009A Sample Final 1, Problem 4 - Revision history

MathAdmin at 16:10, 10 April 2017

2017-04-10T16:10:04Z

MathAdmin at 22:14, 18 April 2016

2016-04-18T22:14:18Z

MathAdmin at 19:06, 18 April 2016

2016-04-18T19:06:50Z

MathAdmin: Created page with " If :::::: $y=x^2+\cos (\pi(x^2+1))$ compute $\frac{dy}{dx}$ and fin..."

2016-03-05T07:08:24Z

Created page with " If ::::::<math>y=x^2+\cos (\pi(x^2+1))</math> compute <math style="vertical-align: -12px">\frac{dy}{dx}</math> and fin..."

New page

If

::::::<math>y=x^2+\cos (\pi(x^2+1))</math>

compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  and find the equation for the tangent line at <math style="vertical-align: -3px">x_0=1</math>. You may leave your answers in point-slope form.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' What two pieces of information do you need to write the equation of a line?
|-
|
::You need the slope of the line and a point on the line.
|-
|'''2.''' What does the Chain Rule state?
|-
|
::For functions <math style="vertical-align: -5px">f(x)</math>  and <math style="vertical-align: -5px">g(x),</math>  <math style="vertical-align: -12px">~\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x).</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we compute  <math style="vertical-align: -13px">\frac{dy}{dx}.</math> We get
|-
|
::<math>\frac{dy}{dx}\,=\,2x-\sin(\pi(x^2+1))(2\pi x).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>  in the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  from Step 1, we get
|-
|
::<math>m=2(1)-\sin(2\pi)2\pi\,=\,2.</math>
|-
|To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math>  into the equation given.
|-
|So, we have  <math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2.</math>
|-
|Thus, the equation of the tangent line is  <math style="vertical-align: -5px">y=2(x-1)+2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>
|-
|
:<math>y=2(x-1)+2</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ 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;"
 !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}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}</math>
 |-
-|'''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>
 |}
 '''Solution:'''
@@ Line 24: / 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 35: / 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
 |-
+|
-::<math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{y_0} & = & \displaystyle{\cos^{-1}\bigg(2\frac{\sqrt{3}}{4}\bigg)}\\
-|Thus, the equation of the tangent line is
+&&\\
 |-
+|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>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 57: / Line 78: @@
 |-
+|
-&nbsp;&nbsp; <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>
 |-
+|
-&nbsp;&nbsp; <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 42: / Line 42: @@
 |To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math>&thinsp; 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>
 |}

009A Sample Final 1, Problem 4 - Revision history

MathAdmin at 16:10, 10 April 2017

MathAdmin at 22:14, 18 April 2016

MathAdmin at 19:06, 18 April 2016

MathAdmin: Created page with " If ::::::y=x^2+\cos (\pi(x^2+1)) compute \frac{dy}{dx} and fin..."

MathAdmin: Created page with " If :::::: $y=x^2+\cos (\pi(x^2+1))$ compute $\frac{dy}{dx}$ and fin..."