Expert: Paul Klarreich - 10/18/2007

Question
A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank. Assume the opposite end of the plank follows a path perpendicular to the wall of the building, and the worker pulls the rope at a rate of 0.15 meters per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?


Answer
Questioner:   Josh
Category:  Advanced Math
Private:  No
 
Subject:  Calc 1 help
Question:  A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank. Assume the opposite end of the plank follows a path perpendicular to the wall of the building, and the worker pulls the rope at a rate of 0.15 meters per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?
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Hi, Josh,

Is this your first attempt at Related-Rate 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, your general knowledge and brilliance, whatever you have to.  This may be the hard part.
4. Differentiate implicitly, THEN substitute the known quantities and rates, and solve for the unknown rate.

In this case, the variables are:

x = distance from base of plank to base of wall.
y = height of top of plank.

The rates are:

dx/dt = rate the end of the plank slides along the ground, TO BE FOUND.
dy/dt = rate at which the top of the plank rises, GIVEN as 0.15 m/sec

The relation:

x^2 + y^2 = 5^2, because the three make a right triangle.

Differentiate:

2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0

Substitute known values to solve for dx/dt.  We need x,y, dy/dt.  We have:

x = 2.5
dy/dt = 0.15

We need y, which we get from:

2.5^2 + y^2 = 5^2

6.25 + y^2 = 25

 y^2 = 18.75 = 3(6.25)

y = 2.5 sqrt(3)

Ready to go now:

x dx/dt + y dy/dt = 0   << repeated

(2.5) dx/dt + 2.5 sqrt(3)(0.15) = 0

dx/dt + sqrt(3)(0.15) = 0

dx/dt = - sqrt(3)(0.15)