A ladder 13 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft. per second,how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 5ft from the wall?
h is height
x is distance from wall
sin T = h/13
cos T = x/13
d cos T /dt = - sin T dT/dt =(1/13)dx/dt
or
dT/dt = -(1/(13 sin T))(dx/dt)
sin t = h/13
at x = 5
h = sqrt (169 - 25) = 12
so
sin T = 12/13
so
dT/dt = -(1/12)dx/dt = -1/12
radians/second
To solve this problem, we need to find the rate of change of the angle between the ladder and the ground. Let's denote the angle as θ.
We know that the ladder is 13 ft long and the bottom of the ladder slides away from the wall at a rate of 1 ft per second. We want to find how fast the angle θ is changing when the bottom of the ladder is 5 ft from the wall.
Let's set up a right triangle to represent the situation:
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13 | \
| \
| \
| \
|____\
5
The ladder forms the hypotenuse of the right triangle, the adjacent side to the angle θ is 5 ft, and the opposite side is the height of the triangle, which we need to find.
Using the Pythagorean theorem, we can find the height of the triangle:
height = √(13^2 - 5^2)
= √(169 - 25)
= √144
= 12 ft
Now, we can differentiate both sides of the equation of the right triangle with respect to time (t):
height = 12
Differentiating both sides:
d(height)/dt = d(12)/dt
d(height)/dt = 0 (since 12 is a constant)
Now, we will differentiate the equation relating the height, the distance from the wall, and the angle θ:
sin(θ) = height/13
Differentiating both sides of the equation with respect to time (t):
d(sin(θ))/dt = d(height)/dt / 13
d(sin(θ))/dt = 0 / 13
d(sin(θ))/dt = 0
So, the rate of change of the angle θ with respect to time is 0. The angle θ is not changing when the bottom of the ladder is 5 ft from the wall.
To find how fast the angle between the ladder and the ground is changing, we can use the concept of related rates. Let's assume that the distance between the bottom of the ladder and the wall is 'x', and the angle between the ladder and the ground is 'θ'.
Given:
- The ladder's length (hypotenuse) is 13 ft.
- The bottom of the ladder is sliding away from the wall at 1 ft per second, so dx/dt = 1 ft/s.
We want to find dθ/dt, the rate at which the angle θ is changing, when the bottom of the ladder is 5 ft from the wall (x = 5 ft).
Using the Pythagorean theorem, we have:
x² + (wall height)² = (ladder length)²
x² + y² = 13²
Differentiating both sides with respect to time(t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in the rate of change of angle θ, we need to express this equation in terms of θ. We can use the trigonometric relationship between θ and x/y:
tan(θ) = y/x
Differentiating both sides with respect to time(t), we have:
sec²(θ)dθ/dt = (1/x)(dy/dt) - y(x²)/x² (using quotient rule)
Now let's substitute the given values:
x = 5 ft
dx/dt = 1 ft/s
From the Pythagorean theorem:
y² = 13² - x²
y² = 13² - 5²
y = √(13² - 5²)
y ≈ 12 ft
Now we can find dy/dt by substituting these values into the first equation we derived:
2x(dx/dt) + 2y(dy/dt) = 0
2(5)(1) + 2(12)(dy/dt) = 0
10 + 24(dy/dt) = 0
24(dy/dt) = -10
dy/dt ≈ -10/24
dy/dt ≈ -5/12 ft/s
Now let's substitute the values we have into the equation we derived for dθ/dt:
sec²(θ)(dθ/dt) = (1/x)(dy/dt) - y(x²)/x²
At x = 5 ft:
sec²(θ)(dθ/dt) = (1/5)(-5/12) - 12(5²)/5²
sec²(θ)(dθ/dt) = -1/12 - 144/25
sec²(θ)(dθ/dt) = (-25 - 288)/300
sec²(θ)(dθ/dt) = -313/300
Finally, solving for dθ/dt:
dθ/dt = (-313/300) * 1/cos²(θ)
Note: We don't have the exact value for θ, so we can't determine the exact value of dθ/dt without knowing the angle. However, we can calculate dθ/dt for a specific angle once we know its value.