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| | <span class="exam"> A lighthouse is located on a small island 3km away from the nearest point <math style="vertical-align: 0px">P</math> 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 <math style="vertical-align: 0px">P?</math> | | <span class="exam"> A lighthouse is located on a small island 3km away from the nearest point <math style="vertical-align: 0px">P</math> 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 <math style="vertical-align: 0px">P?</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009A Sample Final 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |When we see a problem talking about rates, it is usually a '''related rates''' problem.
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| − | |Thus, we treat everything as a function of time, or <math style="vertical-align: -1px">t.</math>
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| − | |We can usually find an equation relating one unknown to another, and then use implicit differentiation.
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| − | |Since the problem usually gives us one rate, and from the given info we can usually find the values of
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| − | variables at our particular moment in time, we can solve the equation
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| − | for the remaining rate.
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| − | '''Solution:''' | + | [[009A Sample Final 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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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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| − | |In the picture, we can consider the distance from the point <math style="vertical-align: 0px">P</math> to the spot the light hits the shore to be the variable <math style="vertical-align: 0px">x.</math>
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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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| − | <math style="vertical-align: 0px">x</math> is related to the angle <math style="vertical-align: 0px">\theta</math> by the equation
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| − | | <math>{\displaystyle \tan\theta\ =\ \frac{\textrm{side~opp.}}{\textrm{side~adj. }}\ =\ \frac{x}{3}.}</math>
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| − | |This gives us a relation between the two variables.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we use implicit differentiation to find
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| − | | <math>{\displaystyle \sec^{2}\theta\cdot\frac{d\theta}{dt}\ =\ \frac{1}{3}\cdot\frac{dx}{dt}.}</math>
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| − | |Rearranging, we have
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| − | | <math>{\displaystyle \frac{dx}{dt}\ =\ 3\sec^{2}\theta\cdot\frac{d\theta}{dt}.}</math>
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| − | |Again, everything is a function of time.
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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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 <math style="vertical-align: 0px">P.</math>
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| − | |This tells us that <math style="vertical-align: -1px">x=1.</math>
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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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| − | |However, <math style="vertical-align: 0px">\theta</math> is measured in radians, and there are <math style="vertical-align: 0px">2\pi</math> radians in a revolution.
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| − | |Thus, we know
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| − | | <math>{\displaystyle \frac{d\theta}{dt}\ =\ 4\cdot2\pi\ =\ 8\pi.}</math>
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| − | |Finally, we require secant. Since we know <math style="vertical-align: -3px">x=1,</math>
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| − | |we can solve the triangle to get that the length of the hypotenuse is
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| − | | <math>\sqrt{1^{2}+3^{2}}=\sqrt{10}.</math>
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| − | |This means that
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| − | | <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:
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| − | |Now, we can plug in all these values to find
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| − | <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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>80\pi\text{ km/min}</math>
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| − | |}
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| | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
