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???
???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???
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???