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

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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
 
|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>  
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|&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>  
 
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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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|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>

Latest revision as of 08:27, 2 April 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 is the equation of the tangent line to a function at a given point. If we are given the point , then we will have the approximation . Note that such an approximation is usually only good "fairly close" to your original point .

 Solution:

Part (a):  
Note that . Since and , we have
    
Similarly, Together, this means that
    
                
Part (b):  
This is simply an exercise in plugging in values. We have

    
                        
                        
                        

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