Question

Using r(t) = acos(t)i+asin(t)j, T(t) = (r' (t))/(||r' (t)||) and curvature is k(t) = (||T' (t) ||)/(||r' (t) ||)
Prove that the curvature of a circle is constant and = 1/a

Answer 1

So you just need to calculate.
I.e., given r, what is r', ||r'||, T and k ?

Just do the calculations; it's very straightforward.

Answer 2

Really? Once I do the calculations you think i'll just see the proof?

Answer 3

Just try it, really. Muster up your courage and do it.

Answer 4

haha, ok, i shall return!

Answer 5

Let's break this down. First, what's r' , ||r'|| and therefore T?

Answer 6

I'm nervous you're typing for so long. Give it to me in pieces. What first is r' and ||r'||?

Answer 7

even more nervous now.

Answer 8

\[r(t)=acos(t)i+asin(t)j\]
\[r'(t)=(a'\cos(t)-asin(t))i+(a'\sin(t)+acos(t))j\]
\[||r'(t)||=\sqrt{(a'cost-asint)^{2}+(a'sint+acost)^{2}} \]
\[=\sqrt{(a'\cos)^{2}-2a'acostsint+(asint)^{2}+(a'sint)^{2}+2a'acostsint+(acost)^{2}}\]
\[=\sqrt{(a'\cos)^{2}+(asint)^{2}+(a'sint)^{2}+(acost)^{2}}\]
\[=\sqrt{a'^{2}\cos^{2}t+a^{2}\sin^{2}t+a'^{2}\sin^{2}t+a^{2}\cos^{2}t}\]

Answer 9

lol, oops just saw your messages

Answer 10

sorry

Answer 11

a is a constant.

Answer 12

I believe T is called the tangent unit

Answer 13

a is a consant, ok

Answer 14

Am I on the right path though?

Answer 15

ah ha!

Answer 16

one sec on the phone. remember that a' =0 everywhere

Answer 17

\[r'(t) = -asin(t)i+acos(t)j\]

Answer 18

yep

Answer 19

so then after calculating ||r'(t)|| I end up with just a, want me to show that work?

Answer 20

no. that's right: ||r'|| = a

Answer 21

Hence T = r' / ||r'|| = ...?

Answer 22

-sin(t)i+cos(t)j

Answer 23

yes

Answer 24

hence T' = ....

Answer 25

ahhh

Answer 26

1/a

Answer 27

You're a genious

Answer 28

T' doesn't = 1/a

Answer 29

But I see how (||T' (t) ||)/(||r' (t) ||) will eventually = 1/a now

Answer 30

T' = -cos t i - sin t j ==> ||T'|| = 1. Therefore

k = ||T'||/||r'|| = 1/a

Answer 31

Right

Answer 32

Thanks so much

Answer 33

Are you a math major?

Answer 34

I was, yes.