Let x = 2t^2, y = (3/2)t^2 t = [3,5]

I have to find to length of the curve from these equations using integration, without eliminating parameters.

Then I need to sketch it having eliminated parameters, and finally, find the length using algebra.

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- Apr 26th 2009, 08:40 PMyuropodCurve Sketching
Let x = 2t^2, y = (3/2)t^2 t = [3,5]

I have to find to length of the curve from these equations using integration, without eliminating parameters.

Then I need to sketch it having eliminated parameters, and finally, find the length using algebra. - Apr 26th 2009, 10:00 PMredsoxfan325
$\displaystyle \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t}{4t} = \frac{3}{4}$.

As $\displaystyle t=[3,5], x=[18,50]$.

Use the arc length formula: $\displaystyle \int_{18}^{50}\sqrt{1+(3/4)^2}\,dx = \int_{18}^{50}\sqrt{25/16}\,dx = \frac{5}{4}\int_{18}^{50}\,dx = \frac{5}{4}\cdot 32 = 40$

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$\displaystyle t=\sqrt{\frac{x}{2}}~~~~~t=\sqrt{\frac{2y}{3}}$

$\displaystyle \sqrt{\frac{x}{2}}=\sqrt{\frac{2y}{3}} \implies \frac{x}{2}=\frac{2y}{3} \implies y=\frac{3x}{4}$ (This is the equation of the line you want to sketch.)

Using $\displaystyle x=[18,50]$ again, we have $\displaystyle y=\left[\frac{27}{2},\frac{75}{2}\right]$

Using the Pythagorean Theorem, we have the length of this line to be $\displaystyle \sqrt{(50-18)^2+(75/2-27/2)^2} = \sqrt{32^2+24^2} = \sqrt{1600}=40$ which agrees with the above answer. - Apr 27th 2009, 12:36 AMyuropod
Thanks a bunch!

- Apr 27th 2009, 01:37 AMSoroban
Hello, yuropod!

Quote:

Let: .$\displaystyle \begin{array}{ccc}x &=&2t^2 \\ y &=&\frac{3}{2}t^2\end{array} \quad t \in [3,5]$

Find the length of the curve*without eliminating parameters.*

We have: .$\displaystyle \begin{array}{ccc}\frac{dx}{dt} &=& 4t \\ \\[-4mm]\frac{dy}{dt} &=& 3t \end{array}$

. . Then: .$\displaystyle \left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2 \:=\:(4t)^2 + (3t)^2 \:=\:16t^2 + 9y^2 \:=\:25t^2$

. . Hence: .$\displaystyle \sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2} \;=\;\sqrt{25t^2} \;=\;5t$

Therefore: .$\displaystyle L \;=\;\int^5_3 5t\,dt \;=\;\tfrac{5}{2}\,t^2\,\bigg]^5_3 \;=\;\tfrac{5}{2}(5^2) - \tfrac{5}{2}(3^2) \;=\;40$

- Apr 27th 2009, 02:24 AMcurvature