009A Sample Final 2, Problem 5 - Revision history

Kayla Murray at 17:16, 1 December 2017

2017-12-01T17:16:09Z

2017-05-26T19:50:56Z

2017-05-26T18:45:20Z

2017-05-26T18:42:09Z

2017-05-26T18:38:18Z

2017-05-26T18:26:27Z

2017-03-06T21:57:03Z

2017-03-06T21:48:03Z

2017-02-19T01:55:21Z

2017-02-18T02:28:09Z

@@ Line 1: / Line 1: @@
 <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;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 2, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 30: / Line 30: @@
 &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
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \tan\theta\ =\ \frac{\textrm{side opp.}}{\textrm{side adj. }}\ =\ \frac{x}{3}.}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \tan\theta\ =\ \frac{\textrm{side~opp.}}{\textrm{side~adj. }}\ =\ \frac{x}{3}.}</math>
 |-
 |This gives us a relation between the two variables.
@@ Line 74: / Line 74: @@
 |This means that
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side~adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
 |}

@@ Line 23: / Line 23: @@
 |We can begin this physical word problem by drawing a picture.
 |-
-|Insert picture
+|[[File:009A_SF2_5_GP.png|center|325px]]
 |-
 |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>

@@ Line 53: / Line 53: @@
 |-
 |We want to know the rate that the beam is moving along the shore when
-we are one km away from the point <math>P.</math>
+we are one km away from the point &nbsp;<math style="vertical-align: 0px">P.</math>
 |-
-|This tells us that <math>x=1.</math>
+|This tells us that &nbsp;<math style="vertical-align: -1px">x=1.</math>
 |-
 |The problem also tells us that the lighthouse beam is revolving at 4 revolutions
 per minute.
 |-
-|However, <math>\theta</math> is measured in radians, and there are
+|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.
 |-
 |Thus, we know
@@ Line 67: / Line 66: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \frac{d\theta}{dt}\ =\ 4\cdot2\pi\ =\ 8\pi.}</math>
 |-
-|Finally, we require secant. Since we know <math>x=1,</math>
+|Finally, we require secant. Since we know &nbsp;<math style="vertical-align: -3px">x=1,</math>
 |-
 |we can solve the triangle to get that the length of the hypotenuse is
@@ Line 74: / Line 73: @@
 |-
 |This means that
+|-
-|-&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
 |}

@@ Line 38: / Line 38: @@
 !Step 2: &nbsp;
 |-
+|
 |-
+|
 |}
@@ Line 47: / Line 99: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>80\pi\text{ km/min}</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 !Foundations: &nbsp;
 |-
+|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 &nbsp;<math style="vertical-align: -1px">t.</math>
 |-
+|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
 |}
@@ Line 19: / Line 21: @@
 !Step 1: &nbsp;
 |-
+|We can begin this physical word problem by drawing a picture.
 |-
+|Insert picture
 |-
+|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>
 |-
+|By drawing a right triangle with the beam as its hypotenuse, we can see that our variable
 |}

← Older revision		Revision as of 21:48, 6 March 2017
Line 1:		Line 1:
−	<span class="exam"> A lighthouse is located on a small island 3km away from the nearest point <math>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>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>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

← Older revision		Revision as of 02:28, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">~~Compute~~	+	<span class="exam"> A lighthouse is located on a small island 3km away from the nearest point <math>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>P?</math>
−
−	~~::<span class="exam">~~a) <math ~~style="vertical-align: -14px"~~>~~\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}~~</math>
−
−	~~::<span class="exam">b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin^2x}{3x}</math>~~
−
−	~~::<span class="exam">c)~~ <math ~~style="vertical-align: -14px"~~>~~\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}~~</math>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"