathematical concept used to find the rate at which one quantity changes with respect to another related quantity. This is typically done by using derivatives and applying the chain rule to determine the relationship between the two rates.

2. How are related rates used in real-world applications?

Related rates are commonly used in physics, engineering, and other scientific fields to analyze and solve problems involving changing quantities. For example, they can be used to determine the rate at which the volume of a balloon is changing as it is being inflated, or the rate at which the distance between two moving objects is changing.

3. What is optimization?

Optimization is a mathematical process used to find the maximum or minimum value of a function. This can be done by finding the critical points of the function and evaluating them to determine which point yields the highest or lowest value.

4. How are optimization problems solved?

Optimization problems can be solved using various techniques such as using derivatives and setting them equal to zero, using graphical methods, or using the first or second derivative test. The specific method used will depend on the given function and the constraints of the problem.

5. What are integrals?

Integrals are mathematical tools used to find the area under a curve or the accumulation of a quantity over a given interval. They are the inverse of derivatives and can be thought of as the sum of infinitesimal changes in a given quantity.

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