Difference between revisions of "009A Sample Midterm 1, Problem 3"

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!Foundations:    
 
!Foundations:    
 
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|-
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|'''1.''' '''Limit Definition of Derivative'''
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|'''1.''' &nbsp;<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
 
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
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|'''2.''' The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 
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|-
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|'''2.''' '''Equation of a tangent line'''
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
 
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|-
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|Let <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
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|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
 
|-
 
|-
 
|Using the limit definition of the derivative, we have
 
|Using the limit definition of the derivative, we have
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|We start by finding the slope of the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math>
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|We start by finding the slope of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
 
|-
 
|-
 
|Using the derivative calculated in part (a), the slope is  
 
|Using the derivative calculated in part (a), the slope is  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1)</math>
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|Now, the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1)</math>
 
|-
 
|-
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|has slope <math style="vertical-align: -13px">m=\frac{3}{2}</math> and passes through the point <math style="vertical-align: -5px">(2,1).</math>
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|has slope &nbsp;<math style="vertical-align: -13px">m=\frac{3}{2}</math>&nbsp; and passes through the point &nbsp;<math style="vertical-align: -5px">(2,1).</math>
 
|-
 
|-
 
|Hence, the equation of this line is
 
|Hence, the equation of this line is

Revision as of 15:50, 26 February 2017

Let  

(a) Use the definition of the derivative to compute   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{dy}{dx}}   for  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 y=\sqrt{3x-5}.}

(b) Find the equation of the tangent line to  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 y=\sqrt{3x-5}}   at  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 (2,1).}


Foundations:  
1.  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)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}
2. The equation of the tangent line to  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)}   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 (a,b)}   is
        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 y=m(x-a)+b}   where  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 m=f'(a).}


Solution:

(a)

Step 1:  
Let  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)=\sqrt{3x-5}.}
Using the limit definition of the derivative, 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3(x+h)-5}-\sqrt{3x-5}}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}.} \end{array}}

Step 2:  
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\sqrt{3x+3h-5}-\sqrt{3x-5})}{h} \frac{(\sqrt{3x+3h-5}+\sqrt{3x-5})}{(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(3x+3h-5)-(3x-5)}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3}{\sqrt{3x+3h-5}+\sqrt{3x-5}}}\\ &&\\ & = & \displaystyle{\frac{3}{\sqrt{3x-5}+\sqrt{3x-5}}}\\ &&\\ & = & \displaystyle{\frac{3}{2\sqrt{3x-5}}.} \end{array}}

(b)

Step 1:  
We start by finding the slope of the tangent line to  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)=\sqrt{3x-5}}   at  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 (2,1).}
Using the derivative calculated in part (a), the slope is
        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 \begin{array}{rcl} \displaystyle{m} & = & \displaystyle{f'(2)}\\ &&\\ & = & \displaystyle{\frac{3}{2\sqrt{3(2)-5}}}\\ &&\\ & = & \displaystyle{\frac{3}{2}.} \end{array}}
Step 2:  
Now, the tangent line to  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)=\sqrt{3x-5}}   at  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 (2,1)}
has slope  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 m=\frac{3}{2}}   and passes through 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 (2,1).}
Hence, the equation of this line is
        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 y=\frac{3}{2}(x-2)+1.}


Final Answer:  
    (a)     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{3}{2\sqrt{3x-5}}}
    (b)     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 y=\frac{3}{2}(x-2)+1}

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