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

the unit tangent vector T at t=4 is T₀ = ( 1/144 , 1/180 , 1)

Step-by-step explanation:

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

then the tangent vector will be r'(t) =dr/dt

r'(t) =dr/dt = 2*t-3 , 4 , 39*t² + 24*t

denoting r₀ , r'₀ as the vectors at t=4

r'₀=r'(4)= (2*4-3 , 4 , 39*4² + 24*4) = (5,4, 720)

the modulus of the vector r'₀ will be

|r'₀| = √(5²+4²+720²) = 720.028 ≈ 720

then the tangent vector will be

T₀ = r'₀/|r'₀| =(1/720) * (5,4, 720) = ( 1/144 , 1/180 , 1)

T₀ = ( 1/144 , 1/180 , 1)

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answer: an 8-foot rope is tied from the top of a pole to a

stake in the ground, as shown in the diagram

below

step-by-step explanation: