Expert: Paul Klarreich - 9/7/2009

Question
As a car transverses a viaduct at the rate of 40 ft/sec, another traveling at a speed of 80 ft/sec on the road 30 ft below and at right anles to the viaduct, passes directly under it. How fast are the two separating at the end of 2 seconds?


Answer
Questioner: Erwin
Category: Calculus
Private: No
Subject: time rates
Question: As a car transverses a viaduct at the rate of 40 ft/sec, another traveling at a speed of 80 ft/sec on the road 30 ft below and at right anles to the viaduct, passes directly under it. How fast are the two separating at the end of 2 seconds?
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Is this your first attempt at R-R problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.

3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is the key step.

4. Now differentiate implicitly, substitute the known quantities and rates, and solve for the unknown rate.

AND, Please check the archives for other Related Rates examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.
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Variables:

x = distance of lower car from intersection point.
y = distance of upper car from intersection point.
r = distance between them.

Rates:

dx/dt = 80
dy/dt = 40
dr/dt TO BE FOUND.

Relation: (Pythagorean Theorem in 3-D)

x^2 + y^2 + 30^2 = r^2

Take it from there.