Question

A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline and its light makes seven revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from P?

Answer 1

Answer:

Light is moving at the speed = 77.78 km/min

Step-by-step explanation:

Given:

Distance between island and point P = 5 km

Moving along the the shoreline when it is 1 km from P.

So, x = 1

From the above statement light makes seven revolutions per minute.

Therefore,              

$\frac{d\theta}{dt}=\frac{7\times 2\pi\ rad}{1\ min} = 14\pi\ rad/min$  

Solution:

From the given figure.

$tan\theta=\frac{x}{5}$

Substitute x = 1

$tan\theta=\frac{1}{5}$  -------------------(1)

Now, first we differentiate both side with respect to t.

$\frac{d(tan\theta)}{dt} = \frac{d}{dt}.(\frac{x}{3})$

Using chain rule.

$\frac{d(tan\theta)}{dt}.\frac{d\theta}{dt} = (\frac{1}{5})\frac{dx}{dt}$

$sec^{2} \theta.\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}$

$(1+tan^{2} \theta).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}$

Substitute $tan\theta=\frac{1}{5}$ from equation 1.

$(1+(\frac{1}{3})^{2} ).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}$

$(1+\frac{1}{9}).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}$

Substitute $\frac{d\theta}{dt}=14\pi$

$\frac{10}{9}.14\pi = \frac{1}{5}.\frac{dx}{dt}$

Applying cross multiplication rule.

$\frac{dx}{dt}= \frac{700\pi }{9}$

$\frac{dx}{dt}= 77.78\pi$

Therefore, light is moving along the shore at the speed of 77.78 km/min