Difference between revisions of "009C Sample Final 2, Problem 10"
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |First, we need to calculate <math style="vertical-align: -14px">\frac{dx}{dt}</math> and <math style="vertical-align: -14px">\frac{dy}{dt}.</math> |
| + | |- | ||
| + | |Since <math style="vertical-align: -14px">x=t^2,~\frac{dx}{dt}=2t.</math> | ||
| + | |- | ||
| + | |Since <math style="vertical-align: -14px">y=t^3,~\frac{dy}{dt}=3t^2.</math> | ||
|- | |- | ||
| − | | | + | |Using the formula in Foundations, we have |
|- | |- | ||
| | | | ||
| + | <math>L=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.</math> | ||
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 2: | !Step 2: | ||
| + | |- | ||
| + | |Now, we have | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{L} & = & \displaystyle{\int_1^2 \sqrt{4t^2+9t^4}~dt}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\int_1^2 \sqrt{t^2(4+9t^2)}~dt}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\int_1^2 t\sqrt{4+9t^2}~dt.}\\ | ||
| + | \end{array}</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 3: | ||
| + | |- | ||
| + | |Now, we use <math>u</math>-substitution. | ||
| + | |- | ||
| + | |Let <math>u=4+9t^2.</math> | ||
| + | |- | ||
| + | |Then, <math>du=18tdt</math> and <math>\frac{du}{18}=tdt.</math> | ||
| + | |- | ||
| + | |Also, since this is a definite integral, we need to change the bounds of integration. | ||
| + | |- | ||
| + | |We have | ||
| + | |- | ||
| + | | <math>u_1=4+9(1)^2=13</math> and <math>u_2=4+9(2)^2=40.</math> | ||
| + | |- | ||
| + | |Hence, | ||
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| + | \displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{1}{27} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\ | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 38: | Line 76: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | | + | | <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math> |
|} | |} | ||
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 21:31, 4 March 2017
Find the length of the curve given by
- 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 x=t^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 y=t^3}
- 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 0\leq t \leq 2}
| Foundations: |
|---|
| The formula for the arc length 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} of a parametric curve with 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 \alpha \leq t \leq \beta } 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 L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.} |
Solution:
| Step 1: |
|---|
| First, we need to calculate 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{dx}{dt}} and 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}{dt}.} |
| Since 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 x=t^2,~\frac{dx}{dt}=2t.} |
| Since 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=t^3,~\frac{dy}{dt}=3t^2.} |
| Using the formula in Foundations, 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=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.} |
| Step 2: |
|---|
| Now, 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{L} & = & \displaystyle{\int_1^2 \sqrt{4t^2+9t^4}~dt}\\ &&\\ & = & \displaystyle{\int_1^2 \sqrt{t^2(4+9t^2)}~dt}\\ &&\\ & = & \displaystyle{\int_1^2 t\sqrt{4+9t^2}~dt.}\\ \end{array}} |
| Step 3: |
|---|
| Now, we use 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 u} -substitution. |
| 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 u=4+9t^2.} |
| Then, 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 du=18tdt} and 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{du}{18}=tdt.} |
| Also, since this is a definite integral, we need to change the bounds of integration. |
| 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 u_1=4+9(1)^2=13} and 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 u_2=4+9(2)^2=40.} |
| Hence, |
| 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{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{27} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\ &&\\ & = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\ \end{array}} |
| Final Answer: |
|---|
| 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{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}} |