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

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~