Difference between revisions of "009A Sample Final 2, Problem 5"

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<span class="exam"> A lighthouse is located on a small island 3 km away from the nearest point &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline on a point 1km away from &nbsp;<math style="vertical-align: 0px">P?</math>
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<span class="exam"> A lighthouse is located on a small island 3km away from the nearest point &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline on a point 1km away from &nbsp;<math style="vertical-align: 0px">P?</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
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|
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|When we see a problem talking about rates, it is usually a '''related rates''' problem.
 
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|-
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|
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|Thus, we treat everything as a function of time, or &nbsp;<math style="vertical-align: -1px">t.</math>
 
|-
 
|-
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|
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|We can usually find an equation relating one unknown to another, and then use implicit differentiation.
 
|-
 
|-
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|
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|Since the problem usually gives us one rate, and from the given info we can usually find the values of
 +
variables at our particular moment in time, we can solve the equation
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for the remaining rate.
 
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|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|
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|We can begin this physical word problem by drawing a picture.
 +
|-
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|[[File:009A_SF2_5_GP.png|center|325px]]
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|-
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|In the picture, we can consider the distance from the point &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; to the spot the light hits the shore to be the variable &nbsp;<math style="vertical-align: 0px">x.</math>
 
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|-
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|
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|By drawing a right triangle with the beam as its hypotenuse, we can see that our variable
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&nbsp;<math style="vertical-align: 0px">x</math>&nbsp; is related to the angle &nbsp;<math style="vertical-align: 0px">\theta</math>&nbsp; by the equation
 
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \tan\theta\ =\ \frac{\textrm{side~opp.}}{\textrm{side~adj. }}\ =\ \frac{x}{3}.}</math>
 
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|-
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|
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|This gives us a relation between the two variables.
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|
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|Now, we use implicit differentiation to find
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec^{2}\theta\cdot\frac{d\theta}{dt}\ =\ \frac{1}{3}\cdot\frac{dx}{dt}.}</math>
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|-
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|Rearranging, we have
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \frac{dx}{dt}\ =\ 3\sec^{2}\theta\cdot\frac{d\theta}{dt}.}</math>
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|-
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|Again, everything is a function of time.
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|}
 +
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
 +
|-
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|We want to know the rate that the beam is moving along the shore when
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we are one km away from the point &nbsp;<math style="vertical-align: 0px">P.</math>
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|-
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|This tells us that &nbsp;<math style="vertical-align: -1px">x=1.</math>
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|-
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|The problem also tells us that the lighthouse beam is revolving at 4 revolutions
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per minute.
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|-
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|However, &nbsp;<math style="vertical-align: 0px">\theta</math>&nbsp; is measured in radians, and there are &nbsp;<math style="vertical-align: 0px">2\pi</math>&nbsp; radians in a revolution.
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|-
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|Thus, we know
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \frac{d\theta}{dt}\ =\ 4\cdot2\pi\ =\ 8\pi.}</math>
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|-
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|Finally, we require secant. Since we know &nbsp;<math style="vertical-align: -3px">x=1,</math>
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|-
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|we can solve the triangle to get that the length of the hypotenuse is
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sqrt{1^{2}+3^{2}}=\sqrt{10}.</math>
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|-
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|This means that
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side~adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 4: &nbsp;
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|-
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|Now, we can plug in all these values to find
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\frac{dx}{dt}} & = & \displaystyle{3\sec^{2}\theta\cdot\frac{d\theta}{dt}}\\
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&&\\
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& = & \displaystyle{3\left(\frac{\sqrt{10}}{3}\right)^{2}(8\pi)}\\
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&&\\
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& = & \displaystyle{3\left(\frac{10}{3}\right)(8\pi)}\\
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&&\\
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& = & \displaystyle{80\pi\text{ km/min.}}
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\end{array}</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>80\pi\text{ km/min}</math>
 
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|}
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:48, 26 May 2017

A lighthouse is located on a small island 3km away from the nearest point    on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline on a point 1km away from  

Foundations:  
When we see a problem talking about rates, it is usually a related rates problem.
Thus, we treat everything as a function of time, or  
We can usually find an equation relating one unknown to another, and then use implicit differentiation.
Since the problem usually gives us one rate, and from the given info we can usually find the values of

variables at our particular moment in time, we can solve the equation for the remaining rate.


Solution:

Step 1:  
We can begin this physical word problem by drawing a picture.
325px
In the picture, we can consider the distance from the point    to the spot the light hits the shore to be the variable  
By drawing a right triangle with the beam as its hypotenuse, we can see that our variable

   is related to the angle  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta}   by the equation

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \tan\theta\ =\ \frac{\textrm{side~opp.}}{\textrm{side~adj. }}\ =\ \frac{x}{3}.}}
This gives us a relation between the two variables.
Step 2:  
Now, we use implicit differentiation to find
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \sec^{2}\theta\cdot\frac{d\theta}{dt}\ =\ \frac{1}{3}\cdot\frac{dx}{dt}.}}
Rearranging, we have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \frac{dx}{dt}\ =\ 3\sec^{2}\theta\cdot\frac{d\theta}{dt}.}}
Again, everything is a function of time.
Step 3:  
We want to know the rate that the beam is moving along the shore when

we are one km away from the point  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P.}

This tells us that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1.}
The problem also tells us that the lighthouse beam is revolving at 4 revolutions

per minute.

However,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta}   is measured in radians, and there are  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi}   radians in a revolution.
Thus, we know
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \frac{d\theta}{dt}\ =\ 4\cdot2\pi\ =\ 8\pi.}}
Finally, we require secant. Since we know  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1,}
we can solve the triangle to get that the length of the hypotenuse is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{1^{2}+3^{2}}=\sqrt{10}.}
This means that
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side~adj.}}\ =\ \frac{\sqrt{10}}{3}.}}
Step 4:  
Now, we can plug in all these values to find

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{dx}{dt}} & = & \displaystyle{3\sec^{2}\theta\cdot\frac{d\theta}{dt}}\\ &&\\ & = & \displaystyle{3\left(\frac{\sqrt{10}}{3}\right)^{2}(8\pi)}\\ &&\\ & = & \displaystyle{3\left(\frac{10}{3}\right)(8\pi)}\\ &&\\ & = & \displaystyle{80\pi\text{ km/min.}} \end{array}}


Final Answer:  
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 80\pi\text{ km/min}}

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