How can i find the tangent equation of a curve

Question

I am so confusing about get the tangent equation for a particular curve, I read many topics about tangent equation but i got many questions to need to answers I read that to get line equation i have to follow that 1- https://i.stack.imgur.com/cwibt.png 2- https://i.stack.imgur.com/Q0raq.png So how can i use the same approach to get tangent for specific curve equation like helix and so on ... i don't want equation or example ... i just want to understand by graph how can i get it .. For example tangent equation for K(t)=(sin3t cost, sin3t sint, 0) at point pi/3 Thanks

Answer 1

Suppose you have a vector curve of the form

$$ K(t)=(x(t),y(t),z(t)) $$

and you want the vector equation of the tangent line $L(t)$ to $K$ at $t=t_0$.

This will be the line which must satisfy the conditions

1. $T(t_0)=K(t_0)$
2. $T^\prime(t_0)=K^\prime(t_0)$

The first condition is that the point of tangency must belong to both the curve $K$ and the line $L$.

The second condition is that the direction vector of the curve $K$ at $t_0$ must be parallel to the direction vector of the line $T$.

Since the direction of a line is constant, it should be parallel to the direction vector of $K$ at $t_0$. The simplest way for this to happen is for the direction vector of $T$ to equal the direction $K^\prime(t_0)$ of $K$ at $t_0$.

So the tangent line $L$ at $t_0$ has vector equation

$$T(t)=K(t_0)+K^\prime(t_0)(t-t_0)\tag{1}$$

For example, let $K(t)=(t\cos(t),\sin(t),t)$. Find the equation of the tangent line to curve $K$ when $t=\frac{\pi}{2}$.

1. $K\left(\frac{\pi}{2}\right)=\left(0,1,\frac{\pi}{2}\right)$
2. $K^\prime\left(t\right)=(\cos(t)-t\sin(t),\cos(t),1)$ so $K^\prime\left(\frac{\pi}{2}\right)=\left(-\frac{\pi}{2},0,1\right)$

Using equation $(1)$ we find that the equation of the tangent line is

\begin{eqnarray} T(t)&=&\left(0,1,\frac{\pi}{2}\right)+\left(-\frac{\pi}{2},0,1\right)\left(t-\frac{\pi}{2}\right)\\ &=&\left(-\frac{\pi}{2}\left(t-\frac{\pi}{2}\right),1,t\right) \end{eqnarray}

Alternately, it can also be written in the form

$$T(t)=\left(\left(\frac{\pi}{2}\right)^2,1,0\right)+\left(-\frac{\pi}{2},0,1\right)t $$ To check that this is correct we see that

1. $T\left(\frac{\pi}{2}\right)=\left(0,1,\frac{\pi}{2}\right)=K\left(\frac{\pi}{2}\right)$
2. $T^\prime\left(t\right)=\left(-\frac{\pi}{2},0,1\right)=K^\prime\left(\frac{\pi}{2}\right)$

Answer 2

Let me assume that you want to evaluate tangent at arbitary point $(x,y)$. Find out $dy/dx$ for the curve and substitute the values of $x$ and $y$ .Then you can find the tangent to curve using the one point form of tangent.