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AgentPangolin

AgentPangolin

23.03.2020 •

Mathematics

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?

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Ответ:

Olaf2018

21.01.2022

The movement of the beam of light is an illustration of distance and rates

The beam of light along the shoreline is moving at 228.8 km per minute when it is 1 km from P

From the question, we have:

$\mathbf{\frac{d\theta}{dt} = 7\ rev\ min^{-1}}$

Convert to radians per mins

$\mathbf{\frac{d\theta}{dt} = 7 \times 2\pi\ rad\ min^{-1}}$

$\mathbf{\frac{d\theta}{dt} = 14\pi\ rad\ min^{-1}}$

Take tan of $\mathbf{\theta}$

$\mathbf{tan(\theta) = \frac{x}{5}}$

Differentiate both sides with respect to t

$\mathbf{sec^2(\theta) \frac{d\theta}{dt}= \frac{1}{5} \frac{dx}{dt}}$

Multiply both sides by 5

$\mathbf{5sec^2(\theta) \frac{d\theta}{dt}= \frac{dx}{dt}}$

Substitute $\mathbf{\frac{d\theta}{dt} = 14\pi\ rad\ min^{-1}}$

$\mathbf{5sec^2(\theta) \times 14\pi= \frac{dx}{dt}}$

$\mathbf{sec^2(\theta) \times 70\pi= \frac{dx}{dt}}$

Next, calculate the hypotenuse (h) side of the triangle, using the following Pythagoras theorem

$\mathbf{h^2 = x^2 + 5^2}$

$\mathbf{h^2 = x^2 +25}$

From the question, x = 1.

So, we have:

$\mathbf{h^2 = 1^2 +25}$

$\mathbf{h^2 = 1 +25}$

$\mathbf{h^2 = 26}$

Take square roots

$\mathbf{h = \sqrt{26}}$

Calculate cos theta

$\mathbf{cos(\theta) = \frac{5}{\sqrt{26}}}$

Take inverse, to calculate secant

$\mathbf{sec(\theta) = \frac{\sqrt{26}}{5}}$

Substitute $\mathbf{sec(\theta) = \frac{\sqrt{26}}{5}}$ in $\mathbf{sec^2(\theta) \times 70\pi= \frac{dx}{dt}}$

$\mathbf{(\frac{\sqrt{26}}{5})^2 \times 70\pi= \frac{dx}{dt}}$

$\mathbf{\frac{26}{25} \times 70\pi= \frac{dx}{dt}}$

$\mathbf{72.8\pi= \frac{dx}{dt}}$

Rewrite as:

$\mathbf{\frac{dx}{dt} = 72.8\pi}$

So, we have:

$\mathbf{\frac{dx}{dt} = 72.8\times \frac{22}{7}}$

$\mathbf{\frac{dx}{dt} = 228.8}$

Hence, the beam of light along the shoreline is moving at 228.8 km per minute when it is 1 km from P

Read more about distance and rates at:

link

A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline

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Ответ:

LaBreeska

21.01.2022

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

A lighthouse is located on a small island 5 km away from the nearest point P on a straight shoreline

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Ответ:

denysmayatskyy

23.11.2022

35/100* 800=35*800/100 this is the equation

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