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

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(Created page with "<br> <span style="font-size:135%"> <font face=Times Roman>8. (a) Find the linear approximation <math style="vertical-align: -14%;">L(x)</math> to the function <math style="ver...")
 
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! Foundations: &nbsp;  
 
! Foundations: &nbsp;  
 
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|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: -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''<span style="font-size:85%"><sub>0</sub></span>.
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|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>.
 
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'''Solution:'''
 
'''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 ''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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|&nbsp;&nbsp;&nbsp;&nbsp; <math>f'(\pi /3) = 2\cdot\frac{\sqrt{3}/2}{1/2} = 2\sqrt{3}. </math>  
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|&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, ''f''(&pi;/3) = sec(&pi;/3) = 2. 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 09:14, 24 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 at the point .
    (b) Use 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 L(x) is the equation of the tangent line to a function at a given point. If we are given the point x0, 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 x0.

Solution:

Part (a):  
Note that f '(x) = sec x tan x. Since sin(π/3) = √3/2 and cos(π/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, f(π/3) = sec(π/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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