Question

4. (10 points)A lighthouse is located on a small island 4 mi away from the nearest point P on a straight shoreline and its light makes ten revolutions per minute. How fast (in mi/min) is the beam of light moving along the shoreline when it is 4 mi from P

Answer 1

Answer:

The beam of the light is moving along the shoreline at the rate of $\mathbf{ 160 \ \pi \ mi/min}$

Step-by-step explanation:

From the given information: Let say x is the distance between the lighthouse and the shoreline. Then, the  tangent of  trigonometry can be written as:

$tan (\theta )=\dfrac{x}{4}$

In respect to time, the expression can be derived as:

$sec^2 (\theta) \dfrac{d \theta}{dt} = \dfrac{1 }{4}\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\dfrac{4}{cos ^2 (\theta) }\dfrac{d \theta}{dt}$

To estimate the value of $\dfrac{dx}{dt}$, we need to know cos²(θ) and D

where;

D = $\sqrt{4^2+4^2}= \sqrt{32}\ mi$

$cos (\theta) = \dfrac{4}{D}$

$cos (\theta) = \dfrac{4}{\sqrt{32}}$

$cos (\theta) = \dfrac{4}{\sqrt{16 \times 2}}$

$cos (\theta) = \dfrac{4}{4 \sqrt{\ 2}}$

$cos (\theta) = \dfrac{1}{ \sqrt{\ 2}}$

$cos ^2 (\theta)= \dfrac{1}{2}$

Finally, we substitute the value of $cos^2(\theta)$ and $\dfrac{d \theta}{dt}$ in the expression derived earlier.

i.e.

$\dfrac{dx}{dt} = \dfrac{4}{\dfrac{1}{2}}\times 10 \ rev/min$

where;

10 rev/min = (10 × 2π) rad/min = 20π rad/min

Then:

$\dfrac{dx}{dt} = \dfrac{4}{\dfrac{1}{2}}\times 20 \ \pi$

$\mathbf{\dfrac{dx}{dt} = 160 \ \pi \ mi/min}$