Question

The curve C is defined by r(t)=<2t, 3sint, 3cost>
Find the length of this curve for -4<=t<=4

Answer 1

\[\int ds\]
sounds familiar

Answer 2

I started some work on this but the answer I was getting was quite messy so i was doubting if it was correct.

\[\int\limits_{-4}^{4} \sqrt{4+9\cos^2-9\sin^2}\]

\[= \int\limits_{-4}^{4} \sqrt{4+9(1-2\sin^2t)}\]

\[ = \int\limits_{-4}^{4} \sqrt{4+9-18\sin^2t}\]

\[= \sqrt{13-18 \sin^2t}\]

Answer 3

would cos(2t) work any better than (1-2sin^2) ?

Answer 4

scratch that last one missed the integral.

so it would be
\[\int\limits_{-4}^{4} \sqrt{13-18\sin^2t}\]
\[=\frac{ 2 }{ 3 } (13-18\sin^2t)^\frac{ 3 }{ 2 }(13t-18(\frac{ 1 }{ 2 }t-\frac{ 1 }{ 4 } \sin2t))\]
then evaluated that last one from -4 to 4

quite messy

Answer 5

possibly ill retry

Answer 6

site went out for me ....

Answer 7

x'^2 + y'^2 + z'^2

there shouldnt be any negative, or subtractions

Answer 8

ok so carrying through where that changes

\[=\int\limits_{-4}^{4} \sqrt{4+9\cos2t}\]
\[=\int\limits_{-4}^{4} ( 4+9\cos2t)^\frac{ 1 }{ 2 }\]
\[=\frac{ 2 }{ 3 }(4+9\cos2t)^\frac{ 3 }{ 2 }(4t+\frac{ 9 }{ 2 }\sin2t)\]
evaluated form -4 to 4

Answer 9

does that look like it would be a correct answer or is something still off prior to evaluating?

Answer 10

\[\sqrt{[(2t)']^2+[(3sin(t))']^2+[(3cos(t))']^2}\]
\[\sqrt{[2]^2+[3cos(t)]^2+[-3sin(t)]^2}\]
\[\sqrt{4+9cos^2(t)+9sin^2(t)}=\sqrt{4+9(cos^2+sin^2)}=\sqrt{13}\]

Answer 11

oh i see ok i had just found where that was in the work.
thanks that is alot better. :)

Answer 12

:) good luck