Question

what is the curvature of r=5sin(2θ)?

Answer 1

this sounds sooo familiar ....

Answer 2

|'x''|/|cube'|

Answer 3

calc 3, has to do with finding double derivative or something

Answer 4

armistre64, thats what I was trying, but idk what r' and r" are

Answer 5

*first and second derivative

Answer 6

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Answer 7

really? derivative of a sin function is prettu standard

Answer 8

Hi everybody, can I manipulate right on r or I have to transfer to x, and y?

Answer 9

\[k=\frac{|r'xr''|||}{|r'|^3}\]

Answer 10

yes, but it's not r' and r" its r(t)' and r(t)"

Answer 11

is r even function of t?

Answer 12

lol, my mistake, i assumed that a single variable was implied

Answer 13

not its not, I'm not sure how to make r into a vector r(t)

Answer 14

r = 5 sin(2t)

r' = 10 cos(2t)

r'' = -20 sin(2t)

Answer 15

hmmm, i see a function format as well

Answer 16

I was trying to use x=rcostheta and y=rsintheta to get
<5sin(2θ)cos(θ), 5sin(2θ)sin(θ), 0 > as my vector

Answer 17

r = 5 sin(2t)

sin (2t) = 2 sin(t) cos(t)

r(t) = 10 x(t) y(t)

Answer 18

\[r = \sqrt{x^2+y^2}\]
\[\sqrt{x^2+y^2} = 10xy\]

Answer 19

I see where you're going with this

Answer 20

so x2+y2 = 100x2y2

Answer 21

yep

Answer 22

so then that equals r, then what do I use as my vector?

Answer 23

oh, I differentiate r = 100x2y2

Answer 24

r = 10xy, r^2 = 100x^2y^2 :)

lets stick with r=10xy and do partials is what im thinking, right? its been awhile

Answer 25

I will have to give that a try, thank you!

Answer 26

ill review on a few sites to clear some cobwebs

Answer 27

thank you!

Answer 28

when r is defined as a function, it looks like\[k=\frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}\]

Answer 29

\[k=\frac{|-20sin(2t)|}{(1+(10cos(2t))^2)^{3/2}}\]

Answer 30

i just remembered how to parametrize this stuff :)

x = r cos(t) = 5sin(2t)cos(t) = 5sin(t) cos^2(t)
y = r sin(t) = 5sin(2t)sin(t) = 5 sin^2(t) cos(t)

Answer 31

therefore;

r = 5sin(t) cos^2(t) [i] + 5 sin^2(t) cos(t) [j]

r' = a little bit of work .... lol

Answer 32

for X hos did you get a cos&2(t)?

Answer 33

and why does the y component have a cos in it?

Answer 34

oh! nvm I see, you used the double angle ID for sin, thanks for the help!

Answer 35

youre welcome ... and if its easier to leave it as 5 sin(2t) then by all means work it ;)