Question

Consider the vector function given below. r(t) = 3t, 5 cos(t), 5 sin(t) (a) find the unit tangent and unit normal vectors t(t) and n(t).'

Answer 1

Well, by definition, the tangent vector is the vector that is the result of differentiating every component of the function. Hence, the tangent vector is T=(3,-5sint,5cost). This has a magnitude of $|r|^2=3^2+(-5sint)^2+(5cost)^2$ which yields |r|=$\sqrt{34}$ (by using the known trigonometric formula $cost^2+sint^2=1$ that holds for all t). Thus, the unit tangent vector t is given by: t=$\frac{T}{\sqrt{34}}$. Now, to find a normal unit vector, we just need to find a normal vector N and scale it appropriately. To find a normal unit vector, we can just find vector that satisfies V*T=0 (dot product of vectors). By inspection, we see that the vector N=(0, cost, sint) satisfies this requirement and that it is also a unit vector (due to the aforementioned trigonometric identity). Hence, a normal unit vector to our vector function is N.