Question

Curvature of the helix r(t)=a cost i +a sint j +bt k (a,b>= 0) and curvature K= a/(a^2 +b^2) . What is the largest value of K for a given value of b

Answer 1

\[k=\frac{|r'xr''|}{|r'|}\]
or am i missing a ^2 underneath?

Answer 2

^3 in the denom

Answer 3

r(t)=<a cost, sint,bt >

r'(t)=<-a sint, cost,b >

r''(t)=<-a cost, -sint, 0>

Answer 4

for curvature we are using the formula K= 1/IVI * ldT/dtl

Answer 5

r' <-a sint, cost,b >
xr''<-a cost, -sint, 0>
--------------------
x = -b sint
-y = (-ab cost)
z = a (sin^2t + cos^2t)

r'xr'' = <-b sint, ab cost, a>

Answer 6

yeah, I believe that simplifies to this |r'xr''|/|r'|^3

Answer 7

http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx

Answer 8

ok

Answer 9

Don't you just have to maximize the curvature they give already by deriving with respect to b? Because from what I can tell, finding the curvature is going to give you just what is already given.

Answer 10

|r'xr''| = sqrt( b^2 sin^2t + (ab)^2 cos^2t + a^2)

Answer 11

yes i am trying to find the maximum value of K with respect to b @ dominusscholae

Answer 12

i gotta get to class, so yall have fun with it ;)

Answer 13

thank amistre

Answer 14

thanks*

Answer 15

dominusscholae can you help me with this question

Answer 16

Yeah. From what I can tell, you just derive K with respect to b to find the critical point. We are just assuming a is constant so that we find the correct b value at which K is maximum.