Find the unit tangent vector t(t) at the point with the given value of the parameter t. r(t) = t2 − 3t, 1 + 4t, 1 3 t3 + 1 2 t2 , t = 4.

T(t) = [1/144 , 1/180 , 1]

Step-by-step explanation:

if the parametric curve is

r(t) = t² − 3t, 1 + 4t, 13*t³ + 12* t²

then r'(t) = [ x'(t) , y'(t) , z'(t) ]

therefore

r'(t) = 2t - 3, 4 , 39* t² + 24t

at t=4

r'(4) = 2*4 -3 , 4 , 39*4² + 24*4 = 5 , 4 , 720

also the modulus of the r'(4) vector is

| r'(4) | = √(5²+4²+720²) ≈ 720

thus

T(t) = r'(4) / | r'(4) | = 5/720 , 4/720 , 720/720 = 1/144 , 1/180 , 1

T(t) = 1/144 , 1/180 , 1

step-by-step explanation :

we know that

the measurement of the outer angle is the semi-difference of the arcs which comprises

in this problem

m< xzy is the outer angle

substitute the values

the measure of angle xoy is equal to

> by central angle

substitute the value