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 to the function at the point .
    (b) Use to estimate the value of .

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 . 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
    
Similarly, f(π/3) = sec(π/3) = 2. Together, this means that
    
                
Part (b):  
This is simply an exercise in plugging in values. We have

    
                        
                        
                        

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