Related Rates: 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 0.9 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?
at the moment in question, we have a 6-8-10 triangle.
You know that
tanθ = y/x = √(100-x^2)/x
or,
x tanθ = √(100-x^2)
So,
tanθ dx/dt + x sec^2θ dθ/dt = -x/y dx/dt
Now just plug in your values to find dθ/dt
To solve this related rates problem, we need to differentiate the given equation with respect to time. Let's denote:
- The length of the ladder as L (which is a constant value of 10 ft).
- The distance between the bottom of the ladder and the wall as x (which is changing with time).
- The angle between the ladder and the ground as θ (which we need to find the rate of change).
Given:
- dx/dt = -0.9 ft/s (since the bottom of the ladder is moving away from the wall, the rate of change is negative).
- We need to find dθ/dt (the rate of change of the angle).
To relate x, L, and θ, we can use the right triangle formed by the ladder, the wall, and the ground. The hypotenuse of the triangle is the ladder, the side adjacent to the angle θ is x, and the side opposite to θ is the height of the wall.
Using trigonometry, we have:
cos(θ) = x / L
To differentiate this equation with respect to time, we will use the chain rule:
-d(sin(θ))/dt = d(x/L)/dt
Differentiating both sides with respect to time:
cos(θ) * dθ/dt = (1/L) * dx/dt
Now, we can substitute the given values:
cos(θ) * dθ/dt = (1/10) * (-0.9)
Simplifying:
dθ/dt = (-0.9) / (10 * cos(θ))
Now, we need to find the value of cos(θ) when the bottom of the ladder is 6 ft from the wall. Again, using the right triangle, when x = 6 ft, we have:
cos(θ) = 6 / 10
cos(θ) = 0.6
Substituting this value of cos(θ) into the equation:
dθ/dt = (-0.9) / (10 * 0.6)
Simplifying:
dθ/dt ≈ -0.15 rad/s
Therefore, the angle between the ladder and the ground is changing at a rate of approximately -0.15 radians per second.