A searchlight revolving once each minute is located at a distance of 1/4 mile from a straight beach. How fast is the light moving along the beach when the beam makes an angle of 60° with the shore line?
x/(1/4) = tanθ
4 dx/dt = sec^2θ dθ/dt
now just plug in your numbers. Careful with the units.
Thank you so much Sir!
To find the speed of the searchlight along the beach, we can start by drawing a diagram. Let's call the distance along the beach as x and the distance from the light to the point where the beam hits the shore as y.
Since the searchlight revolves once each minute, the angular speed (ω) is given by:
ω = (2π radians) / (60 seconds)
= π/30 radians/second
The given information is that when the beam makes an angle of 60° with the shore line, the distance from the light to the shore (y) is 1/4 mile. We can use this information to find x in terms of y using trigonometry.
Using the sine function, we have:
sin(60°) = y / x
Simplifying, we get:
1/2 = y / x
1 = 2y / x
x = 2y
Since the searchlight is moving along the beach, we need to find dx/dt, the rate of change of x with respect to time (t). To do this, we can differentiate both sides of the equation x = 2y with respect to t:
dx/dt = d(2y)/dt
= 2dy/dt
Now, we need to find dy/dt, the rate of change of y with respect to t. We can start by differentiating sin(60°) = y / x with respect to t:
d(sin(60°))/dt = d(y / x)/dt
Using the chain rule, we have:
cos(60°) * d(60°)/dt = (dy/dt * x - y * dx/dt) / x^2
Since d(60°)/dt = 0 (since the angle is constant), and cos(60°) = 1/2, the equation simplifies to:
1/2 * 0 = (dy/dt * x - y * dx/dt) / x^2
0 = (dy/dt * (2y) - y * dx/dt) / (2y)^2
0 = (dy/dt * 2y - y * dx/dt) / 4y^2
0 = (2y * dy/dt - y * dx/dt) / 4y^2
0 = (2dy/dt - dx/dt) / 4y
Next, we can solve for dx/dt:
2dy/dt - dx/dt = 0
dx/dt = 2dy/dt
Since we know that dx/dt = 2dy/dt, the rate at which x is changing is twice the rate at which y is changing.
Now, we need to find dy/dt when y = 1/4 mile (since the distance from the light to the shore is given as 1/4 mile).
Since the searchlight revolves once each minute, dy/dt is the rate at which the distance between the searchlight and the point where the beam hits the shore changes. In one minute, the searchlight beam will have covered an angle of 360° (2π radians).
Using this information, we can calculate dy/dt:
dy/dt = (360°)/(1 minute)
= (2π radians)/(1 minute)
Finally, substituting dy/dt into the equation for dx/dt:
dx/dt = 2(2π radians)/(1 minute)
= 4π radians/minute
Therefore, the light is moving along the beach at a speed of 4π radians per minute when the beam makes an angle of 60° with the shore line.
To find the speed at which the light is moving along the beach, we can use trigonometry.
Let's denote the distance the light has traveled along the beach by y, and the distance from the light to the point on the shore line directly below the beam by x. We want to find dy/dt, the rate of change of y with respect to time.
First, we can find a relationship between x and y using the given information. Since the light is located 1/4 mile from the shore, we can apply the Pythagorean theorem:
x² + y² = (1/4)²
x² + y² = 1/16
Next, we can differentiate both sides of the equation with respect to time t using implicit differentiation:
2x(dx/dt) + 2y(dy/dt) = 0
We are given that the beam makes an angle of 60° with the shore line. This means that the tangent of the angle is equal to the ratio of y to x:
tan(60°) = y / x
√3 = y / x
Now we have two equations:
x² + y² = 1/16 (equation 1)
√3 = y / x (equation 2)
To solve for x and y, we can divide equation 2 by x:
√3 / x = y / x²
Substituting this expression for y into equation 1, we get:
x² + (√3 / x)² = 1/16
Simplifying:
x⁴ + 3 = (1/16) * x²
Multiply both sides by 16x²:
16x⁴ + 48x² - 1 = 0
Now we can solve this quartic equation for x. There may be different methods to solve quartic equations, such as using the rational root theorem or numerical methods. However, since it can be quite complex, we can use a software or calculator to find the numerical solution.
Once we have the value of x, we can find y = √(1/16 - x²) and substitute it into equation 2 to find the value of dy/dt.