Find the unit tangent vector T(t). r(t) = t^2 i + tj + 16/3 k, p(16, 4, 16/3) T(4) = Find a set of parametric equations for the line tangent to the space curve at point P. (Enter your answers at a comma-separated list. Use t for the variable of parameterization.

Therefore the unit tangent vector is

Therefore the parametric form is

Step-by-step explanation:

Given

To find the unit tangent we need to find r'(t) and |r'(t)|.

The unit tangent vector

Differentiate with respect to t

Therefore the unit tangent vector is

.......(1)

The parametrization  equation

x(t) = t² , y(t)=t and z(t)

Given point is

Therefore,

  ,       and      

Then t =4.

The unit tangent vector at t= 4

[Putting t = 4 in equation (1)]

The parametric equation is

x= x₁+at

y =y₁+bt

z=z₁+ct

Here a=8 , b=1 and c=0

 ,     and  

Therefore the parametric form is

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)

jml angle lkj & angle mj = lm