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 1.3 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? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)
To solve this problem, we can use related rates. We need to find the rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 8 ft from the wall.
Let's denote the length of the ladder as L, the distance of the bottom of the ladder from the wall as x, and the angle between the ladder and the ground as θ.
According to the problem, dx/dt = 1.3 ft/s, which represents the rate at which x is changing. We need to find dθ/dt, the rate at which the angle θ is changing.
Using trigonometry, we can see that sin(θ) = x/L.
Differentiating both sides with respect to t (time) gives:
d/dt(sin(θ)) = d/dt(x/L).
Applying the chain rule, we have:
cos(θ) * dθ/dt = (1/L) * dx/dt.
We know L = 10 ft and dx/dt = 1.3 ft/s. We need to find dθ/dt when x = 8 ft.
Substituting the known values, we have:
cos(θ) * dθ/dt = (1/10) * 1.3.
To find cos(θ), we can use the right triangle formed by the ladder against the vertical wall. Since the ladder is 10 ft long and the bottom is 8 ft away from the wall, the height (opposite side of the angle θ) can be found using the Pythagorean theorem:
height = √(L^2 - x^2) = √(10^2 - 8^2) = √(100 - 64) = √36 = 6 ft.
cos(θ) = adjacent/hypotenuse = (8 ft)/(10 ft) = 4/5.
Substituting these values, we have:
(4/5) * dθ/dt = (1/10) * 1.3.
Dividing both sides by (4/5), we get:
dθ/dt = (1/10) * 1.3 * (5/4) = 0.1625 rad/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 a rate of approximately 0.1625 rad/s.
To find the rate of change of the angle, we can use trigonometric ratios. Let's denote the angle between the ladder and the ground as θ.
Given:
- Length of the ladder (hypotenuse) = 10 ft
- Rate of change of the distance between the bottom of the ladder and the wall (x) = 1.3 ft/s
- Distance between the bottom of the ladder and the wall (x) = 8 ft
Let's draw a right-angled triangle to visualize the situation:
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x | / |
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|/ | 10 ft
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Using the Pythagorean theorem, we can find the height of the triangle:
height^2 + x^2 = 10^2
height^2 + 8^2 = 10^2
height^2 = 100 - 64
height^2 = 36
height = 6 ft
Now we can differentiate both sides of the equation with respect to time (t):
d(height)/dt = d(6)/dt
Since the height is not changing, its derivative with respect to time is 0:
0 = d(6)/dt
0 = 0
Next, we can differentiate the equation height^2 + x^2 = 10^2 with respect to time (t):
2(height)(d(height)/dt) + 2(x)(d(x)/dt) = 0
Substituting the given values:
2(6)(0) + 2(8)(1.3) = 0
Simplifying:
16.8 = 0
This is not possible, so let's check our calculation.
Indeed, since the height is not changing, the rate of change of the angle between the ladder and the ground is 0 when the bottom of the ladder is 8 ft from the wall.
we have
x^2 + y^2 = 100
when x=8, y=6
Now, consider
tan θ = y/x
sec^2 θ dθ/dt = (x dy/dt - y dx/dt)/x^2
Now just plug in your numbers and solve for dθ/dt