009A Sample Final A, Problem 8 - Revision history

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8. (a) Find the linear approximation $L(x)$ to the function
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Created page with " 8. (a) Find the linear approximation <math style="vertical-align: -14%;">L(x)</math> to the function <math style="ver..."

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 8. (a) Find the linear approximation <math style="vertical-align: -14%;">L(x)</math> to the function <math style="vertical-align: -14%;">f(x)=\sec x</math> at the point <math style="vertical-align: -14%;">x=\pi/3</math>.
 
    (b) Use <math style="vertical-align: -14%;">L(x)</math> to estimate the value of <math style="vertical-align: -14%;">\sec\,(3\pi/7)</math>. 
 

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|Recall that the linear approximation ''L''(''x'') is the equation of the tangent line to a function at a given point. If we are given the point ''x''0, then we will have the approximation <math style="vertical-align: -25%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>. Note that such an approximation is usually only good "fairly close" to your original point ''x''0.
|}
'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (a):  
|-
|Note that ''f'' '(''x'') = sec ''x'' tan ''x''. Since sin (π/3) = √3/2 and cos (π/3) = 1/2, we have
|-
|     <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{1/2} = 2\sqrt{3}. </math>
|-
|Similarly, ''f''(π/3) = sec (π/3) = 2. Together, this means that
|-
|     <math>L(x) = f'(x_0)\cdot (x-x_0)+f(x_0) </math>
|-
|                 <math>= 2\sqrt{3}(x-\pi/3)+2.</math>
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (b):  
|-
|This is simply an exercise in plugging in values. We have
|-
|      <math>L\left(\frac{3\pi}{7}\right)=2\sqrt{3}\left(\frac{3\pi}{7}-\frac{\pi}{3}\right)+2</math>
|-
|                        <math>=2\sqrt{3}\left(\frac{9\pi-7\pi}{21}\right)+2</math>
|-
|                        <math>= 2\sqrt{3}\left(\frac{2\pi}{21}\right)+2</math>
|-
|                        <math>= \frac{4\sqrt{3}\pi}{21}+2.</math>
|}

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

@@ Line 17: / Line 17: @@
 |Note that <math style="vertical-align: -17%;">f'(x) = \sec x \tan x</math>.  Since <math style="vertical-align: -20%;">\sin(\pi/3)=\sqrt{3}/2</math> and <math style="vertical-align: -20%;">\cos(\pi/3)=1/2</math>, we have
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. </math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(\pi /3) \,\,=\,\, \sec (\pi/3) \tan (\pi/3) \,\,=\,\, \frac {1}{1/2}\cdot\frac{\sqrt{3}/2}{\,\,1/2} \,\,=\,\, 2\sqrt{3}. </math>
 |-
 |Similarly, <math style="vertical-align: -22%;">f(\pi/3) = \sec(\pi/3) = 2.</math> Together, this means that

@@ Line 19: / Line 19: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. </math>
 |-
-|Similarly, <math style="vertical-align: -22%;">f(\pi/3) = \sec(\pi/3) = 2</math>. Together, this means that
+|Similarly, <math style="vertical-align: -22%;">f(\pi/3) = \sec(\pi/3) = 2.</math> Together, this means that
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>L(x) =  f'(x_0)\cdot (x-x_0)+f(x_0) </math>

@@ Line 10: / Line 10: @@
 |Recall that the linear approximation ''L''(''x'') is the equation of the tangent line to a function at a given point. If we are given the point ''x''<span style="font-size:85%"><sub>0</sub></span>, then we will have the approximation <math style="vertical-align: -20%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>.  Note that such an approximation is usually only good "fairly close" to your original point  ''x''<span style="font-size:85%"><sub>0</sub></span>.
 |}
-'''Solution:'''
+&nbsp;'''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 ! Foundations: &nbsp;
 |-
-|Recall that the linear approximation <math style="vertical-align: -25%;">L(x)</math> is the equation of the tangent line to a function at a given point. If we are given the point <math style="vertical-align: -15%;">x_0</math>, then we will have the approximation <math style="vertical-align: -20%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>.  Note that such an approximation is usually only good "fairly close" to your original point  <math style="vertical-align: -15%;">x_0</math>.
+|Recall that the linear approximation <math style="vertical-align: -22%;">L(x)</math> is the equation of the tangent line to a function at a given point. If we are given the point <math style="vertical-align: -12%;">x_0</math>, then we will have the approximation <math style="vertical-align: -20%;">L(x) = f'(x_0)\cdot (x-x_0)+f(x_0)</math>.  Note that such an approximation is usually only good "fairly close" to your original point  <math style="vertical-align: -12%;">x_0</math>.
 |}
 &nbsp;'''Solution:'''
@@ Line 15: / Line 15: @@
 !Part (a): &nbsp;
 |-
-|Note that ''f'' '(''x'') = sec ''x'' tan ''x''.  Since sin(&pi;/3) = &radic;<span style="text-decoration:overline">3</span>/2 and cos(&pi;/3) = 1/2, we have
+|Note that <math style="vertical-align: -17%;">f'(x) = \sec x \tan x</math>.  Since <math style="vertical-align: -20%;">\sin(\pi/3)=\sqrt{3}/2</math> and <math style="vertical-align: -20%;">\cos(\pi/3)=1/2</math>, we have
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. </math>
 |-
-|Similarly, ''f''(&pi;/3) = sec(&pi;/3) = 2. Together, this means that
+|Similarly, <math style="vertical-align: -22%;">f(\pi/3) = \sec(\pi/3) = 2</math>. Together, this means that
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>L(x) =  f'(x_0)\cdot (x-x_0)+f(x_0) </math>