Parametric equations of the tangent line to the vector function

Defining the parametric equations of a vector function

The parametric equations of the tangent line of a vector function $r(t)=\langle{r(t)_1,r(t)_2,r(t)_3}\rangle$ are

$x=x_1+r'(t_0)_1t$

$y=y_1+r'(t_0)_2t$

$z=z_1+r'(t_0)_3t$

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$x_1$, $y_1$ and $z_1$ come from the point $P(x_1,y_1,z_1)$, which is the point of tangency.

You find $t_0$ by plugging $P(x_1,y_1,z_1)$ into the vector function.

Then you find $r'(t_0)_1$, $r'(t_0)_2$ and $r'(t_0)_3$ by plugging $t_0$ into the derivative of the vector function.

How to find the parametric equations of the tangent line of a vector function

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Step-by-step example for finding the parametric equations of the line tangent to a vector at a specific point

Example

Find the parametric equations of the tangent line to the vector at $P(1,0,0)$.

$x=e^t$

$y=-t\cos{t}$

$z=\sin{t}$

First, since the point of tangency is $P(1,0,0)$, we can plug that point into the formulas for the parametric equation of the tangent line from above, and they become

$x=1+r'(t_0)_1t$

and

$y=0+r'(t_0)_2t$

$y=r'(t_0)_2t$

and

$z=0+r'(t_0)_3t$

$z=r'(t_0)_3t$

Now we’ll find a value for $t_0$. We’ll use $x=e^t$, change $t$ to $t_0$ and plug $x=1$ (from $P(1,0,0)$) into the equation and get

$1=e^{t_0}$

$\ln1=\ln{e^{t_0}}$

$t_0=0$

Plugging $t_0=0$ and $y=0$ (from $P(1,0,0)$) into $y=-t\cos{t}$ and get

$0=-0\cos{0}$

$0=0$

Since this equation is true, $t_0=0$ works for $y=-t\cos{t}$ as well as $x=e^t$. Now we’ll plug $t_0=0$ and $z=0$ (from $P(1,0,0)$) into $z=\sin{t}$ and get

$0=\sin{0}$

$0=0$

Since this equation is true, we’ve now shown that $t_0=0$ satisfies $x=e^t$, $y=-t\cos{t}$ and $z=\sin{t}$, so $0$ is the value we want to use for $t_0$. Therefore, the parametric equations of the tangent line become

$x=1+r'(0)_1t$

$y=r'(0)_2t$

$z=r'(0)_3t$

First, since the point of tangency is P(1,0,0), we can plug that point into the formulas for the parametric equation of the tangent line.

Next we need to find the derivative of the vector function. The original function is

$r(t)=\langle{e^t,-t\cos{t},\sin{t}}\rangle$

so its derivative is

$r'(t)=\left\langle e^t,(-1)(\cos{t})+(-t)(-\sin{t}),\cos{(t)}\right\rangle$

$r'(t)=\left\langle e^t,-\cos{t}+t\sin{t},\cos{t}\right\rangle$

$r'(t)=\left\langle e^t,t\sin{t}-\cos{t},\cos{t}\right\rangle$

Plugging $t_0=0$ into the derivative, we get

$r'(0)=\left\langle e^0,0\sin{0}-\cos{0},\cos{0}\right\rangle$

$r'(0)=\left\langle1,0-1,1\right\rangle$

$r'(0)=\left\langle1,-1,1\right\rangle$

We’ll take these three values, plug them into our parametric equations, and the parametric equations become

$x=1+1t$

$y=-1t$

$z=1t$

and these simplify to

$x=1+t$

$y=-t$

$z=t$

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