A 5m long ladder js leaning against a wall. When the person holding it let's go the ladder starts sliding down the wall increasing it's speed as it falls. If the top of he ladder is moving at 1m/s when the bottom is 4m away from the wall, how fast is the lower end moving away from the wall at that time?
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Triangle:
5 hypotenuse
y opposite
x adjacent.
implicitly differnetiate the pythag thorem with respect to time t.
x^2 + y^2 = 25
2x(dx/dt) + 2y(dy/dt) = 0
Now they give you info:
dy/dt = -1 (sliding down)
And, you have 5m, and 4m, so the other side must be 3m. (3-4-5 triangle)
So, since they give you x = 4, y has to be 3. Now solve for dx/dt (should be positive)
2x(dx/dt) + 2y(dy/dt) = 0
2(4)(dx/dt) + 2(3)(-1) = 0
8(dx/dt) - 6 = 0
dx/dt = 6/8
dx/dt = 3/4
Thus, the lower end is moving away from the wall at 3/4 m/s
Hope this helps :D
5 hypotenuse
y opposite
x adjacent.
implicitly differnetiate the pythag thorem with respect to time t.
x^2 + y^2 = 25
2x(dx/dt) + 2y(dy/dt) = 0
Now they give you info:
dy/dt = -1 (sliding down)
And, you have 5m, and 4m, so the other side must be 3m. (3-4-5 triangle)
So, since they give you x = 4, y has to be 3. Now solve for dx/dt (should be positive)
2x(dx/dt) + 2y(dy/dt) = 0
2(4)(dx/dt) + 2(3)(-1) = 0
8(dx/dt) - 6 = 0
dx/dt = 6/8
dx/dt = 3/4
Thus, the lower end is moving away from the wall at 3/4 m/s
Hope this helps :D
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x^2 = 25 - y^2 , at y = 4 , x = 3 , dy/dt = -1 , then it's easy to work out dx/dt