Question

Find a tangent vector of unit length at the point with the given value of the parameter t. r(t) = (7 + t^2)i + t^2j t = 1

Answer 1

First find the derivative at the given point and later divide by the magnitude of this derivative.

Derivative: r' (t) = 2t i+ 2j t

Magnitude:


`\sqrt{ (2t)^(2) + (2t)^(2) } = √(8t^2) =2 √(2) t`

Now use the value t =1

Derivative: 2(1)i + 2(1)j = 2i + 2j
Magnitude: 2√2

Unit vector: [2i + 2j] /[2√2] = (√2)/2 i + (√2)/2 j