Difference between revisions of "009A Sample Midterm 1, Problem 3"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |Let <math>f(x)=\sqrt{3x-5}.</math> | ||
|- | |- | ||
|Using the limit definition of the derivative, we have | |Using the limit definition of the derivative, we have | ||
| Line 57: | Line 59: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |We start by finding the slope of the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1).</math> |
|- | |- | ||
| − | | | + | |Using the derivative calculated in part (a), the slope is |
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| − | + | \displaystyle{m} & = & \displaystyle{f'(2)}\\ | |
| − | + | &&\\ | |
| + | & = & \displaystyle{\frac{3}{2\sqrt{3(2)-5}}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{3}{2}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 69: | Line 75: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1)</math> |
|- | |- | ||
| − | | | + | |has slope <math>m=\frac{3}{2}</math> and passes through the point <math>(2,1).</math> |
|- | |- | ||
| − | | | + | |Hence, the equation of this line is |
|- | |- | ||
| − | | | + | | <math>y=\frac{3}{2}(x-2)+1.</math> |
|} | |} | ||
| Line 81: | Line 87: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | '''(a)''' <math> | + | | '''(a)''' <math>\frac{3}{2\sqrt{3x-5}}</math> |
|- | |- | ||
| − | |'''(b)''' | + | | '''(b)''' <math>y=\frac{3}{2}(x-2)+1</math> |
|} | |} | ||
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 12:29, 16 February 2017
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 y=\sqrt{3x-5}.}
- 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. Limit Definition of Derivative |
| 2. Tangent line equation |
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} |