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 2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 5 ft from the wall? Evaluate your answer numerically. (Round the answer to three decimal places.)
let x be the horizontal distance from wall to ladder.
cosineTheta=x/10
10 cosTheta=x
-10sinTheta dTheta/dt= dx/dt
so you are looking for dTheta/dt
when x=5, theta = arccos.5, or 60 deg, and sin of 50 = .877
dTheta/dt= dx/dt / (10*.877)
check all that.
.228
To solve this problem, we can use trigonometry and related rates.
Let's draw a diagram to represent the situation:
|\
| \
10| \ x
| \
| \
|_____\
5 ft
In the diagram above, the ladder is represented by the hypotenuse of the right triangle, the distance from the bottom of the ladder to the wall is 5 ft, and the distance along the ground is represented by x ft.
We are given that the bottom of the ladder is sliding away from the wall at a rate of 2 ft/s, which means dx/dt = 2 ft/s. We need to find dθ/dt, the rate at which the angle between the ladder and the ground is changing when x = 5 ft.
To solve this, we can use the trigonometric relationship between the angle θ and the sides of the triangle:
sin(θ) = opposite/hypotenuse
In this case, the opposite side is x and the hypotenuse is 10 ft. Therefore, we have:
sin(θ) = x/10
Now, we can differentiate both sides of this equation with respect to time (t) using the chain rule:
d/dt(sin(θ)) = d/dt(x/10)
cos(θ) * dθ/dt = (1/10) * dx/dt
Since dx/dt = 2 ft/s, we can substitute the values into the equation:
cos(θ) * dθ/dt = (1/10) * 2
cos(θ) * dθ/dt = 1/5
Now, we need to find the value of cos(θ) when x = 5 ft. To do this, we can use the right triangle and the Pythagorean theorem:
cos(θ) = adjacent/hypotenuse
cos(θ) = (10 - x)/10
cos(θ) = 5/10
cos(θ) = 1/2
Substituting this value into the equation:
(1/2) * dθ/dt = 1/5
To solve for dθ/dt, we can multiply both sides by 2:
dθ/dt = (1/5) * 2
dθ/dt = 2/5
Therefore, when the bottom of the ladder is 5 ft from the wall, the angle between the ladder and the ground is changing at a rate of 2/5 radians per second.
Numerically, this is approximately equal to 0.400 radians per second when rounded to three decimal places.