Respuesta :
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, Â Â Â Â Â Â Â
[tex]\frac{d\theta}{dt}=\frac{7\times 2\pi\ rad}{1\ min} = 14\pi\ rad/min[/tex] Â
Solution:
From the given figure.
[tex]tan\theta=\frac{x}{5}[/tex]
Substitute x = 1
[tex]tan\theta=\frac{1}{5}[/tex] Â -------------------(1)
Now, first we differentiate both side with respect to t.
[tex]\frac{d(tan\theta)}{dt} = \frac{d}{dt}.(\frac{x}{3})[/tex]
Using chain rule.
[tex]\frac{d(tan\theta)}{dt}.\frac{d\theta}{dt} = (\frac{1}{5})\frac{dx}{dt}[/tex]
[tex]sec^{2} \theta.\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}[/tex]
[tex](1+tan^{2} \theta).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}[/tex]
Substitute [tex]tan\theta=\frac{1}{5}[/tex] from equation 1.
[tex](1+(\frac{1}{3})^{2} ).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}[/tex]
[tex](1+\frac{1}{9}).\frac{d\theta}{dt} = \frac{1}{5}.\frac{dx}{dt}[/tex]
Substitute [tex]\frac{d\theta}{dt}=14\pi[/tex]
[tex]\frac{10}{9}.14\pi = \frac{1}{5}.\frac{dx}{dt}[/tex]
Applying cross multiplication rule.
[tex]\frac{dx}{dt}= \frac{700\pi }{9}[/tex]
[tex]\frac{dx}{dt}= 77.78\pi[/tex]
Therefore, light is moving along the shore at the speed of 77.78 km/min

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:
[tex]\mathbf{\frac{d\theta}{dt} = 7\ rev\ min^{-1}}[/tex]
Convert to radians per mins
[tex]\mathbf{\frac{d\theta}{dt} = 7 \times 2\pi\ rad\ min^{-1}}[/tex]
[tex]\mathbf{\frac{d\theta}{dt} = 14\pi\ rad\ min^{-1}}[/tex]
Take tan of [tex]\mathbf{\theta}[/tex]
[tex]\mathbf{tan(\theta) = \frac{x}{5}}[/tex]
Differentiate both sides with respect to t
[tex]\mathbf{sec^2(\theta) \frac{d\theta}{dt}= \frac{1}{5} \frac{dx}{dt}}[/tex]
Multiply both sides by 5
[tex]\mathbf{5sec^2(\theta) \frac{d\theta}{dt}= \frac{dx}{dt}}[/tex]
Substitute [tex]\mathbf{\frac{d\theta}{dt} = 14\pi\ rad\ min^{-1}}[/tex]
[tex]\mathbf{5sec^2(\theta) \times 14\pi= \frac{dx}{dt}}[/tex]
[tex]\mathbf{sec^2(\theta) \times 70\pi= \frac{dx}{dt}}[/tex]
Next, calculate the hypotenuse (h) side of the triangle, using the following Pythagoras theorem
[tex]\mathbf{h^2 = x^2 + 5^2}[/tex]
[tex]\mathbf{h^2 = x^2 +25}[/tex]
From the question, x = 1.
So, we have:
[tex]\mathbf{h^2 = 1^2 +25}[/tex]
[tex]\mathbf{h^2 = 1 +25}[/tex]
[tex]\mathbf{h^2 = 26}[/tex]
Take square roots
[tex]\mathbf{h = \sqrt{26}}[/tex]
Calculate cos theta
[tex]\mathbf{cos(\theta) = \frac{5}{\sqrt{26}}}[/tex]
Take inverse, to calculate secant
[tex]\mathbf{sec(\theta) = \frac{\sqrt{26}}{5}}[/tex]
Substitute [tex]\mathbf{sec(\theta) = \frac{\sqrt{26}}{5}}[/tex] in [tex]\mathbf{sec^2(\theta) \times 70\pi= \frac{dx}{dt}}[/tex]
[tex]\mathbf{(\frac{\sqrt{26}}{5})^2 \times 70\pi= \frac{dx}{dt}}[/tex]
[tex]\mathbf{\frac{26}{25} \times 70\pi= \frac{dx}{dt}}[/tex]
[tex]\mathbf{72.8\pi= \frac{dx}{dt}}[/tex]
Rewrite as:
[tex]\mathbf{\frac{dx}{dt} = 72.8\pi}[/tex]
So, we have:
[tex]\mathbf{\frac{dx}{dt} = 72.8\times \frac{22}{7}}[/tex]
[tex]\mathbf{\frac{dx}{dt} = 228.8}[/tex]
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:
https://brainly.com/question/7225176
