009A Sample Final 1, Problem 4 - Revision history

Kayla Murray at 22:46, 2 December 2017

2017-12-02T22:46:41Z

Kayla Murray at 20:14, 18 March 2017

2017-03-18T20:14:16Z

Kayla Murray at 17:09, 27 February 2017

2017-02-27T17:09:18Z

Kayla Murray at 16:48, 27 February 2017

2017-02-27T16:48:39Z

Kayla Murray at 01:47, 19 February 2017

2017-02-19T01:47:16Z

Kayla Murray at 01:43, 19 February 2017

2017-02-19T01:43:28Z

Grad at 18:45, 4 March 2016

2016-03-04T18:45:07Z

Grad: /* 1 */

2016-03-04T18:44:51Z

Grad: /* 2 */

2016-03-04T18:44:16Z

Grad: /* 1 */

2016-03-04T18:41:35Z

@@ Line 3: / Line 3: @@
 <span class="exam">You may leave your answers in point-slope form.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |-
-|'''2.'''
+|'''2.''' Recall
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d}{dx}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}</math>

@@ Line 6: / Line 6: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' What two pieces of information do you need to write the equation of a line?
+|'''1.''' '''Chain Rule'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
-::You need the slope of the line and a point on the line.
+|-
 |-
-|'''2.''' What does the Chain Rule state?
+|'''3.''' The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
 |}
@@ Line 23: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|First, we compute&thinsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math> We get
+|First, we compute &nbsp;<math style="vertical-align: -13px">\frac{dy}{dx}.</math>
 |-
+|
-::<math>\frac{dy}{dx}\,=\,2x-\sin(\pi(x^2+1))(2\pi x).</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 34: / Line 42: @@
 |To find the equation of the tangent line, we first find the slope of the line.
 |-
-|Using <math style="vertical-align: -3px">x_0=1</math>&thinsp; in the formula for &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; from Step 1, we get
+|Using &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; in the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; from Step 1, we get
 |-
+|
-::<math>m=2(1)-\sin(2\pi)2\pi\,=\,2.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
-|To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math>&thinsp; into the equation given.
+|To get a point on the line, we plug in &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math>&nbsp; into the equation given.
 |-
-|So, we have&thinsp; <math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2.</math>
+|So, we have
 |-
-|Thus, the equation of the tangent line is&thinsp; <math style="vertical-align: -5px">y=2(x-1)+2.</math>
+|Thus, the equation of the tangent line is &nbsp; <math style="vertical-align: -14px">y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}.</math>
 |}
@@ Line 51: / Line 78: @@
 |-
+|
-:<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{-2}{\sqrt{1-4x^2}}</math>
 |-
+|
-:<math>y=2(x-1)+2</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> If
+<span class="exam"> If &nbsp;<math style="vertical-align: -5px">y=\cos^{-1} (2x)</math> compute &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; and find the equation for the tangent line at &nbsp;<math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}.</math>
-::<math>y=x^2+\cos (\pi(x^2+1))</math>
+<span class="exam">You may leave your answers in point-slope form.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 18: / Line 18: @@
 ::For functions  <math style="vertical-align: -5px">f(x)</math>&thinsp; and <math style="vertical-align: -5px">g(x),</math>&nbsp; <math style="vertical-align: -12px">~\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x).</math>
 |}
 '''Solution:'''
@@ Line 46: / Line 47: @@
 |Thus, the equation of the tangent line is&thinsp; <math style="vertical-align: -5px">y=2(x-1)+2.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 51: / Line 51: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>
 |-
-|&nbsp;&nbsp; <math>y=2(x-1)+2</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]