Find a tangent vector of unit length at the point with the given value of the parameter t?

r(t) = 2 sin(t)i + 3 cos(t)j

Plug pi/6 into the derivative or r(t)
2cos(pi/6); – 3sin(pi/6)
To find the magnitude, use the Pythagorean theorem with the results from plugging pi/6 into r'(t).
sqrt((sqrt(3))^2 + (3/2)^2) = sqrt(3+(9/4))
Find a common denominator (should be 4) and solve from there.
sqrt( 12/4 + 9/4) = sqrt(21)/2
Then, simply place the sqrt part of it in the denominator and multiply the constant in front of the derivative functions
2*sqrt(3)/sqrt(21)i -3/sqrt(21)j

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