Question

Find the curvature of r(t)=2ti+2tj+k.

Answer 1

@wio

Answer 2

The curvature is calculated for every point with:
\[k=\left| \frac{f'(t)\times f''(t)}{(f'(t))^3}\right|\]

Answer 3

where f(t) is your r(t)

Answer 4

I know that r'(t)=2i+2j and r"(t)=0.

Answer 5

correct

Answer 6

So how to do r'(t)xr"(t) and (r'(t))^3?

Answer 7

(r'(t))^3=(2i+2j)^3?

Answer 8

i think i should have written the equation more precisely, that should be:
\[k=\frac{\left|f'(t)\times f''(t)\right|}{\left|f'(t)\right|^3}\]

Answer 9

got it now?

Answer 10

wait im confused

Answer 11

No. How to do abs(r'(t)xr"(t)) and abs((r'(t))^3)?

Answer 12

@iambatman @wio

Answer 13

if you have a vactor r(t) = ai + bj + ck, then
\(\left| r(t) \right| = \sqrt{a^2 + b^2 + c^2}\)

Answer 14

Okay~, I got it. Thanks for the help.

Answer 15

for r'(t) x r"(t) , use determinant rule to find it out, then, take norm of the vector product

Answer 16

the same thing with the denominator, that is you take the norm of r'(t), it is |r'(t)| then ^3 it

Answer 17

What's the determinant rule?

Answer 18

the x in r'(t) x r"(t) is cross product, right? how to do it? using the way you calculate determinant, right?

Answer 19

ok, r(t) = <2t, 2t, k>
r'(t) = <2, 2, 0>
r"(t) =<0,0,0>
hence your numerator is <2,2,0> x <0,0,0> =<0,0,0>
no need to find denominator, the curvature =0

Answer 20

Thank you, @Loser66.