what is the curvature of r=5sin(2θ)?

this sounds sooo familiar ....

|'x''|/|cube'|

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

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

*first and second derivative

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really? derivative of a sin function is prettu standard

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

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

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

is r even function of t?

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

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

r = 5 sin(2t) r' = 10 cos(2t) r'' = -20 sin(2t)

hmmm, i see a function format as well

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

r = 5 sin(2t) sin (2t) = 2 sin(t) cos(t) r(t) = 10 x(t) y(t)

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

I see where you're going with this

so x2+y2 = 100x2y2

yep

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

oh, I differentiate r = 100x2y2

r = 10xy, r^2 = 100x^2y^2 :) lets stick with r=10xy and do partials is what im thinking, right? its been awhile

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

ill review on a few sites to clear some cobwebs

thank you!

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

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

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)

therefore; r = 5sin(t) cos^2(t) [i] + 5 sin^2(t) cos(t) [j] r' = a little bit of work .... lol

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

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

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

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

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