Find a tangent vector of unit length at the point with the given value of the parameter t. r(t) = 2 sin(t)i + 7 cos(t)j t = π/6

Applying derivatives, the tangent vector of unit length at the point given is:

The vector function is:

The tangent vector is it's derivative, which is given by:

At point , we have that:

The norm of the vector is:

The unit vector is given by each component divided by the norm, thus:

A similar problem is given at link

The tangent vector to r(t) at any t in the domain is

At t = π/6, the tanget vector is

To get the unit tangent, normalize this vector by dividing it by its magnitude:

So the unit tangent at the given point is

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i