Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 1 + 2√t, y = t3 - t, z = t3 + t; (3,0,2)

Solution :

Given parametric equation for :

The point is (3, 0, 2)

The vector equation is equal to :

Solving for r'(t) by differentiating each of the components of r(t) w.r.t. to t,

The parameter value corresponding to (3, 0, 2) is t = 1. Putting in t=1 into r'(t) to solve for r'(t), we get

We know that parametric equation for  line through the point  and parallel to the direction vector <a, b, c > are

Now substituting the     = (3, 0, 2) and  <a, b, c > into x, y and z, respectively to solve for the parametric equation of the tangent line to the curve, we get:

x  = 3 + t

y = (0) + (2)t

y = 2t

z = (2) + (4)t

z = 2 + 4t

a) justify

step-by-step explanation:

justifying is just part of everyday life. justify kind of means proof so i think the answer is a) justify.

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step-by-step explanation: