Question

Compute the arc lenth for the curve traced out by the vector valued function r(t)= 5sin^3 (t), 3cos^3 (t), t...

HELP!!!

Answer 1

@RolyPoly ...@TuringTest

Answer 2

i get..15sin^2 (t) cos(t) - 9 cost^2(t) sin (t) + 1

Answer 3

finding the norm gives me...225sin^4 (t) cos^2 (t) + 9cos^4(t) sin^2 (t) +1

Answer 4

and den i am blank.

Answer 5

\[L=\int_{a}^{b} |r'(t)|=\int _{a}^{b} \sqrt{(x')^2+(y')^2+(z')^2}dt\]
right?

Answer 6

5sin^3 (t), 3cos^3 (t), t...
\[\sqrt{15^2 cos^4 + (-9)^2sin^4 + 1}\]

Answer 7

wait wait..wheres cos^2 t and sin ^2 t ?

Answer 8

changing cos to sin or sin to cos might be useful

Answer 9

[15cos^2]^2 ....

Answer 10

the derivative of the parts are squared then added

Answer 11

15sin^2 (t) cos(t) - 9 cost^2(t) sin (t) + 1...dont we squre this ??

Answer 12

cos^2t+sin^2t =1...i understand..but how do i show that here ?

Answer 13

(cos^2 = 1-sin^2)^2

225(1- 2sin^2 + sin^4)

225- 450sin^2 + 225sin^4
+1 +81sin^4
-------------------------
224 -450sin^2 + 306sin^4

Answer 14

since your sin and cos part dont have equal coeffs in front of them, you cant factor them into the form cos^2+sin^2

Answer 15

i know, so isnt ||r't|| = 225sin^4 (t) cos^2 (t) + 9cos^4(t) sin^2 (t) +1

Answer 16

im not sure how your trying to get that

Answer 17

finding the norm of r't !

Answer 18

arclength is measured by tangents, not norms that im aware of

\[r'=<15^2 cos^4, (-9)^2sin^4, 1>\]

Answer 19

i forgot to unsquare the copy i did :)

Answer 20

sorry to bug you with lots of questions...but just simply..what is the differentiation of 5sin^3t ??

Answer 21

5*3 cos^2(t) ..... i see. sigh, getting old bites :)

Answer 22

fine, we can do it the proper way if you want :)

Answer 23

not a prob bro..happens..integration..sick stuff :(

Answer 24

oh no..this is not integration..messing my head up..finals tomrrow :(

Answer 25

r = 5sin^3 (t) , 3cos^3 (t) , t

r' = 15sin^2(t)cos(t) , -9cos^2(t)sin(t), 1

that should be better

Answer 26

indeed ! :D

Answer 27

\[225sin^4cos^2+81sin^2cos^4+1\]
im wondering if the polar method would be any easier

Answer 28

not sure..this is where i am stuck !

Answer 29

by changing the variables to suit a polar method, might make it simpler to calculate, cause at the moment it looks like a bear unless you can determine a suitable substitution

http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx

Answer 30

wolfram can't do the indefinite integral but here is arc length from 0 to 2pi

http://www.wolframalpha.com/input/?i=integrate+sqrt%28%2815sin%28t%29^2cos%28t%29%29^2+%2B+%28-9sin%28t%29cos%28t%29^2%29^2+%2B1%29+dt+from+0+to+2pi

Answer 31

insane....

Answer 32

http://phdmath.wordpress.com/2010/02/24/arc-length-2/

this at least has the formula for arclength in different formats

Answer 33

unless youve actually covered these other formats in class, it might not be a good idea to try them yet

Answer 34

i cant see i way to simplify it at the moment either ....

Answer 35

i can get it to\[\sqrt{sin^2(t)(12.5^2sin^2(2t)+1)+cos^2(t)(4.5^2sin^2(2t)+1)}\]
if it wasnt for those different coeffs, it would factor

Answer 36

the curve at least looks like this :)

Answer 37

@TuringTest i believe this is your mess somehow :)