Mr. Cook was painting Arvada West High School, standing on top of a 25�]foot ladder. Why? We will never
know. However, he was horrified to find the ladder being pulled away by the same notorious student that
threw a snowball at Mrs. Evans. The base of the ladder was moving from the school at a constant rate of 2
feet per second.
9) At what rate was the angle between the building and the ladder changing when the base of the ladder was
17 feet away from AWest?
To find the rate at which the angle between the building and the ladder is changing, we can use trigonometry. Let's assume that the ladder makes an angle "θ" with the ground.
First, let's draw a diagram to visualize the situation:
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-------------------------------- School
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In this diagram, the ladder is represented by the line inclined at angle θ to the ground, and the base of the ladder is at point AWest (17 feet away from the school).
Now, we know that the base of the ladder is moving away from the school at a constant rate of 2 feet per second. The rate at which the angle between the building and the ladder is changing can be found by taking the derivative of the angle θ with respect to time.
Let's assume that t represents time in seconds.
To solve the problem, we need to establish a relationship between the angle θ and the distance x between the base of the ladder and the school. We can use trigonometry to do this.
In a right-angled triangle with the ladder as the hypotenuse, we have:
sin(θ) = opposite/hypotenuse
cos(θ) = adjacent/hypotenuse
In this case, the opposite side is the height of the wall (which remains constant at 25 feet), and the adjacent side is the distance between the base of the ladder and the school (x).
From the information given, we know that the base of the ladder is moving away at a rate of 2 feet per second. We can express this as:
dx/dt = 2 feet per second
Now, let's differentiate the trigonometric equation with respect to time t:
Differentiating sin(θ) = opposite/hypotenuse with respect to t, we get:
cos(θ) * dθ/dt = -dh/dt
Differentiating cos(θ) = adjacent/hypotenuse with respect to t, we get:
-sin(θ) * dθ/dt = dx/dt
We can solve these equations to find the rate at which the angle θ is changing with respect to time.
Since we want to find dθ/dt, let's rearrange the equations:
dθ/dt = -dh/dt / cos(θ) (Equation 1)
dθ/dt = -dx/dt / sin(θ) (Equation 2)
We have dx/dt = 2 feet per second and we need to find dθ/dt when the base of the ladder is 17 feet away from AWest (17 feet).
Using Equation 2 and substituting the given values:
dθ/dt = -2 / sin(θ)
To find the value of sin(θ), we can use the right triangle formed by the ladder, where the opposite side is height (25 feet) and the adjacent side is the distance between the base of the ladder and the school (17 feet).
Using the Pythagorean theorem, we have:
hypotenuse^2 = opposite^2 + adjacent^2
ladder^2 = 25^2 + 17^2
Simplifying:
ladder^2 = 625 + 289
ladder^2 = 914
Taking the square root of both sides:
ladder = √914
ladder ≈ 30.23 feet
Now, we can find sin(θ):
sin(θ) = opposite/hypotenuse = 25/30.23
sin(θ) ≈ 0.825
Substituting this value into Equation 2:
dθ/dt = -2 / 0.825
dθ/dt ≈ -2.42 degrees per second
Therefore, when the base of the ladder is 17 feet away from AWest and moving away at a rate of 2 feet per second, the angle between the building and the ladder is changing at a rate of approximately -2.42 degrees per second. The negative sign indicates that the angle is decreasing or getting smaller.