A searchlight is 120 ft from a straight wall. As the beam moves along the wall, the angle between the beam and the perpendicular to the wall is increasing at the rate of 1.5 degrees divided by s. How fast is the length of the beam increasing when it is 130 ft long?
To find the rate at which the length of the beam is increasing, we need to find an expression relating the length of the beam to the angle between the beam and the perpendicular to the wall.
Let's denote the length of the beam as L and the angle as θ. Based on the given information, we have a right triangle formed by the searchlight, the wall, and the perpendicular from the searchlight to the wall.
Using trigonometry, we know that the length of the beam can be expressed as:
L = 120 / cos(θ)
To find the rate at which L is changing when it is 130 ft long (L = 130), we need to take the derivative of L with respect to time (t) and solve for dL/dt.
dL/dt = d(120 / cos(θ))/dt
We know that dθ/dt = 1.5 degrees/s, but we need to convert it to radians/s since cosine uses radians.
dθ/dt = 1.5 degrees/s * (π / 180 degrees) = (π / 120) radians/s
Now let's differentiate the expression for L with respect to time using the chain rule.
dL/dt = d(120 / cos(θ))/dθ * dθ/dt
To find d(120 / cos(θ))/dθ, we need to apply the quotient rule. Let's differentiate the numerator and denominator separately.
dL/dt = (cos(θ) * 0 - 120 * (-sin(θ)))/cos^2(θ) * dθ/dt
Simplifying further:
dL/dt = 120sin(θ)/cos^2(θ) * dθ/dt
Now we have an expression for dL/dt in terms of θ.
To find θ when L = 130, we can plug it into the expression for L:
130 = 120 / cos(θ)
cos(θ) = 120 / 130
θ = arccos(120 / 130)
Now we can substitute the values into the expression for dL/dt to find the rate at which the length of the beam is increasing when it is 130 ft long:
dL/dt = 120sin(arccos(120 / 130))/cos^2(arccos(120 / 130)) * (π / 120) radians/s
Calculating the numerical value:
dL/dt ≈ 8.726 ft/s
To solve this problem, we can use the concept of related rates.
Let's denote the length of the beam as "x" and the angle between the beam and the perpendicular to the wall as "θ." We are given that the beam is 120 ft from the wall and the rate at which the angle is changing is 1.5 degrees per second (1.5°/s).
We are asked to find the rate at which the length of the beam is increasing when it is 130 ft long. Let's denote the rate of change of the length of the beam as dx/dt.
Since the searchlight forms a right triangle with the wall and the beam, we can use trigonometry to relate the length of the beam and the angle θ. In particular, we can use the tangent function:
tan(θ) = x/120
Differentiating both sides of this equation with respect to time, we get:
sec²(θ) dθ/dt = dx/dt / 120
Now, we are given dθ/dt = 1.5°/s. Let's convert this to radians per second since the trigonometric functions typically use radians:
dθ/dt = 1.5°/s * π/180° = π/120 radians/s
Plugging this into the equation above, we have:
sec²(θ) * (π/120 radians/s) = dx/dt / 120
Now we need to find the value of sec²(θ). Since we know the length of the beam is 130 ft, we can use that to find θ using the Pythagorean theorem:
x² + 120² = 130²
x² = 130² - 120²
x = √((130² - 120²))
Now we can use this value of x to find the value of θ:
tan(θ) = x/120
θ = atan(x/120)
Now, substituting the known values into the equation, we can find sec²(θ):
sec²(θ) = sec²(atan(x/120))
Finally, we can solve for dx/dt:
sec²(θ) * (π/120 radians/s) = dx/dt / 120
Multiply both sides of the equation by 120:
sec²(θ) * (π/120 radians/s) * 120 = dx/dt
Simplifying the equation further, we get:
π/120 * sec²(θ) = dx/dt
Now we can plug in the value of θ and solve for dx/dt:
dx/dt = π/120 * sec²(atan(x/120))
Substituting x = 130 ft into the equation, we can calculate the value of dx/dt.
1.5° = 0.026 radians
so, if the length of the beam is x, then
cosθ = 120/x
when x = 130, sinθ = 5/13
-sinθ dθ/dt = -120/x^2 dx/dt
Now just plug in the numbers and solve for dx/dt