A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)
To solve this problem, we can use trigonometry and related rates.
Let's denote the angle between the ladder and the ground as θ(t), where t represents time.
We are given that the ladder is 10 ft long and the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s. This means that the rate of change of the distance x between the wall and the bottom of the ladder is dx/dt = -1.2 ft/s (negative sign because x is decreasing).
We are asked to find how fast the angle θ is changing (dθ/dt) when the bottom of the ladder is 8 ft from the wall (x = 8 ft).
Using trigonometry, we can relate the angle θ to the distance x and the length of the ladder using the sine function:
sin(θ) = x / 10
To differentiate both sides of the equation with respect to time t, we obtain:
cos(θ) * dθ/dt = (1/10) * dx/dt
Now, substitute the given values:
cos(θ) * dθ/dt = (1/10) * (-1.2 ft/s)
We can rearrange this equation to solve for dθ/dt (the rate at which the angle θ is changing):
dθ/dt = (-1.2 ft/s) / (10 ft * cos(θ))
To find the value of cos(θ), we can use the Pythagorean theorem:
cos(θ) = √(1 - sin^2(θ))
Since we know x = 8 ft and the ladder is 10 ft long, we can substitute x and 10 into this equation:
cos(θ) = √(1 - (8/10)^2)
cos(θ) = √(1 - 0.64)
cos(θ) = √0.36
cos(θ) = 0.6
Now substitute this value of cos(θ) into the equation for dθ/dt:
dθ/dt = (-1.2 ft/s) / (10 ft * 0.6)
dθ/dt ≈ -0.02 radians/s
Therefore, when the bottom of the ladder is 8 ft from the wall, the angle between the ladder and the ground is changing at approximately -0.02 radians/s.
Use the Pyth theorem:
10^2=base^2 + height^2 You know the base, find the height.
Now, to work the rates...
take the derivative of the pyth theorem
0=2b db/dt + 2h dh/dt
so dh/dt=-b/h db/dt
Now you know b, h, and db/dt solve for dh/dt.