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

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

Revision as of 22:19, 26 March 2015


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

Foundations:  
Recall that the linear approximation 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)} is the equation of the tangent line to a function at a given point. If we are given the 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} , then we will have the approximation 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)} . Note that such an approximation is usually only good "fairly close" to your original point .

 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.}

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