Question

Find the curvature κ(t) of the curve X (t) = ( −4 sint )I + ( −4 sint ) j + ( 3 cost ) K

Answer 1

I found a similar question and its answer....
QUES:Find the curvature κ(t ) of the curve X(t ) =
(3 sin t ) I + (3 sin t ) j + (−5 cos t ) K
ANSWER:
The curvature of space curve in parametrical form like:
X(t) = x(t) · i + y(t) · j + z(t) · k
is given by

κ(t ) = sqrt[ (z''·y' - y''·z')² + (x''·y' - y''·x')² + (x''·z' - z''·x')² ] / (x'² + y'² + z'²) ^(3/2)

( ' denotes derivative with respect to t)

x(t)= 3·sint
→
x'(t) = 3·cost
x''(t) = - 3·sint

y(t)= 3·sint
→
y'(t) = 3·cost
y''(t) = - 3·sint

z(t)= -5·cost
→
z'(t) = 5·sint
z''(t) = 5·cost

The curvature is:

κ(t ) = sqrt[ (5·cost·3·cost + 3·sint·5·sint)² + (-3·sint·3·cost + 3·sint·3·cost)² +
(- 3·sint·5·sint - 5·cost·3·cost)² ] / (9·cos²t + 9·cos²t + 25·sin²t) ^(3/2)
→
κ(t ) = sqrt[ (15·cos²t + 15·sin²t)² + (0)² + (- 15·sin²t· - 15·cos²t)² ] / (18·cos²t + 25·sin²t) ^(3/2)

With the Pythagorean identity
sin²t + cos²t =1 ←→ cos²t = 1- sin²t
this simplifies to

κ(t ) = sqrt[ (15)² + (- 15)² ] / (18 ·(1 - sin²t) + 25·sin²t) ^(3/2)
= 15 · sqrt(2) / (18 + 7·sin²t)^(3/2)

Answer 2

it said its wrong