Difference between revisions of "009A Sample Final A, Problem 8"

Revision as of 21:19, 26 March 2015

8. (a) Find the linear approximation $L(x)$ to the function $f(x)=\sec x$ at the point $x=\pi /3$ .
(b) Use $L(x)$ to estimate the value of $\sec \,(3\pi /7)$ .

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

L(x)=f'(x_{0})\cdot (x-x_{0})+f(x_{0})

. Note that such an approximation is usually only good "fairly close" to your original point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0} .

Solution:

Part (a):
Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x) = \sec x \tan x} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(\pi/3)=\sqrt{3}/2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\pi/3)=1/2} , we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{\,\,1/2} = 2\sqrt{3}. }
Similarly, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\pi/3) = \sec(\pi/3) = 2} . Together, this means that
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(x) = f'(x_0)\cdot (x-x_0)+f(x_0) }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\sqrt{3}(x-\pi/3)+2.}

Part (b):
This is simply an exercise in plugging in values. We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L\left(\frac{3\pi}{7}\right)=2\sqrt{3}\left(\frac{3\pi}{7}-\frac{\pi}{3}\right)+2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =2\sqrt{3}\left(\frac{9\pi-7\pi}{21}\right)+2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\sqrt{3}\left(\frac{2\pi}{21}\right)+2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{4\sqrt{3}\pi}{21}+2.}

Return to Sample Exam

@@ 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>

Difference between revisions of "009A Sample Final A, Problem 8"

Revision as of 21:19, 26 March 2015

Navigation menu

Search