Question

Calculus. find the length of the curve. details inside

Answer 1

use \[L= \int\limits_{a}^{b}\sqrt{r^2+(\frac{ dr }{ d \theta })} d \theta\]
when\[r= 3 \sin \theta\]\[0\le \theta \le \frac{ \pi }{ 4 }\]

Answer 2

my work =\[L= \int\limits_{0}^{\frac{ \pi }{ 4 }}\sqrt{(9\sin^2 \theta)-(9\cos^2 \theta)}d \theta\]

Answer 3

typ0** the (dr/ d theta) should be squared as well.

Answer 4

\[L= \int\limits_{0}^{\frac{ \pi }{ 4 }}\sqrt{9 (\sin^2 \theta-\cos^2\theta)}d \theta\]\[L= 3\int\limits_{0}^{\frac{ \pi }{ 4 }}\sin^2 \theta- \cos^2 \theta\]

Answer 5

this is where i get lost. does sin^2 theta - cos^2 theta = -1?

Answer 6

no it doesn't ... and you should also take a look at the formula you applied originally.

it is helpful to know where the formula comes from - Pythagoreas, basically - and that might prevent it being misapplied right at the start.


if you want to work this through, especially the derivation of the arc length formula, please shout out.

good luck!

Answer 7

arc length in polar is given by
\[s=\int\sqrt{r^2+(\frac{dr}{d\theta })^2}d\theta \]

Answer 8

this is what i keep getting ...\[3\int\limits_{0}^{\frac{ \pi }{ 4 }}\theta* d theta\]=\[3[(\frac{ \pi }{ 4 })-0)= \frac{ 3\pi }{ 4 }\]

Answer 9

hi x! yea, that was what i meant by i had a typo above

Answer 10

why negative?

Answer 11

you put negative for what reason?

Answer 12

should be 9sin^2theta+9cos^2theta=9

Answer 13

oh.. i guess I caught myself on that one... made it plus and then integrated it into a theta

Answer 14

ok so you got that right where did you stuck?

Answer 15

did you do it
i got \[\frac{3\pi}{4}\]

Answer 16

im sorry, i got kicked off! thank you so much! i was just wondering if i was right, thats all