Question

Find the curve the length of the curve define by r= <2t^3/2, cos 2t, sin 2t >, 0= 13 years ago

Answer 1

differentiate the r with respect to t let me know what you get .

r'(t) = ???

Answer 2

i think is r' = <1/2 t^1/2, -2sin 2t, 2 cos 2t>?

Answer 3

check it again i think first term shuld be 3t^1/2

Answer 4

yes is 3 ^1/2 cuz 2 3/2 t^ 3/2 -1

Answer 5

now magnitude?

Answer 6

ok good
take mangitude of r'(t) .

Answer 7

sqrt(3 t^1/2)^2+(2sin 2t)^2+ (2 cos 2t)^2
(9 t^1/4)+4sin^2 2t+ 4cos^2 2t

Answer 8

this part is were im stuck

Answer 9

so far i know that sin^2 + cos^2=1 but 4sin^2 2t+ 4cos^2 2t

Answer 10

still cos^2(2t)+sin^2(2t) =1 just take out 4 as common !

Answer 11

ok i see @sami-21
4 (cos^2(2t)+sin^2(2t))

Answer 12

yes .

Answer 13

sqrt (9 t^1/4)+4= 3 t^1/2+2

Answer 14

check it again !!! you cannot take square root like that .

Answer 15

k

Answer 16

is 3 t^1/8 + 2?

Answer 17

\[\Large |r'(t)|= \sqrt{(3t^\frac{1}{2})^2+(2\sin(2t)^2+(2\cos(2t)^2}\]

simplify
\[\Large |r'(t)|=\sqrt{9t+4}\]

Answer 18

what happen to the 1/2?

Answer 19

it is squared !

Answer 20

one minute 1/2 is the same as sqrt

Answer 21

yes it is same .

Answer 22

got it
so | r'(t)| =3t+4

Answer 23

there is sqrt as well :)

Answer 24

now integrate |r'(t)| fom 0..1

Answer 25

\[\Large Length=\int\limits_{0}^{1} \sqrt{3t+4}dt\]

Answer 26

is \[\sqrt{3t+2}?\]

Answer 27

cuz 4 is perfect squared

Answer 28

sorry i did mistake .

Answer 29

because |r'(t)|= sqrt{9t+4}
\[\Large Length=\int\limits_{0}^{1}\sqrt{9t+4}dt\]
can you integrate this

Answer 30

chain rule?

Answer 31

yes

Answer 32

1/2 (9t +4)^-1/2 * 9= 9/ 2 sqrt (9t+4)

Answer 33

now how inegrate this