Vector Valued Functions...find the curvature, K, of the plane curve at the given point.?

Question

r (t) = - 2t i + (7/t) j , t = 1

is it...
21, 25, 28, 31, or 36 over / (53 sqrt(53)) ???

Answer 1

r(t) = (-2t, 7/t)

The tangent vector is given by:

r'(t) = (-2, -7/t²)

And the unit tangent vector is given by:

T(t) = r'(t)/||r'(t)|| = (-2, -7/t²)/√(4+49/t⁴) = (-2/√(4+49/t⁴), -7/√(4t⁴+49))

The derivative of the unit tangent is:

T'(t) = ((4+49/t⁴)^(-3/2) * (-196t⁻⁵), 7/2 (4t⁴+49)^(-3/2) * (16t³))

The curvature is ||T'(t)||/||r'(t)||. So plugging this in:

T'(1) = ((4+49)^(-3/2) * (-196), 7/2 (4+49)^(-3/2) * 16) = (-196/(53√53), 56/(53√53))

||T'(1)|| = 1/(53√53) √((-196)² + 56²) = √41552/(53√53) = √784√53/(53√53) = 28/53

r'(1) = (-2, -7)
||r'(1)|| = √(4+49) = √53

So the curvature is 28/(53√53) -- which you will note is not on your list. Check to see that it is correct.