Rates of Change

Last week, I was busy studying for (and then taking) the MFE Exam through the Society of Actuaries.  I hope I passed.  That was the reason I did not post.

Next week, I plan on posting about my experiences truly taking advantage of a break when there is no travel planned.

Until then, I solve problems.

Consider the following two problems.

  1. A vehicle travels 30 mph for 1 hour, and then 60 mph for an hour. What is the vehicle’s average speed?
  2. While traveling an unknown distance, a car travels half the distance at 30 mph and the other half at 60 mph. What is the vehicles average speed?

While the first problem is pretty easy to solve, the second one can be a bit more challenging.

First, recognize that mph stands for miles per hour, or in mathematical speak, mi/hr.

We do this so that we can see that when we multiply the speed by the time, we get the distance:

30 \text{mi}/\text{hr} \cdot 1 \text{hr} = 30 \text{mi}

For problem 1, we traveled 30 miles in the first hour, and 60 miles in the second hour for a total of 90 miles in 2 hours.  This means our average speed was 90/2=45 mph.  Like I said, that one was easy.

When we’re working with distances and speed, to get the time you need to take distance divided by speed.  To see this, remember dividing by a fraction is the same as multiplying by the reciprocal:

\frac{\text{mi}}{\text{mi}/\text{hr}} = \text{mi}\cdot \frac{\text{hr}}{\text{mi}} = \text{hr}

 

For problem 2, we don’t know the distance. But that is OK.  We don’t need to. It turns out, we could pick any distance we wanted to solve the problem.  To make it easier, we will do that.  Let’s pick 120 miles.

If the vehicle travels 30 mph for 60 miles (half of 120 miles), then it will take the vehicle 60/30=2 hours to do so. If the vehicle travels 60 mph for the second 60 miles, it will only take the vehicle 60/60=1 hour to do so. Thus, the total travel time is 3 hours.

If the vehicle traveled 120 miles in 3 hours, then this means the average speed is 120/3=40 mph.

Now that you know the answer, see if you can follow along by exploring any distance, we’ll call it d.

If the vehicle travels 30 mph for d/2 miles, then it will take the vehicle (d/2)/30=d/60 hours to do so. If the vehicle travels 60 mph for the second d/2 miles, it will only take the vehicle (d/2)/60=d/120 hours to do so. Thus, the total travel time is d/60+d/120 hours.  Getting a common denominator to add these fractions, we get a total travel time of

\frac{d}{60}+\frac{d}{120}=\frac{2d}{120}+\frac{d}{120}=\frac{3d}{120}=\frac{d}{40}.

If the vehicle traveled d miles in d/40 hours, then this means the average speed is d/(d/40)=d\cdot \frac{40}{d} = 40 mph.

See, the distance doesn’t matter.

Riddler Express on March 2, 2018

The problem as stated on FiveThirtyEight was a rate problem:

Andrea and Barry both exercise every day on their lunch hour on a path that runs alongside a parkway. Andrea walks north on the path at a steady 3 mph, while Barry bikes south on the path at a consistent 15 mph, and each travels in their original direction the whole time — they never turn around and go back the other way. The speed limit on the parkway is the same in both directions and vehicle traffic flows smoothly in both directions exactly at the speed limit.

In order to pass the time while they exercise, both Andrea and Barry count the number of cars that go past them in both directions and keep daily statistics. After several months of keeping such stats, they compare notes.

Andrea says: “The ratio of the number of cars that passed me driving south on the parkway to the number of cars that passed me driving north was 35-to-19.”

Barry retorts: “I think you’re way off. The ratio for me was 1-to-1 — the number of cars that passed me going south was the same as the number that passed me going north.”

Assuming Andrea and Barry are both very good at stats, what is the speed limit on the parkway?

Although this problem may be too difficult to solve for many of you, some of you may be able to follow how it is set up.

We eventually want to find the speed limit on the parkway. I’ll call that v, for velocity.

There are a few other unknowns in this problem.  We are not given the flow of traffic northbound and southbound.  I’ll call these values R_N and R_S.  An example value might be 30 cars per hour.  In this case, you would know that a car is passing a stationary object once every 1/30th of an hour (or 2 minutes).

First, let’s focus on the southbound traffic.  From a stationary point, the flow is R_S as stated above.  However, if you travel south at a velocity of v (the same speed as the cars), the flow at which you see cars pass you that are going southbound reduces to 0.  This is a linear relationship.  If y is the perceived flow, and x is the velocity at which you travel, then

\displaystyle y= -\frac{R_S}{v}x+R_S

is the equation that describes this relationship. For a stationary object, x=0, and y=R_S, which is what we expect.  If we travel south at speed v, we perceive traffic flowing by us at y=-\frac{R_S}{v}(v)+R_S = -R_S+R_S=0 cars per hour.

Relative to Andrea, who is traveling north along with the northbound traffic, the flow of traffic that is going northbound is reduced to -\frac{R_N}{v}(3)+R_N.  From the perspective of the southbound traffic, she is traveling at -3 mph!  This increases the flow from which she sees the southbound traffic pass her to -\frac{R_S}{v}(-3)+R_S = \frac{3R_S}{v}+R_S.

We are given that the ratio of the second equation to the first is 35:19.  This gives us the complicated mess

\displaystyle \frac{3R_S/v+R_S}{-3R_N/v+R_N} = \frac{35}{19},

which is equivalent to

\displaystyle 57R_S+19R_Sv = -105R_N+35R_Nv

after a little algebra (first, cross multiplying, and then, multiplying by v on both sides).

Using a similar argument for Barry, the flow of the traffic going southbound reduces to -\frac{R_S}{v}(15)+R_S and the flow of the traffic going northbound increases to \frac{R_N}{v}(15)+R_N.  Since he witnesses a 1:1 ratio, this gives us the complicated mess

\displaystyle \frac{-15R_S/v+R_S}{15R_N/v+R_N} = 1,

which simplifies to

\displaystyle -15R_S+R_Sv = 15R_N+R_Nv

after a little algebra (cross multiplying while thinking 1=1/1, and then multiplying both sides by v).

We’re in a dilemma right now as we have two equations and 3 unknowns.  In order to be able to solve for a number of different variables, we need to have the same number of equations as we do unknowns.

We can get around this issue since we do not really care about the values of R_S and R_N.  What we will do, is divide both the equations above by R_N, to create one unknown R = \frac{R_S}{R_N}.

Now, the two equations become

\displaystyle 57R+19Rv = -105+35v

\displaystyle -15R+Rv = 15+v .

This seems like a headache to solve, so let’s use our friend WolframAlpha.  This computational engine really likes x’s and y’s, so let’s plug

57x+19xy=-105+35y, -15x+xy=15+y

into that website and see what we get.

It provides us with the two solutions x = -\frac{21}{19}, y = \frac34, and x=\frac53, y = 60.  The second solution is the only one that makes sense, which suggests that the cars are traveling at 60 miles per hour.

Less important, but we might as well address it, are the flow rates.  We’re only able to determine the ratio of the flow rate of the southbound traffic to the northbound traffic.  If the southbound traffic has about 5000 cars per hour, we know the northbound traffic has about 3000 cars per hour since x=5/3.

 

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