Expert: Paul Klarreich - 5/29/2008

Question
Hi, I'm having a problem with 2 related rates word problems.

In the heat of the sun, a sheet of aluminum in the shape of an equilateral triangle is expanding so that its side length increases by 1 millimeter per hour. When the side length is 100 millimeters, how is the area increasing?

and

A street light is at the top of a 11 ft tall pole. A man  6.2 ft tall walks away from the pole with a speed of 4.5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 31 feet from the pole?

Answer
Questioner:   Kiana
Category:  Calculus
Private:  No
 
Subject:  related rates
Question:  Hi, I'm having a problem with 2 related rates word problems.
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Hi, Kiana,

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, then 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.  In fact, I am sure you will find your second one -- I know I have done it for someone.
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In the heat of the sun, a sheet of aluminum in the shape of an equilateral triangle is expanding so that its side length increases by 1 millimeter per hour. When the side length is 100 millimeters, how is the area increasing?

1. Variables:

s = one side of the triangle.
A = area of the triangle.

2. Rates:

ds/dt = rate of incr of a side, given = 1 mm/hr
dA/dt = rate of incr of area, TO BE FOUND.

3. Relation:

A = s^2 sqrt(3)/4  << look in any geometry book, if you don't remember this one.

4. Diff imp:

dA/dt = 2s ds/dt sqrt(3)/4

dA/dt = s ds/dt sqrt(3)/2

Subst: ds/dt = 1,  s = 100

dA/dt = (100) (1) sqrt(3)/2

You can handle the rest.  Don't forget to write  mm^2/hour as the units.
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and

A street light is at the top of a 11 ft tall pole. A man  6.2 ft tall walks away from the pole with a speed of 4.5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 31 feet from the pole?
 
I'll let you find this one among the archived answers.  It's probably there at least three times.