Difference between revisions of "009C Sample Final 2, Problem 10"

Revision as of 21:18, 4 March 2017

Find the length of the curve given by

x=t^{2}

y=t^{3}

0\leq t\leq 2

ExpandFoundations:
The formula for the arc length $L$ of a parametric curve with $\alpha \leq t\leq \beta$ is
$L=\int _{\alpha }^{\beta }{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}~dt.$

Solution:

ExpandStep 2:

ExpandFinal Answer:

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|The formula for the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)</math>&nbsp; with &nbsp;<math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math>&nbsp; is
+|The formula for the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a parametric curve with &nbsp;<math style="vertical-align: -4px">\alpha \leq t \leq \beta </math>&nbsp; is
 |-
 |
-&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
 |}