Question

find the tangent line of the curve at the points t=+1,-1,0
r(s)=(s^2-1,s(s^2-1),0)

Answer 1

Answer on Question #37128 - Math - Differential Geometry

Question.

Find the tangent line of the curve at the points $s = +1, -1, 0$

$$r(s) = (s^2 - 1, s(s^2 - 1), 0)$$

Solution.

The tangent line of the curve

$$x = x(s)$$ $$y = y(s)$$ $$z = z(s)$$

in the point $s_0$ is the line

$$\frac{x - x_{0i}}{x_{0i}'} = \frac{y - y_{0i}}{y_{0i}'} = \frac{z - z_{0i}}{z_{0i}'}, \quad i = 1, 2, 3$$

In our case, we have:

$$x(s) = s^2 - 1$$ $$y(s) = s(s^2 - 1)$$ $$z(s) = 0$$ $$x_{01} = s^2 - 1_{s=1} = 0, \quad y_{01} = 0, \quad z_{01} = 0$$ $$x_{02} = s^2 - 1_{s=-1} = 0, \quad y_{02} = 0, \quad z_{02} = 0$$ $$x_{03} = s^2 - 1_{s=0} = -1, \quad y_{03} = 0, \quad z_{03} = 0$$ $$x'(s) = 2s, \quad y'(s) = 3s^2 - 1, \quad z'(s) = 0$$ $$x'_{01} = 2s_{s=1} = 2, \quad y'_{01} = 2, \quad z'_{01} = 0$$ $$x'_{02} = 2s_{s=-1} = -2, \quad y'_{02} = 2, \quad z'_{02} = 0$$ $$x'_{03} = 2s_{s=0} = 0, \quad y'_{03} = -1, \quad z'_{03} = 0$$

So, the tangent lines:

$s_{01} = 1$:

$$\frac{x - 0}{2} = \frac{y - 0}{2} = \frac{z - 0}{0}$$

$s_{02} = -1$:

$$\frac{x - 0}{-2} = \frac{y - 0}{2} = \frac{z - 0}{0}$$

$s_{03} = 0$:

$$\frac{x + 1}{0} = \frac{y - 0}{-1} = \frac{z - 0}{0}$$