A 15foot ladder is leaning against a house. The base of
the ladder is pulled away from the house at a rate of 2 feet per second.
How fast is the top of the ladder moving down the wall when the base
of the ladder is 9 feet from the house.
Foundations:

Like most geometric word problems, you should start with a picture. This will help you declare variables and write meaningful equation(s). In this case, we will have to use implicit differentiation to arrive at our related rate.

Solution:
Step 1:

Write the Basic Equation: From the picture, we can see that the ladder forms a right triangle with the wall and the ground, so we can treat our variables as

 $x^{2}+y^{2}\,\,=\,\,15^{2}\,\,=\,\,225,$

where $x$ is the height of the ladder on the wall, and $y$ is the distance between the wall and the base of the ladder.

Step 2:

Use Implicit Differentiation: We take the derivative of the equation from Step 1 to find

 $2x{\frac {dx}{dt}}+2y{\frac {dy}{dt}}\,\,=\,\,0,$

or

 ${\frac {dy}{dt}}\,\,=\,\,{\frac {x}{y}}\cdot {\frac {dx}{dt}},$

Step 3:

Evaluate and Solve: At the particular moment we care about,

 $x=9,\quad y=12,\quad dx/dt=2.$

From this, we can simply plug in to find

 ${\frac {dy}{dt}}\,\,=\,\,{\frac {x}{y}}\cdot {\frac {dx}{dt}}\,\,=\,\,{\frac {9}{12}}\cdot 2\,\,=\,\,{\frac {3}{2}}$

With units, we have that the ladder is sliding down the wall at $3/2$ feet per second.

Final Answer:

With units, we have that the ladder is sliding down the wall at $3/2$ feet per second.

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