CalculusCheck please?
posted by Alaina .
Fabio stands atop his 16 foot ladder when he realizes that the ladder is slipping down the side of the building. He decides that the base of the ladder is moving away from the bottom of the building at a rate of 2 feet per second when it is 3 feet from the bottom of the building. How fast is Fabio falling at that instant?
my answer is 10.477 or 10.48 but it's wrong. please give guidance. thanks

CalculusCheck please? 
I hope that this helps for future use.
Your answer was obtained by multiplying the derivatives incorrectly. No worries!
1. Draw a rectangular box with a straight line at the bottom poking out. This line represents the 3 ft from the building that the ladder is at.
*Because the ladder is moving away from the building at a rate of 2 ft/sec, then this is the derivative of the 3 ft leg.
2. Draw a diagonal line from the building to an intersection of the straight line. This is both the ladder and the hypotenuse of the invisible triangle.
3. Find the second leg of the invisible triangle by using the Pythagorean Theorem. This will yield sqrt(247) ft.
So, to rehash:
Ladder 16 ft
Distance between ladder and building 3 ft
Rate that ladder is moving on ground 2 ft
Height of ladder sqrt(247) ft
The last step is to determine dh/dt where h=sqrt(247)
Using the Pythagorean Theorem again, the derivative is taken:
2l*(dl/dt)+2h*(dh/dt)=0 (The zero represents the constant 256.)
Substitute:
2(3)(2)+2(sqrt(247))(dh/dt)=0
Solve for dh/dt:
6/sqrt(247)=dh/dt=.38177...
So, it is falling at a rate of nearly 0.38 ft/sec.
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