I have a parametric curve $$x=t-\sin(t)$$ $$y=1-\cos(t)$$ $$z=4\sin\left(\frac{t}{2}\right)$$ I need to find the Tangent and Normal on $$M_0\left(\frac{\pi}2-1;1;2\sqrt{2}\right)$$
I know the formula in $\mathbb{R^2}$ space: $$y=f'(x_0)(x-x_0)+f(x_0)$$ And $$y_x'=\frac{y_t'}{x_t'}$$ But how to work in $\mathbb{R^3}$ space?
$$t=\frac{\pi}2$$ $$r=<t-sin(t),1-cos(t),4sin(\frac t2)>$$ $$r'=<1-cos(t),sin(t),2cos(\frac t2)>$$ $$r'(\frac{\pi}2)=<1,1,0>$$ $$x=\frac{\pi}2-1+t$$ $$y=1+t$$ $$z=2\sqrt2$$ What to do with normal line equation?
Here's a hint: The equation of the tangent line can be given in parametric form
$$ \vec{r}(t) = (x_0 + x'(t_0)t, y_0 + y'(t_0)t, z_0 + z'(t_0)t) $$
where $M_0(x_0,y_0,z_0)$ is the given point on the curve and $(x'(t_0),y'(t_0),z'(t_0))$ is the tangent vector evaluated at that point.
For the normal, because you're working in 3D, the normal line will not be unique. Instead you have an equation for the normal plane that's given by
$$ (x-x_0,y-y_0,z-z_0)\cdot (x'(t_0),y'(t_0),z'(t_0)) = 0 $$