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\centerline
{
\bf Dr. Z's Math251 Handout \#13.2 [Derivates and Integrals of Vector
Functions]
}

By Doron Zeilberger

{\bf Problem Type 13.2a}: 
Find a parametric equation for the tangent line to the curve
with the given parametric equation at the specified point
$$
x=f_1(t) \, , \, y=f_2(t) \, , \, z=f_3(t) \quad ; \quad P(p_1,p_2,p_3)
$$


{\bf Example Problem 13.2a}: 
Find a parametric equation for the tangent line to the curve
with the given parametric equation at the specified point
$$
x=t^2-1 \quad , \quad y=t^2+1 \quad , \quad z=t+1 \quad ; \quad (-1,1,1)
$$


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the relevant $t$ at the designated
point. Let's call it $t_0$. Solve the equations
$f_1(t_0)=p_1,f_2(t_0)=p_2,f_3(t_0)=p_3$.
If there is no solution refuse to do the problem.
&
{\bf 1.} 
$-1=t_0^2-1 \, , \, 1=t_0^2+1 \, , \, 1=t_0+1$
means $t_0=0$.
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{\bf 2.} 
Putting ${\bf r} (t)=\langle x(t),y(t),z(t) \rangle$, take
the derivative ${\bf r}'(t)$ by doing it component-by-component
$$
{\bf r}' (t)=\langle x'(t),y'(t),z'(t) \rangle \quad.
$$
&
{\bf 2.} 
%%%Specific Step 2
$$
{\bf r}(t)= \langle \, t^2-1 \, , \, t^2+1 \, , \, t+1 \, \rangle
$$
$$
{\bf r}'(t)= \langle \, (t^2-1)' \, , \, (t^2+1)' \, , \, (t+1)' \, \rangle
= \langle \, 2t \, , \, 2t \, , \, 1  \, \rangle \quad .
$$

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{\bf 3.} Plug-in the specific value ($t=t_0$),
that you found in step 1, to get
${\bf r}'(t_0)$, which is the {\bf direction vector},
let's call it {\bf D}.
and using the given point $P$ as the {\bf starting point},
the {\bf parametric equation} of the line is
$\langle x,y,z \rangle =P+tD $. Finally, spell out
the expressions for $x,y,z$.
&
{\bf 3.} The direction vector is
$\langle \, 2t \, , \, 2t \, , \, 1 \, \rangle$ plugged-in at $t=0$,
it is $\langle 0 \, , \, 0 \, , \, 1 \, \rangle$, and since the point is
$\langle \, -1 \, , \, 1 \, , \, 1 \,\rangle$ the equation of the tangent line,
in vector-form is
$$
\langle \, -1 \, , \, 1 \, , \, 1 \, \rangle+
t\langle \, 0 \, , \, 0 \,, \, 1 \, \rangle=
\langle \, -1 \, , \, 1 \, , \, 1+t \, \rangle \quad,
$$
and spelling it out
$$
x=-1 \quad ,\quad y=1 \quad , \quad z=1+t \quad.
$$
This is the {\bf Ans.}.
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{\bf Problem Type 13.2b}: 
Find ${\bf r}(t)$ if 
$$
{\bf r}'(t)=f_1(t) \, {\bf i} +f_2(t) \, {\bf j}+f_3(t) \,{\bf k} \quad
$$
and 
$$
{\bf r}(t_0)=a \, {\bf i} +b \, {\bf j}+c \, {\bf k} \quad
$$
%%General Problem


{\bf Example Problem 13.2b}: 
Find ${\bf r}(t)$ if 
$$
{\bf r}'(t)=2t \, {\bf i} +3t^2 \, {\bf j}+4t^3 \,{\bf k} \quad
$$
and 
$$
{\bf r}(1)={\bf i} +4 \, {\bf j}+5 \,{\bf k} \quad
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the indefinite integral by
integrating every component and not
forgetting to add {\bf an arbitrary constant}
that now is a {\bf vector}.

&
{\bf 1.} 
$$
{\bf r}(t)=
\int ( 2t \, {\bf i} +3t^2 \, {\bf j}+4t^3 \, {\bf k} ) \, dt
$$
$$
= t^2 \, {\bf i} + t^3 \, {\bf j}+t^4 \,{\bf k} \, + \,{\bf C}
\quad .
$$
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{\bf 2.} Plug-in $t=t_0$ and solve for ${\bf C}$.
&
{\bf 2.} 
$$
{\bf r}(1)= 1^2 \, {\bf i} + 1^3 \, {\bf j}+1^4 \,{\bf k} + {\bf C}=
$$
$$
{\bf i} + {\bf j}+{\bf k}  + {\bf C}=
{\bf i} +4 \, {\bf j}+5 \,{\bf k} \quad .
$$
Solving for ${\bf C}$ gives ${\bf C}=3\, {\bf j}+4\, {\bf k}$
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{\bf 3.} Go back to step 1 and incorporate the
specific ${\bf C}$ that you've found in step 2.
&
{\bf 3.} 
$$
{\bf r}(t) = t^2 \, {\bf i} + t^3 \, {\bf j}+t^4 \, {\bf k} 
+3 \, {\bf j}+4 \, {\bf k}
$$
$$
=t^2 \, {\bf i} \, + \, (t^3+3) \, {\bf j} \,+ \,(t^4+4) \,{\bf k} 
\quad .
$$
This is the {\bf Ans.}.
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