find length of loop of x=6t-2t^3, y=6t^2 (please check ans.)

[XBOX999]

Please Help me as fast as possible by checking my answer

Find the length of the loop of the given curve.
x= 6t-2t^3
y=6t^2
My answer is
I Differentiate both:
x '(t) = 6 - 6t2
y '(t) = 12t

Square and add both
(6-6t2)(6-6t2) + 144t2 = 36 -72t2+36t4 + 144 t2= 36 + 72t2+ 36t4= 36(1+2t2+t4)

Integrate:
36(t + 2t3/3 + t5/5)[ba

I want to know how can I know a and b.

 

[Deleted member 4993]

Re: Please Help me as fast as possible by checking my answer

Find the length of the loop of the given curve.
x= 6t-2t^3
y=6t^2
My answer is
I Differentiate both:
x '(t) = 6 - 6t2
y '(t) = 12t

Square and add both
(6-6t2)(6-6t2) + 144t2 = 36 -72t2+36t4 + 144 t2= 36 + 72t2+ 36t4= 36(1+2t2+t4)

Integrate:
36(t + 2t3/3 + t5/5)[ba

I want to know how can I know a and b.

by setting x = y and solving for 't'.<<< This is incorrect - Arthur below has described the correct procedure.

 

[fasteddie65]

Re: Please Help me as fast as possible by checking my answer

Looking at the graph, the curve appears to intersect to form a loop on the y-axis, near the origin and near y = 15 on the y-axis. Let's set x = 0 and solve for t.

6t - 2t^3 = 0
2t(3 - t^2) = 0
2t = 0 or 3 - t^2 = 0
t = 0 3 = t^2
±?3 = t

I would use t = 0 and t = ?3, and double the result, because of symmetry.

 

[soroban]

Re: Please Help me as fast as possible by checking my answer

\(\displaystyle \text{Hello, XBOX999!}\)


\(\displaystyle \text{You didn't complete the formula . . .}\)

. . \(\displaystyle L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt\)

\(\displaystyle \text{Find the length of the loop of the curve: }\;\begin{array}{ccc} x&=& 6t-2t^3 \\y&=&6t^2\end{array}\)


\(\displaystyle \text{My answer is:}\)

\(\displaystyle \text{Differentiate both: }\;\begin{array}{ccc}x't)&=&6 - 6t^2 \\ y'(t)&=&12t\end{array}\)

\(\displaystyle \text{Square and add:}\)

. . \(\displaystyle (6-6t^2)^2+ 144t^2 \:=\; 36 -72t^2+36t^4 + 144t^2\;=\; 36 + 72t^2+ 36t^4\;=\; 36(1+2t^2+t^4)\)

\(\displaystyle \text{We want the }square\;root\text{ of that: }\;\sqrt{36(1 + 2t^2 + t^4)}\;=\;\sqrt{36(1+t^2)^2} \;=\;6(1 + t^2)\)

\(\displaystyle \text{Then we integrate: }\;L \;=\;6\int^b_a(1 + t^2)\,dt\)


\(\displaystyle \text{Now we must determine }a\text{ and }b.\)


. . . \(\displaystyle \sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\qquad\sim\)


\(\displaystyle \text{For two distinct values of }t,\;x\text{ and }y\text{ will be equal.}\)

\(\displaystyle \text{Let the two values be }a\text{ and }b.\)

\(\displaystyle \text{Then: }\;\begin{array}{cccccc}x(a) \:=\:x(b)\!:& 6a - 2a^3 \:=\: 6b - 2b^3 & [1] \\ y(a) \:=\:y(b)\!: & 6a^2 \:=\:6b^2 & [2]\end{array}\)



\(\displaystyle \text{From [1]: }\;6a-6b \:=\:2a^3 - 2b^3 \quad\Rightarrow\quad 6(a-b)\:=\:2(a^3-b^3)\)

. . . . . \(\displaystyle 2(a-b) \;=\;2(a-b)(a^2+ab+b^2)\)

\(\displaystyle \text{Since }a \neq b,\;(a-b) \neq 0\text{, and we have: }\;a^2+ab+b^2 \:=\:3\;\;[3]\)



\(\displaystyle \text{From [2]: }\;6(a^2-b^2) \:=\:0 \quad\Rightarrow\quad 6(a-b)(a+b) \:=\:0\)

\(\displaystyle \text{Since }(a-b) \neq 0\text{, we have: }\;a+b \:=\:\quad\Rightarrow\quad b \:=\:\text{-}a\)



\(\displaystyle \text{Substitute into [3]: }\;a^2 + a(\text{-}a) + (\text{-}a)^2 \:=\:3 \quad\Rightarrow\quad a^2 \:=\:3 \quad\Rightarrow\quad a \:=\:\pm\sqrt{3}\)

. . \(\displaystyle \text{Hence: }\;b \:=\:\mp\sqrt{3}\)

\(\displaystyle \text{And we have found our limits!}\)


\(\displaystyle \text{Therefore: }\;L \;=\;6\int^{\sqrt{3}}_{\text{-}\sqrt{3}}(1+t^2)\,dt\)