009C Sample Final 3, Problem 7 - Revision history

Kayla Murray at 23:25, 3 December 2017

2017-12-03T23:25:25Z

Kayla Murray at 01:51, 3 December 2017

2017-12-03T01:51:15Z

Kayla Murray at 23:24, 5 March 2017

2017-03-05T23:24:25Z

Kayla Murray at 23:21, 5 March 2017

2017-03-05T23:21:10Z

Kayla Murray at 23:00, 5 March 2017

2017-03-05T23:00:45Z

Kayla Murray at 23:00, 5 March 2017

2017-03-05T23:00:04Z

Kayla Murray at 01:21, 5 March 2017

2017-03-05T01:21:13Z

Kayla Murray at 00:59, 5 March 2017

2017-03-05T00:59:35Z

Kayla Murray at 19:01, 18 February 2017

2017-02-18T19:01:29Z

Kayla Murray: Created page with "Compute ::a) $\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}$ ::b)
2017-02-18T02:51:54Z

Created page with "Compute ::a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math> ::b) <mat..."

New page
Compute

::a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math>

::b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>

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'''Solution:'''

'''(a)'''

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'''(b)'''

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[[009C_Sample_Final_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar coordinates by
-::<math>r=1+\cos^2(2\theta)</math>
+::<math>r=1+\cos^2(2\theta).</math>
 <span class="exam">(a) Show that the point with Cartesian coordinates &nbsp;<math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>&nbsp; belongs to the curve.

@@ Line 9: / Line 9: @@
 <span class="exam">(c) In Cartesian coordinates, find the equation of the tangent line at &nbsp;<math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 56: / Line 56: @@
 \end{array}</math>
 |-
-|So, &nbsp;<math>\theta=\frac{\pi}{4}.</math>
+|So, &nbsp;<math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
 |-
-|Now, this point in polar is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
+|Now, this point in polar is &nbsp;<math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
 |}
@@ Line 64: / Line 64: @@
 !Step 2: &nbsp;
 |-
-|Now, we plug in &nbsp;<math>\theta=\frac{\pi}{4}</math>&nbsp; into our polar equation.
+|Now, we plug in &nbsp;<math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math>&nbsp; into our polar equation.
 |-
 |We get
@@ Line 76: / Line 76: @@
 \end{array}</math>
 |-
-|So, the point &nbsp;<math>(x,y)</math>&nbsp; belongs to the curve.
+|So, the point &nbsp;<math style="vertical-align: -5px">(x,y)</math>&nbsp; belongs to the curve.
 |}
@@ Line 117: / Line 117: @@
 !Step 2: &nbsp;
 |-
-|Now, recall from part (a) that the given point in polar coordinates is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
+|Now, recall from part (a) that the given point in polar coordinates is &nbsp;<math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
 |-
 |Therefore, the slope of the tangent line at this point is
@@ Line 129: / Line 129: @@
 \end{array}</math>
 |-
-|Therefore, the equation of the tangent line at the point &nbsp;<math>(x,y)</math>&nbsp; is
+|Therefore, the equation of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(x,y)</math>&nbsp; is
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>

@@ Line 34: / Line 34: @@
 !Step 1: &nbsp;
 |-
+|First, we need to convert this Cartesian point into polar.
 |-
+|So, &nbsp;<math>\theta=\frac{\pi}{4}.</math>
 |-
+|Now, this point in polar is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
 |}
@@ Line 44: / Line 64: @@
 !Step 2: &nbsp;
 |-
+|Now, we plug in &nbsp;<math>\theta=\frac{\pi}{4}</math>&nbsp; into our polar equation.
 |-
+|So, the point &nbsp;<math>(x,y)</math>&nbsp; belongs to the curve.
 |}
@@ Line 63: / Line 93: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |-
+|
 |-
+|
@@ Line 74: / Line 117: @@
 !Step 2: &nbsp;
 |-
+|Now, recall from part (a) that the given point in polar coordinates is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
 |}
@@ Line 83: / Line 138: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See above.
 |-
-|&nbsp;&nbsp; '''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See above.
 |-
-|&nbsp;&nbsp; '''(c)'''
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 48: / Line 48: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!(b) &nbsp;
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+|Insert graph
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+|

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar coordinates by
-::::::<math>r=1+\cos^2(2\theta)</math>
+::<math>r=1+\cos^2(2\theta)</math>
-::<span class="exam">a) Show that the point with Cartesian coordinates <math>(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math> belongs to the curve.
+<span class="exam">(a) Show that the point with Cartesian coordinates &nbsp;<math>(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>&nbsp; belongs to the curve.
-::<span class="exam">b) Sketch the curve.
+<span class="exam">(b) Sketch the curve.
-::<span class="exam">c) In Cartesian coordinates, find the equation of the tangent line at <math>\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
+<span class="exam">(c) In Cartesian coordinates, find the equation of the tangent line at &nbsp;<math>\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 22: / Line 22: @@
+|
 |}
 '''Solution:'''
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+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 71: / Line 93: @@
 |-
 |&nbsp;&nbsp; '''(b)'''
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]