**Grade 12 Calculus, Related Rates Problem?**

* *

*The top of a 5 m wheeled ladder rests against a vertical wall. If the bottom of the ladder rolls away from the base of the wall at a rate of 1/3 meters per second, how fast is the top of the ladder sliding down the wall when it is 3 m above the base of the wall? *

Let x = the horizontal distance of the base to the bottom of the ladder, and y = height of ladder up the wall.

Then dx/dt = 1/3 and you are finding dy/dt when y = 3. But when y = 3, x = 4 (Pythag. triple).

x^2 + y^2 = 25 by the Pythag. thm.

2x*dx/dt + 2y*dy/dt = 0 when you differentiate.

2*4*(1/3) + 2*3*dy/dt = 0

8/3 + 6dy/dt = 0

6dy/dt = -8/3

dy/dt = -8/18 = -4/9 m/sec

For speed take the abs. value of the velocity, so it is moving downward at a rate of 4/9 m/sec.

**Ladder rate-of-change problem**

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