A lighthouse is located on a small island 3 km away from the nearest point P on a straight shoreline and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?
Consider a right angled triangle,
tan(A)=x/3
cos(A)=3/sqrt(x^2+9)
sec^2(A)=(x^2+9)/3^2
dA/dt=4 rev/min = 2*pi/15 rad/s
Also,
x=3 tan(A)
dx/dt=3 sec^2(A) dA/dt
dx/dt=(x^2+9)/3 * 0.419
At x=1, dx/dt=1.396 km/s or 5027 km/hr.
The above attempt confused sec^2(A) with 1/sec^2(A), and 1/cos^2(A) with cos^2(A).
To find the speed of the beam of light along the shoreline when it is 1 km from point P, we can use the chain rule from calculus.
Let's denote the distance between the lighthouse and the point P as x (in km). We want to find the rate of change of x with respect to time (dx/dt) when x = 1 km.
Given that the lighthouse makes four revolutions per minute, we can determine the rate of change of the angle θ (in radians) between the beam of light and the shoreline. Since there are 2π radians in one complete revolution, the angular speed of the beam is 4 revolutions/minute * 2π radians/revolution = 8π radians/minute.
Now, let's consider a right triangle with the lighthouse at vertex L, the point P on the shoreline, and the position of the beam of light at a given time at vertex B. The side of the triangle opposite to angle θ is x, and the side adjacent to angle θ is the distance y (in km) along the shoreline between the point P and the position of the beam of light.
Using the Pythagorean theorem, we have: x^2 + y^2 = (3 km)^2.
Differentiating both sides of the equation with respect to time t, we get: 2x(dx/dt) + 2y(dy/dt) = 0.
Since we want to find dx/dt when x = 1 km, we can substitute this value into the equation.
1^2 + y^2 = (3 km)^2,
y^2 = (3 km)^2 - 1^2 = 9 km^2 - 1 km^2 = 8 km^2,
y = sqrt(8 km^2) = 2sqrt(2) km.
Plugging these values into the differentiated equation, we have:
2(1 km)(dx/dt) + 2(2sqrt(2) km)(dy/dt) = 0.
We know that dy/dt is the rate of change of y with respect to time, and since the beam of light is moving along the shoreline, dy/dt is equal to the speed of the beam along the shoreline.
Solving the equation for dx/dt:
2(1 km)(dx/dt) = -2(2sqrt(2) km)(dy/dt),
dx/dt = -(2sqrt(2) km)(dy/dt) / (2(1 km)).
Simplifying:
dx/dt = -sqrt(2) km/min * dy/dt.
Since dy/dt represents the speed of the beam along the shoreline, dx/dt is equal in magnitude but opposite in direction to dy/dt.
Therefore, when the beam of light is 1 km from point P, it is moving along the shoreline at a speed of sqrt(2) km/min.
To find how fast the beam of light is moving along the shoreline when it is 1 km from point P, we can start by drawing a diagram.
Let's call the point where the lighthouse is located L, and the point 1 km from P as Q. We know that the distance from L to P is 3 km. So, LQ represents the distance from the lighthouse to Q, which is 1 km.
Since the light makes four revolutions per minute, it completes 4 full circles around the lighthouse in one minute. This means that the light travels a distance equal to the circumference of a circle with a radius of 3 km.
The formula for the circumference of a circle is C = 2πr, where r is the radius. In this case, the radius is 3 km, so the circumference is C = 2π(3) = 6π km.
Now, we can determine how fast the beam of light is moving along the shoreline when it is 1 km from P. To do that, we need to find the rate at which the beam sweeps out distance along the shoreline. This is essentially the rate at which the angle is changing.
Since the light makes four revolutions per minute, and each revolution corresponds to a full circle, the angle swept out by the light in one minute is 360 degrees (or 2π radians). Therefore, the rate at which the angle is changing is 360 degrees per minute (or 2π radians per minute).
To find how fast the light is moving along the shoreline, we can use the chain rule of differentiation. Let's differentiate the equation of the distance from the lighthouse to the point Q with respect to time:
d/dt (LQ) = d/dt(√(LP^2 + PQ^2))
Since the distance from the lighthouse to P is constant at 3 km, the equation simplifies to:
d/dt (LQ) = d/dt(√(9 + PQ^2))
Now, we can differentiate the equation using the chain rule, where PQ is the distance at which we want to find the rate of change:
d/dt (LQ) = (1/2)(9 + PQ^2)^(-1/2) * 2PQ * (dPQ/dt)
Simplifying further:
d/dt (LQ) = PQ / (9 + PQ^2) * (dPQ/dt)
To find dPQ/dt, we know that PQ = 1 km and LQ = 3 km. Using the Pythagorean theorem, we can find the value of PQ:
PQ^2 + LQ^2 = LP^2
1^2 + 3^2 = LP^2
1 + 9 = LP^2
LP = √10 km
Now, let's substitute these values back into the equation:
d/dt (LQ) = (1 km) / (9 + (1 km)^2) * (dPQ/dt)
d/dt (LQ) = 1 km / (9 + 1 km^2) * (dPQ/dt)
Finally, we need to find dPQ/dt. Since PQ is moving along the shoreline, we can consider it as the horizontal component of a right-angled triangle, where the vertical component is LQ and the hypotenuse is LP. So, we can define:
sin(θ) = PQ / LP
Taking the derivative with respect to time, we have:
cos(θ) * (dθ/dt) = (dPQ/dt) / LP
Since LP = √10 km, we can solve for (dPQ/dt):
(dPQ/dt) = LP * cos(θ) * (dθ/dt)
Substituting this into the original equation, we get:
d/dt (LQ) = 1 km / (9 + 1 km^2) * (LP * cos(θ) * (dθ/dt))
Now, we have all the components to find how fast the beam of light is moving along the shoreline:
- Angle (θ) = 2π radians per minute
- LP = √10 km
- Cos(θ) = cos(2π) = 1
- d/dt (LQ) = (1 km) / (9 + (1 km)^2) * (LP * cos(θ) * (dθ/dt))
Substituting these values into the equation, we can calculate the speed at which the beam of light is moving along the shoreline when it is 1 km from point P.
A is angle in radians between perpendicular to shore at P (length 3 ) and light at shore (x which is 1 at start).
tan A = x/3
dA/dt = w = rad/sec = 4* 2 pi/60 = .419 rad/s
dx/dt = 3 d tan A /dt
dx/dt = (3/sec^2 A) dA/dt
dx/dt = 1.26 /sec^2 A = 1.26 cos^2 A
at x = 1, tan A = 1/3
so A = 18,43 deg
so cos A = .949
1.26 cos^2 A = 1.13 km/s