A lighthouse is on a small island 3 km away from the nearest point $P$ on a straight shoreline, and its light makes 4 revolutions per minute. At what rate is the beam of light moving along the shoreline when it is 1 km away from $P$ ?
A) $20 \pi \mathrm { km } / \mathrm { h }$

B) $\frac { 70 \pi } { 3 } \mathrm {~km} / \mathrm { h }$
C) $\frac { 80 \pi } { 3 } \mathrm {~km} / \mathrm { h }$
D) $\frac { 100 \pi } { 3 } \mathrm {~km} / \mathrm { h }$
E) $\frac { 110 \pi } { 3 } \mathrm {~km} / \mathrm { h }$
F) $40 \pi \mathrm { km } / \mathrm { h }$

G) $\frac { 130 \pi } { 3 } \mathrm {~km} / \mathrm { h }$

H) $\frac { 140 \pi } { 3 } \mathrm {~km} / \mathrm { h }$

Question

A lighthouse is on a small island 3 km away from the nearest point $P$ on a straight shoreline, and its light makes 4 revolutions per minute. At what rate is the beam of light moving along the shoreline when it is 1 km away from $P$ ?
A) $20 \pi \mathrm { km } / \mathrm { h }$

B) $\frac { 70 \pi } { 3 } \mathrm {~km} / \mathrm { h }$ 
C) $\frac { 80 \pi } { 3 } \mathrm {~km} / \mathrm { h }$ 
D) $\frac { 100 \pi } { 3 } \mathrm {~km} / \mathrm { h }$ 
E) $\frac { 110 \pi } { 3 } \mathrm {~km} / \mathrm { h }$
F) $40 \pi \mathrm { km } / \mathrm { h }$

G) $\frac { 130 \pi } { 3 } \mathrm {~km} / \mathrm { h }$

H) $\frac { 140 \pi } { 3 } \mathrm {~km} / \mathrm { h }$

Quizplus · Accepted Answer

To find the rate at which the beam of light is moving along the shoreline, we can use the relationship between the angular velocity of the light beam and its linear velocity at a point on the shoreline. The angular velocity is $4$ revolutions per minute, which is $4 \times 2\pi = 8\pi$ radians per minute. When the beam is 1 km away from $P$, we can form a right-angled triangle with the island as the vertex, the point on the shoreline where the beam is currently pointing as one point, and the point $P$ as the other. The distance from the island to $P$ is 3 km, forming the adjacent side of the triangle, and the distance from $P$ to where the beam is currently pointing is 1 km, forming the opposite side. Using trigonometry, specifically the tangent function, we can find the angle $\theta$ at which the beam is currently pointing relative to $P$. However, to find the rate at which the beam is moving along the shoreline, we use the fact that the linear velocity $v$ of the beam along the shoreline is related to the angular velocity $\omega$ and the distance $r$ from the lighthouse to the point on the shoreline by the formula $v = r\omega$.Given that the beam is 1 km away from $P$, we find the hypotenuse of the triangle using Pythagoras' theorem, which is $\sqrt{3^2 + 1^2} = \sqrt{10}$ km. The linear speed of the beam along the shoreline is then the product of the angular speed and this distance, $v = 8\pi \times \sqrt{10}$. However, we need to find the component of this velocity that is parallel to the shoreline, which involves projecting this velocity onto the shoreline. This calculation simplifies under the assumption that the question asks for the rate when the beam makes a right angle with $P$ and the point 1 km away, which is a misunderstanding of the setup.The correct approach involves recognizing that the distance from the lighthouse to the point on the shoreline where the beam is pointing changes as the beam moves, and the rate of change of this distance with respect to the angle is key to finding the rate at which the beam moves along the shoreline. However, the provided solution skips these steps, leading to a direct application of the angular velocity to find the linear velocity without adequately accounting for the geometry of the situation.The correct calculation involves relating the angular velocity of the beam to its linear velocity along the shoreline using the geometry of the situation. When the beam is 1 km away from $P$, the distance from the lighthouse to this point on the shoreline can be found using the Pythagorean theorem, but the critical step involves differentiating with respect to time to find the rate of change of the position of the light along the shoreline.Given the oversight in the explanation, the correct answer involves recognizing that the angular velocity of the light (in radians per minute) needs to be converted into a linear velocity at the specific point on the shoreline, taking into account the distance from the lighthouse to the point on the shoreline and the rate at which this distance changes as the beam moves. The calculation should involve the relationship between angular velocity and linear velocity ($v = r\omega$), where $r$ is the distance from the lighthouse to the point on the shoreline where the beam is pointing, and $\omega$ is the angular velocity of the beam. However, the correct application of these principles to arrive at one of the provided answer choices requires a detailed calculation that accurately reflects the geometry of the situation, which is not fully addressed in the response.

A Lighthouse Is on a Small Island 3 Km Away PPP

A Lighthouse Is on a Small Island 3 Km Away $P$