Calculus with parametric curves

[ineedhelpnow]

#1 find the length of the curve $x=3t^2$, $y=2t^3$, $0\le t \le 3$

$L=\int_{\alpha}^{\beta} \ \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$$\frac{dx}{dt}=6t$

$\frac{dy}{dt}=6t^2$$L=\int_{0}^{3} \ \sqrt{(6t)^2+(6t^2)^2}dt$
$=\int_{0}^{3} \ \sqrt{6t^2+6t^4}dt$
$=\int_{0}^{3} \ \sqrt{6t^2(1+t^2)}dt$
$=\int_{0}^{3} \ \sqrt{6t^2}dt + \int_{0}^{3} \ \sqrt{1+t^2}dt$$=(\frac{\sqrt{6}*t^2}{2})^3_0 + (\frac{\ln\left({\sqrt{t^2+1}+t}\right)}{2}+\frac{t\sqrt{t^2+1}}{2})^3_0$

did i do it right?

 

[Euge]

No. In the second line of the calculation of $L$ -- $6^2 = 36$, not 6. So you should have

$L = \int_0^3 6\sqrt{t^2 (1 + t^2)} = 6 \int_0^3 t\sqrt{1+t^2}\, dt$.

Then you can use the u-sub $u = 1 + t^2$ to finish the rest.

 

[ineedhelpnow]

oh that makes it way more simpler. thanks

 

[ineedhelpnow]

$6 \int_{0}^{3} \ t\sqrt{1+t^2}dt$

$u=1+t^2$

$du=2t dt$

$dt=\frac{1}{2} du$$6 \int_{0}^{3} \ \frac{t\sqrt{u}}{2t}du$

$3 \int_{0}^{3} \ \sqrt{u}}du$

$3[\frac{2u^{3/2}}{3}]^3_0$

$3[\frac{2(1+t^2)^{3/2}}{3}]^3_0$

is that right?

 

[Prove It]

$6 \int_{0}^{3} \ t\sqrt{1+t^2}dt$

$u=1+t^2$

$du=2t dt$

$dt=\frac{1}{2} du$$6 \int_{0}^{3} \ \frac{t\sqrt{u}}{2t}du$

$3 \int_{0}^{3} \ \sqrt{u}}du$

$3[\frac{2u^{3/2}}{3}]^3_0$

$3[\frac{2(1+t^2)^{3/2}}{3}]^3_0$

is that right?

Yes, although there's no reason to keep those common factors there to make the calculation more difficult, write it as $\displaystyle \begin{align*} 2 \left[ \left( 1 + t^2 \right) ^{\frac{3}{2}} \right] _0^3 \end{align*}$...

 

[MarkFL]

$6 \int_{0}^{3} \ t\sqrt{1+t^2}dt$

$u=1+t^2$

$du=2t dt$

$dt=\frac{1}{2} du$$6 \int_{0}^{3} \ \frac{t\sqrt{u}}{2t}du$

$3 \int_{0}^{3} \ \sqrt{u}}du$

$3[\frac{2u^{3/2}}{3}]^3_0$

$3[\frac{2(1+t^2)^{3/2}}{3}]^3_0$

is that right?

I would change the limits too in accordance with the substitution, and then be done with $t$ altogether. :D