A 5m long ladder rests on a vertical wall. if the bottom of the ladder slides away from the wall at 1m/s, how fast is the angle between the lader and the ground changing when the bottom of the ladder is 3m from the wall?
Let X be the distance of the bottom from the wall. Let A be the angle
X = 5 cos A
dX/dt = (dX/dA)*(dA/dt)
= -5 sin A* dA/dt
dA/dt = (1 m/s)(-1/5)(1/sinA)
In this case, sin A = 4/5 when X =3 m.
dA/dt will be in radians per second.
To find the rate at which the angle between the ladder and the ground is changing, we need to use trigonometry and related rates.
Let's consider a right triangle formed by the ladder, the wall, and the ground. We can label the hypotenuse (ladder) as "L," the distance from the wall to the bottom of the ladder as "x," and the angle between the ladder and the ground as θ.
From the given information, we know that L = 5m and dx/dt = 1m/s (the rate at which x is changing with respect to time). We are asked to determine dθ/dt (the rate at which θ is changing with respect to time) when x = 3m.
Using trigonometry, we know that sin θ = x / L. Taking the derivative of both sides with respect to time (t), we get:
d/dt (sin θ) = d/dt (x / L)
Using the chain rule, we can rewrite this as:
cos θ * dθ/dt = (1/L) * (dx/dt)
Rearranging the equation:
dθ/dt = (1/L) * (dx/dt) / cos θ
Since L = 5m and dx/dt = 1m/s, we need to calculate cos θ when x = 3m.
To find cos θ, we can use the Pythagorean Theorem. In our right triangle, the base (x) is 3m, and the hypotenuse (L) is 5m. Using the formula:
cos θ = adjacent / hypotenuse = x / L = 3m / 5m = 3/5
Now, we can substitute the values into the equation:
dθ/dt = (1/5m) * (1m/s) / (3/5)
Simplifying:
dθ/dt = (1/5m) * (5/3s)
The units cancel out, leaving us with:
dθ/dt = 1/3 rad/s
Therefore, when the bottom of the ladder is 3m from the wall, the angle between the ladder and the ground is changing at a rate of 1/3 radian per second.