Remember = dx(sub 2/dt) – dx(sub 1/dt), how

Remember the classic problem of math horror stories everywhere? You know, where one train leaves Kentucky at 2 p.

m. and another leaves Sacramento at 4 p. m.? In this lesson, tame the horror and learn how to solve these problems using differentiation and related rates.

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Two Trains Problem

Plugging the train velocities into the problem shows the distance is changing at a rate of -75 mph
Train Problem 1

Remember that problem that everybody always makes fun of: If one train leaves Kentucky at 2 p.m.

and another leaves Sacramento at 4 p.m., and the alignment of the moons is such that everything is going 5 mph, when will they cross? We’re going to do a problem that’s similar, but a little more reasonable. Imagine that you have two trains heading toward one another. Train 1 is going eastbound at 30 mph, and Train 2 is headed westbound at 45 mph. They’re going to start 100 miles apart.

How far apart will they be in one hour? When will they pass one another?Because we want to know how far apart they’re going to be and when they pass each other (when the distance between them is zero), we want to write that the distance between them is equal to the position of Train 2 minus the position of Train 1, l = x sub 2 – x sub 1. Now we’ve got the distance between them, which I’ll call l. Here we’re going to use a related rate. We’re going to relate the distance between them to each one of their speeds. I’m going to take the derivative of both the left and right side of this equation, and I’ll get dl/dt (how their distance changes as a function of time) = dx(sub 2/dt) – dx(sub 1/dt), how fast Train 2 is going minus how fast Train 1 is going.

If I plug in the velocity of Train 2 and the velocity of Train 1, I find that the distance is changing at a rate of -75 mph.If I use a separation of variables to solve this differential equation – if I get all the t variables on the right side and all the l variables on the left side – I get dl=-75dt. If I integrate both sides, I find that l=-75t + C. Because they start out when time equals zero at a distance 100 miles apart, I can solve this l equation for C. So 100=(-75)(0) + C, that means C=100 miles. What I end up with is an equation for how far apart they are as a function of time: l=-75t + 100.

After one hour, t=1, I find that l is 25, and they’re 25 miles apart. They pass each other when l=0, when t=1.3 hours.

In the second train problem, velocity 1 is negative because train 1 is moving to the left
The equation shows how the height is changing with respect to how fast the fire truck is moving away
fire ladder problem

Fire Ladder Problem

Let’s say you’re helping out the firemen. You’re at the scene of a fire, and you’re up a 10-meter ladder; you’re at the very top. The ladder is attached to the fire truck, and the fire truck is 3 m away from the building.

So it’s pretty close to the building. If the fire truck starts moving away at 2 m/sec, how fast are you going to slide down the wall?You notice that this is a right triangle, and the hypotenuse is 10 m. You could always write x^2 + y^2 = 10^2. Now you want a related rate between dx/dt (how fast the fire truck is moving away) and dy/dt (how fast you’re sliding down this wall).

Let’s take the derivative of both sides of this Pythagorean theorem. I get: 2x(dx/dt) + 2y(dy/dt) = 0. I can divide both sides by 2 and subtract y(dy/dt), and I find that x(dx/dt)=-y(dy/dt). This how my height (dy/dt) is changing with respect to how fast the fire truck is moving away (dx/dt). Let’s see how fast I’m sliding down the wall. The fire truck is currently moving away from the wall at 2 m/sec.

That means dx/dt is 2 m/sec, and x is 3 m, because that’s how far away from the wall the fire truck is. My equation reduces to 6=-y(dy/dt).I still have one equation and two unknowns. What is y and what is dy/dt? I really want to know what dy/dt is, but in order to solve this equation, I first have to figure out what y is. I’m again going to use the Pythagorean theorem. Remember, x^2 + y^2 =10^2 in this case.

If x=3, then this equation becomes 9 + y^2 = 100. I can solve this for y. It will equal the square root of 91.

So I am the square root of 91 m up in the air. If I plug this in to my related rate equation, that is 6 = –y(dy/dt), then I find that I’m sliding down the wall at the rate of -6 / the square root of 91. So I’m sliding down this wall at a rate of .

06 m/sec.

The truck is 3 m from the wall and is moving away from it at a speed of 2 m/sec
Fire Ladder Problem Visual

Lesson Summary

Let’s review related rates again. In related rates, you’re going to take a relationship that you know. In one problem it is my height as a function of the distance the fire truck is away from the wall.

Another relationship is the distance between two trains, depending on the position of these two trains. Then you’re going to differentiate this relationship to find how these variables are changing with respect to one another.When I differentiate the equation that relates my height with the distance of the fire truck from the wall, I can find out how my height is going to change as the fire truck drives away from the building. Similarly, when I differentiate the train equation with the distance between the trains, I can find how the distance changes with respect to the train velocities.

These are called related rates.

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