Respuesta :
Solution :
Given parametric equation for :
[tex]$x=1+2 \sqrt t$[/tex]
[tex]$y=t^3-t$[/tex]
[tex]z=t^3+t[/tex]
The point is (3, 0, 2)
The vector equation is equal to :
[tex]$r(t) = \left<1+2 \sqrt t, t^3 -t, t^3+t \right>$[/tex]
Solving for r'(t) by differentiating each of the components of r(t) w.r.t. to t,
[tex]$r'(t)= \left< \frac{1}{\sqrt t}, \ 3t^2-1, \ 3t^2+1 \right>$[/tex]
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
[tex]$r'(1) = \left< \frac{1}{\sqrt 1}, \ 3(1)^2-1, \ 3(1)^2+1 \right>$[/tex]
We know that parametric equation for line through the point [tex]$(x_0, y_0, z_0)$[/tex] and parallel to the direction vector <a, b, c > are
[tex]$x=x_0+at$[/tex]
[tex]$y=y_0+bt$[/tex]
[tex]z=z_0+ct[/tex]
Now substituting the [tex]$(x_0, y_0, z_0)$[/tex] = (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:
[tex]$x=3+(1)t$[/tex]
x = 3 + t
y = (0) + (2)t
y = 2t
z = (2) + (4)t
z = 2 + 4t
