Related Rates:Train/Highway

[Messagehelp]

A long level highway bridge passes over a railroad track that is 100 feet below it and at right angles to it. If an automobile is traveling 66feet per second is directly about a train 88 fps, how fast will they be separating 10 seconds later?

This is my work so far:

I have a triangle, with one leg(a) it 100(distance from track to bridge) and another leg(b) is 220 feet(since 10 seconds later the train is 220 feet ahead) That makes the hypotenuse(c) sqrt[58400]. am i trying to find dc/dt or db/dt. either way i don't get the right answer, since it is supposed to be 100sqrt[122]

a^2 + b^2 = c^2
2b * db/dt = 2c * dc/dt
b * db/dt = c * dc/dt
220*22= sqrt[54800] * dc/dt
20.02=dc/dt

which is not right?

 

[tkhunny]

You've more than that going on.

For convenience, let's move the car North and the train East.

Time starts at t = 0 when the car is directly over the from of the train.

d(0) = 100 ft -- Vertical distance only.

Let's just move the car. car(t) = (66 f/s)*t, making (100 ft)^2 + (car(t))^2 = [d(t)]^2. d is a line up from the train track to the car road. It is pointing due North.

Now let's move the train. train(t) = (88 f/s)*t, making (d)^2 + (train(t))^2 = [D(t)]^2. D is a line up from the train track to the car road. It is pointing North Westerly direction.

Showing it all together. D^2 = (100 ft)^2 + (car(t))^2 + (train(t))^2

First Derivative: 2*D(t)*D'(t) = 2*car(t)*car'(t) + 2*train(t)*train'(t) ==> 2*D(t)*D'(t) = 2*car(t)*car'(t) + 2*train(t)*train'(t)

By design, car'(t) and train'(t) are constant. I wonder if D'(t) is? It's always good to think and postulate as you go.

car'(t) = 66 f/s
train'(t) = 88 f/s
car(10 sec) = 660 ft
train(10 sec) = 880 ft
D(10 sec) = D^2 = (100 ft)^2 + (660 ft)^2 + (880 ft)^2

So, D'(t) = [car(t)*car'(t) + train(t)*train'(t)]/D(t)

It's so close. You do the final step or two.

Notes:

1) Generalize.
2) Write things down.
3) Let the notation help you.
4) Always think ahead and behind.