The problem was on the Verazanno-Narrows bridge, whose span is 4200 feet and whose road-height is 200 feet. You have a boat approaching midspan 1000 feet away at the same time as a runner enters the span. The boat goes 20 feet per second, the runner goes 10 feet per second. How fast is the distance separating them decreasing?
Quicly going through the first few of our seven questions: the
independent variable is again time. For dependent variables,
you have to set up a coordinate system. There are several
alternatives that will work, and I will pick just one (you
may have used a different one and still come up with a correct
answer). I pick the origin as the point on the water directly
under midspan. The bridge is parallel to the x-axis,
and the boat is traveling along the y-axis. That
gives us the dependent variable, y, as the boat's
position, and it gives us the dependent variable, x,
as the runner's position. The z coordinates of both
the runner and the boat are constant (the runner is at
Now we get to question
4, which is to come up with the relationships
among the variables. The
s2 = x2 + y2 + 2002where s is the distance separating them. As for the other relationship, you have to expand what you did in the second taxicab problem when you gave the westbound taxicab that head start.
Let's say that the x axis of our coordinate system puts
Staten Island on the negative side and Brooklyn on the positive
side. Then the runner starts out in negative x territory
and runs toward
In this problem neither the runner nor the boat have head starts. They both have hind starts (starting in back of the origin instead of in front of it). Recall that when you gave the westbound taxicab a head start you subtracted its head start from its east-west position when you made up the relationship between that and its north-south position. Likewise, when there is a hind start, you will add it to the position along the appropriate axis. Hence:
x + 2100 y + 1000The 2100 comes from the span being 4200 feet and midspan being where=10 20
2x - 3200 = y
Question 5 asks you to take the derivatives of the relationships. With the Pythagorean relationship you get
ds dx dy
2s = 2x + 2y
dt dt dt
from which you can cancel the factor of 2. And the other relationship gives
dx dy
2 =
dt dt
which just says that the boat's speed is twice the runner's speed, and
we already knew that.
Question 6 asks you to consider what the problem is asking for. The separation distance is s, and the problem is asking how fast that distance changes, which is ds/dt. So that is what you have to solve for. You have an equation above that has ds/dt in it, and you have another equation that suggests a way to substitute for s. So you have:
ds dx dy
Öx2 + y2 + 2002 = x + y
dt dt dt
When you ask yourself question 7, you find that the problem gives you numbers for x, y, dx/dt, and dy/dt. The way we set up the coordinates we have:
x = -2100 feet y = -1000 feet dxPuting all those numbers into the last equation and solving for ds/dt you should get= 10 ft/sec dt dy= 20 ft/sec dt
s = 2334.5235 feet dsObserve that ds/dt is negative, indicating that the distance separating the runner and the boat, s is decreasing. That is as expected. We also expect that the distance should be closing at a rate that is in the vicinity of the 10 or 20 feet per second with which the objects are moving, so you can see that the answer is in the right ballpark (if you had come up with 100 feet per second or 0.1 feet per second you would have had to view your answer with suspicion).= -17.56247 ft/sec dt
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