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jason1242

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A 13-foot ladder leans against a vertical wall so that the ladder, ground and the wall form a right triangle. The top of the ladder starts to slide down the wall at a constant rate of 6 feet per second.
a. How fast is the bottom of the ladder sliding away from the wallwhen the ladder is 5 feet away from the wall?
b. At this moment, how fast is the area of the triangle formed byladder, wall and ground changing?
a^2 + b^2 = c^2
13^2-5^2 = y^2 y= 12
2xx' + 2yy' = 0
y' = -2xx'/2y
=14.4 <----- is this correct for part a
part b = 30 ??
 
You have the right idea for part a), although I would use y for the vertical and x for the horizontal, it doesn't matter which variables you use, you have made an error in substituting the values in.

For part b) that is not the correct time rate of change of the area of the described triangle, show your work so we can see what you did.
 
MarkFL said:
You have the right idea for part a), although I would use y for the vertical and x for the horizontal, it doesn't matter which variables you use, you have made an error in substituting the values in.

For part b) that is not the correct time rate of change of the area of the described triangle, show your work so we can see what you did.
Click to expand...
It would be useful if you showed me and directed me towards the right path instead of giving such a blunt answer. Well thank you anyways.
 
You asked if your results were correct, and I answered you, and told you why part a) is incorrect, you made an error plugging the values of x, y and dx/dt into the expression representing the time rate of change of the distance of the base of the ladder from the wall.

For part b), without knowing what you did, I cannot tell you why your answer is wrong, and so I asked to see what you did so I could see where you went wrong. I honestly wasn't trying to be blunt; I was actually trying to help.
 
Last edited:
jason1242 said:
A 13-foot ladder leans against a vertical wall so that the ladder, ground and the wall form a right triangle. The top of the ladder starts to slide down the wall at a constant rate of 6 feet per second.
a. How fast is the bottom of the ladder sliding away from the wall when the ladder is 5 feet away from the wall?
b. At this moment, how fast is the area of the triangle formed by ladder, wall and ground changing?
a^2 + b^2 = c^2
13^2-5^2 = y^2, y= 12 ft
From this it appears that "y" is vertical. The given rate is y' = - 6 ft/s
2xx' + 2yy' = 0
y' = -2xx'/2y But what you need is x' = (y/x)(-y') = ...
=14.4 ft/s = x' <----- is this correct for part a
part b = 30 ??
Click to expand...
I agree with your part (a) answer - at least if the units are shown. And lets be sure we know which is x and which is y.

For part (b), what are the units on your "30"? Since it is asking for change of area with time, you need to wind up with ft^2/s. I can guess what you did, but the more work you show us, the less we have to guess.

A = (1/2)xy
A' = ...
 
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