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1.Homework Statement
[tex]y=x^{\frac}{1}{2}}- \frac{1}{3}x^{\frac{3}{2}}+\lambda[/tex]
For [itex]0 \leq x \leq 3[/itex]
Show that the arc length,s=[itex]2\sqrt{3}[/itex]
[tex]s=\int_{x_1} ^{x_2} \sqrt{1+ (\frac{dy}{dx})^2} dx[/tex]
[tex]\frac{dy}{dx}=\frac{1}{2\sqrt{x}} - \frac{1}{2}x^{\frac{1}{2}}[/tex]
[tex] (\frac{dy}{dx})^2=\frac{1}{4x}-\frac{1}{2}+\frac{x}{2}[/tex]
[tex]s=\int_{0} ^{3} \sqrt{\frac{1}{4x}+\frac{x}{2}+\frac{1}{2}} dx[/tex]
and now that integral seems a bit too odd to get the answer [itex]2\sqrt{3}[/itex]
[tex]y=x^{\frac}{1}{2}}- \frac{1}{3}x^{\frac{3}{2}}+\lambda[/tex]
For [itex]0 \leq x \leq 3[/itex]
Show that the arc length,s=[itex]2\sqrt{3}[/itex]
Homework Equations
[tex]s=\int_{x_1} ^{x_2} \sqrt{1+ (\frac{dy}{dx})^2} dx[/tex]
The Attempt at a Solution
[tex]\frac{dy}{dx}=\frac{1}{2\sqrt{x}} - \frac{1}{2}x^{\frac{1}{2}}[/tex]
[tex] (\frac{dy}{dx})^2=\frac{1}{4x}-\frac{1}{2}+\frac{x}{2}[/tex]
[tex]s=\int_{0} ^{3} \sqrt{\frac{1}{4x}+\frac{x}{2}+\frac{1}{2}} dx[/tex]
and now that integral seems a bit too odd to get the answer [itex]2\sqrt{3}[/itex]