a ladder 10 ft long rests against a vertical wal. if the bottom of the ladder slides away from the wall at a rate of 0.7 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft fro the wall?
To find the rate at which the angle between the ladder and the ground is changing, we can use trigonometry. Let's denote the angle between the ladder and the ground as θ.
Given:
- The ladder is 10 ft long.
- The bottom of the ladder slides away from the wall at a rate of 0.7 ft/s.
- We want to find the rate at which θ is changing when the bottom of the ladder is 6 ft from the wall.
First, we can draw a diagram to visualize the situation:
|
|\
| \
| \ <- Ladder
| \
| \
| \
| θ \
-------------------------
Wall Ground
Now, let's label the sides of the triangle formed by the ladder, the wall, and the ground using the Pythagorean theorem:
c^2 = a^2 + b^2
The ladder forms the hypotenuse of the right triangle, so c = 10 ft.
The vertical distance from the bottom of the ladder to the ground is b = 6 ft.
Using this information, we can solve for the length of the side adjacent to the angle θ, which is a:
a^2 = c^2 - b^2
a = sqrt(c^2 - b^2)
a = sqrt(10^2 - 6^2)
a = sqrt(100 - 36)
a = sqrt(64)
a = 8 ft
We now have the values of a and b, and we know that the ladder is sliding away from the wall at a rate of 0.7 ft/s. We are interested in finding dθ/dt, the rate at which the angle θ is changing when the bottom of the ladder is 6 ft from the wall.
To find dθ/dt, we can take the derivative of both sides of the equation relating the sides of the triangle:
a^2 + b^2 = c^2
Differentiating both sides with respect to time t:
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
Since we are interested in dθ/dt, we need to relate it to da/dt and db/dt:
sin(θ) = b/c
Differentiating both sides with respect to time:
cos(θ)(dθ/dt) = (db/dt)/c
Rearranging the equation to solve for dθ/dt:
dθ/dt = (db/dt)/(c * cos(θ))
Now, substitute the known values into the equation:
- db/dt = 0.7 ft/s (rate at which the bottom of the ladder slides)
- c = 10 ft (length of the ladder)
- cos(θ) = a/c = 8/10 = 0.8
dθ/dt = (0.7 ft/s) / (10 ft * 0.8)
dθ/dt = 0.0875 rad/s
Therefore, when the bottom of the ladder is 6 ft from the wall, the angle θ is changing at a rate of 0.0875 rad/s.