Find the curvature of the curve?

Question

Find the curvature of the curve r(t)=(t+1)i+2(t^2-1)j+(t-2)k.

r'(t)=i+4tj+k

sqrt(1^2+(4t)^2+1^2)=sqrt(2+16t^2)

r''(t)=4j

r'(t)xr''(t)=?

Please show your work step by step from here to get the answer.

Answer 1

One way you can do this is to use the formula κ = Norm(v×a) / Norm(v)³.

You already have part of it established, the fact that v(t) = r'(t) = i + 4t j + k and a(t) = r''(t) = 4j.

For the cross product, wee just need to set up the determinant.

v(t) × a(t)

   |   i      j      k    |

= |   1    4t     1    |

   |   0     4      0   |

= (4t*0 - 1*4) i + (1*0 - 1*0) j + (1*4 - 4t*0) k

= -4 i + 4 k

Thus Norm(v×a) = 4√2.

Also, Norm(v) = √(1² + (4t)² + 1²) = √(2 +16t²)

Thus the curvature is

κ(t) = Norm(v×a) / Norm(v)³ = (4√2) / (2+16t²)^3/2 = (4√2) / [(2√2)(1+8t²)^3/2] = 2 / (1+8t²)^3/2.