009C Sample Final 1, Problem 10 - Revision history

Kayla Murray at 21:02, 3 December 2017

2017-12-03T21:02:50Z

Kayla Murray at 17:59, 3 December 2017

2017-12-03T17:59:05Z

Kayla Murray at 01:17, 3 December 2017

2017-12-03T01:17:12Z

Kayla Murray at 19:13, 18 March 2017

2017-03-18T19:13:32Z

Kayla Murray at 23:31, 26 February 2017

2017-02-26T23:31:48Z

Kayla Murray at 19:10, 26 February 2017

2017-02-26T19:10:39Z

Kayla Murray at 00:15, 26 February 2017

2017-02-26T00:15:58Z

Kayla Murray at 01:42, 19 April 2016

2016-04-19T01:42:37Z

Kayla Murray at 18:46, 29 February 2016

2016-02-29T18:46:40Z

Kayla Murray at 22:46, 23 February 2016

2016-02-23T22:46:45Z

@@ Line 6: / Line 6: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute the equation of the tangent line at &nbsp;<math>t_0=\frac{\pi}{4}</math>.
+<span class="exam">(b) Compute the equation of the tangent line at &nbsp;<math style="vertical-align: -14px">t_0=\frac{\pi}{4}</math>.
 <hr>

@@ Line 6: / Line 6: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute the equation of the tangent line at &nbsp; <math>t_0=\frac{\pi}{4}</math>.
+<span class="exam">(b) Compute the equation of the tangent line at &nbsp;<math>t_0=\frac{\pi}{4}</math>.
 <hr>

@@ Line 8: / Line 8: @@
 <span class="exam">(b) Compute the equation of the tangent line at &nbsp; <math>t_0=\frac{\pi}{4}</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 10 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 41: / Line 41: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4\sin t}{3\cos t}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{4\sin t}{3\cos t}.</math>
 |-
 |So, at &nbsp;<math>t_0=\frac{\pi}{4},</math>&nbsp; the slope of the tangent line is
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>m=\frac{-4\sin\bigg(\frac{\pi}{4}\bigg)}{3\cos\bigg(\frac{\pi}{4}\bigg)}=\frac{-4}{3}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>m=-\frac{4\sin\bigg(\frac{\pi}{4}\bigg)}{3\cos\bigg(\frac{\pi}{4}\bigg)}=-\frac{4}{3}.</math>
 |}
@@ Line 71: / Line 71: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-\frac{4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}.</math>
 |}
@@ Line 80: / Line 80: @@
 |&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above.
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -14px">y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -14px">y=-\frac{4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math>.
+<span class="exam">(b) Compute the equation of the tangent line at &nbsp; <math>t_0=\frac{\pi}{4}</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 19: / Line 19: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;The slope is <math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;The slope is &nbsp;<math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.</math>
 |}
@@ Line 38: / Line 38: @@
 |First, we need to find the slope of the tangent line.
 |-
-|Since <math style="vertical-align: -14px">\frac{dy}{dt}=-4\sin t</math> and <math style="vertical-align: -14px">\frac{dx}{dt}=3\cos t,</math> we have
+|Since &nbsp; <math style="vertical-align: -14px">\frac{dy}{dt}=-4\sin t</math> &nbsp; and &nbsp; <math style="vertical-align: -14px">\frac{dx}{dt}=3\cos t,</math>&nbsp; we have
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4\sin t}{3\cos t}.</math>
 |-
-|So, at <math>t_0=\frac{\pi}{4},</math> the slope of the tangent line is
+|So, at &nbsp;<math>t_0=\frac{\pi}{4},</math>&nbsp; the slope of the tangent line is
 |-
+|
@@ Line 54: / Line 54: @@
 |Since we have the slope of the tangent line, we just need a find a point on the line in order to write the equation.
 |-
-|If we plug in <math>t_0=\frac{\pi}{4}</math> into the equations for <math style="vertical-align: -5px">x(t)</math> and <math style="vertical-align: -5px">y(t),</math> we get
+|If we plug in &nbsp; <math>t_0=\frac{\pi}{4}</math>&nbsp; into the equations for &nbsp;<math style="vertical-align: -5px">x(t)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y(t),</math>&nbsp; we get
 |-
+|
@@ Line 62: / Line 62: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>y\bigg(\frac{\pi}{4}\bigg)=4\cos\bigg(\frac{\pi}{4}\bigg)=2\sqrt{2}.</math>
 |-
-|Thus, the point <math>\bigg(\frac{3\sqrt{2}}{2},2\sqrt{2}\bigg)</math> is on the tangent line.
+|Thus, the point &nbsp;<math>\bigg(\frac{3\sqrt{2}}{2},2\sqrt{2}\bigg)</math>&nbsp; is on the tangent line.
 |}
@@ Line 68: / Line 68: @@
 !Step 3: &nbsp;
 |-
-|Using the point found in Step 2, the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math> is
+|Using the point found in Step 2, the equation of the tangent line at &nbsp; <math>t_0=\frac{\pi}{4}</math>&nbsp; is
 |-
+|
@@ Line 78: / Line 78: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -14px">y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 24: / Line 24: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!(a) &nbsp;
 |-
 |Insert sketch of curve

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar parametrically by
-::::::<span class="exam"><math>x(t)=3\sin t</math>
+::<span class="exam"><math>x(t)=3\sin t</math>
-::::::<span class="exam"><math>y(t)=4\cos t</math>
+::<span class="exam"><math>y(t)=4\cos t</math>
-::::::<span class="exam"><math>0\leq t \leq 2\pi</math>
+::<span class="exam"><math>0\leq t \leq 2\pi</math>
-::<span class="exam">a) Sketch the curve.
+<span class="exam">(a) Sketch the curve.
-::<span class="exam">b) Compute the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math>.
+<span class="exam">(b) Compute the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 14: / Line 14: @@
 |-
+|
-::You need the slope of the line and a point on the line.
+&nbsp; &nbsp; &nbsp; &nbsp;You need the slope of the line and a point on the line.
 |-
 |'''2.''' What is the slope of the tangent line of a parametric curve?
 |-
+|
-::The slope is <math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;The slope is <math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.</math>
 |}
 '''Solution:'''
@@ Line 42: / Line 43: @@
 |-
+|
-::<math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4\sin t}{3\cos t}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4\sin t}{3\cos t}.</math>
 |-
 |So, at <math>t_0=\frac{\pi}{4},</math> the slope of the tangent line is
 |-
+|
-::<math>m=\frac{-4\sin\bigg(\frac{\pi}{4}\bigg)}{3\cos\bigg(\frac{\pi}{4}\bigg)}=\frac{-4}{3}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>m=\frac{-4\sin\bigg(\frac{\pi}{4}\bigg)}{3\cos\bigg(\frac{\pi}{4}\bigg)}=\frac{-4}{3}.</math>
 |}
@@ Line 58: / Line 59: @@
 |-
+|
-::<math>x\bigg(\frac{\pi}{4}\bigg)=3\sin\bigg(\frac{\pi}{4}\bigg)=\frac{3\sqrt{2}}{2}</math> and
+&nbsp; &nbsp; &nbsp; &nbsp;<math>x\bigg(\frac{\pi}{4}\bigg)=3\sin\bigg(\frac{\pi}{4}\bigg)=\frac{3\sqrt{2}}{2}</math> and
 |-
+|
-::<math>y\bigg(\frac{\pi}{4}\bigg)=4\cos\bigg(\frac{\pi}{4}\bigg)=2\sqrt{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>y\bigg(\frac{\pi}{4}\bigg)=4\cos\bigg(\frac{\pi}{4}\bigg)=2\sqrt{2}.</math>
 |-
 |Thus, the point <math>\bigg(\frac{3\sqrt{2}}{2},2\sqrt{2}\bigg)</math> is on the tangent line.
@@ Line 72: / Line 73: @@
 |-
+|
-::<math>y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' See Step 1 above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above for the graph.
 |-
-|&nbsp;&nbsp; '''(b)''' <math style="vertical-align: -14px">y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -14px">y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|Review tangent lines of polar curves
+|'''1.''' What two pieces of information do you need to write the equation of a line?
 |}