Question

Anybody know how to maximize the curvature of a helix

Answer 1

r(t)=(acost)i+(asint)j+btk (a,b>or = to 0) to be k=(a)/(a^2+b^2)

Answer 2

what is the largest value "k" can have for a given value of b

Answer 3

should i get get the first derivative of that kappa or k function

Answer 4

I am sorry to bother you Zarkon, but you were such a great help with the last question, can you help me here @Zarkon

Answer 5

lol i was about to call Zarkon too haha

Answer 6

yeah, i got no idea on this one

Answer 7

you found k=(a)/(a^2+b^2) to be your curvature?

Answer 8

yes that is the curvature of the helix r(t)

Answer 9

ok...I agree...so do you want to maximize this function of a,b?

Answer 10

trying to make sure I understand what you want

Answer 11

i guess we would have to maximize it as a function of b, right? given that they as us to for the largest value k can hav efor a given value of b....not really sure

Answer 12

that is easy then


let b be zero. that will make the fraction as large as possible.

Answer 13

here is the full question, when you back sir, its #19

Answer 14

the answer in the book is 1/2b

Answer 15

1/(2b)

Answer 16

I understand...you are fixing a value of b and you are maximizing over a...then you will get a=b and thus k=1/(2b)

Answer 17

do you understand?

Answer 18

can you explain it in another way, i dont understand

Answer 19

treat \(k\) as a function of \(a\)

then \[k(a)=\frac{a}{a^2+b^2}\]
find \[k'(a)\]

and solve \[k'(a)=0\text{ for }a\]

Answer 20

ahh, i understand now. Thank you

Answer 21

reply if you have any problems

Answer 22

do i use the quotient rule for this

Answer 23

@myininaya

Answer 24

yes

Answer 25

b is a constat though right, so its derivaitve goes to 0, while the derivative of a goes to 1

Answer 26

yes

Answer 27

\[k'(a)=\frac{(a^2+b^2)-a(2a)}{(a^2+b^2)^2}=0\]

\[\Rightarrow (a^2+b^2)-a(2a)=0\]

\[\Rightarrow (a^2+b^2)-2a^2=0\]

\[\Rightarrow b^2-a^2=0\]

\[\Rightarrow a^2=b^2\]

\[\Rightarrow a=\pm b\]

but we know that \(a,b\ge 0\)

so \(a=b\)

Answer 28

so replace a with b

\[k(a)=\frac{a}{a^2+b^2}\]

\[k(b)=\frac{b}{b^2+b^2}=\frac{b}{2b^2}=\frac{1}{2b}\]

Answer 29

Thank you so much for woking it out for me.

Answer 30

no problem...I hope you understand.

Answer 31

at some point you should check that you indeed found the max...using the first or second derivative test.

Answer 32

basically i beleive we are trying to find the maximum curvature "k" we can obtain for some value of "b". And inorder to do this we have to use the first derivative test to determine at what xvalue (in this case a value) that maximum occurs. Then we plug that value back into the original curvature function inorder to find that maximum value