A 17 foot ladder is leaning against a wall. The bottom of the ladder is moving out away from the wall at 0.6 feet per second. The top of the ladder then begins sliding down the wall. How fast is the top of the ladder going when the bottom is 8 feet away from the wall?
x^2 + y^2 = 17
solve for y when x=8
2x dx/dt + 2y dy/dt = 0
Now just plug in the given values and solve for dy/dt
To find the speed at which the top of the ladder is sliding down the wall, we need to use related rates.
Let's say the distance between the bottom of the ladder and the wall is x, and the distance between the top of the ladder and the ground is y.
We are given that dx/dt = 0.6 ft/s (the change rate of x), and we need to find dy/dt (the change rate of y).
Using Pythagoras' theorem, we know that x² + y² = 17² (since the ladder is 17 feet long). Differentiating both sides with respect to t (time), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Substituting the given values, we have:
2(8)(0.6) + 2y(dy/dt) = 0
Simplifying:
9.6 + 2y(dy/dt) = 0
2y(dy/dt) = -9.6
dy/dt = -4.8/y
Now we need to find y when x = 8. Since x² + y² = 17², we can substitute x = 8 into this equation:
8² + y² = 17²
64 + y² = 289
y² = 225
y = 15 feet
Substituting y = 15 into the expression dy/dt = -4.8/y, we get:
dy/dt = -4.8/15
Therefore, when the bottom of the ladder is 8 feet away from the wall, the top of the ladder is sliding down the wall at a speed of -4.8/15 ft/s.