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\centerline
{
\bf Dr. Z's Math251 Handout \#13.3 [Arc Length and Curvature]
}

By Doron Zeilberger

{\bf Problem Type 13.3a}: Find the length of the curve
${\bf r}(t)= x(t) {\bf i}+ y(t) {\bf j}+ z(t) {\bf k} \,\,$,
$t_0 \leq t \leq t_1$.

{\bf Example Problem 13.3a}: Find the length of the curve
${\bf r}(t)= t^2 \, {\bf i}+2t \, {\bf j}+\ln t \, {\bf k} \,\,$,
$1 \leq t \leq e$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute the derivative
$$
{\bf r}'(t)= x'(t) {\bf i}+ y'(t) {\bf j}+ z'(t) {\bf k} \quad .
$$
&
{\bf 1.} 
$$
{\bf r}'(t)= (t^2)' \, {\bf i}+(2t)'\, {\bf j}+(\ln t)'\, {\bf k}
$$
$$
= 2t \, {\bf i}+ 2 \, {\bf j}+ {{1} \over {t}}\, {\bf k} \quad .
$$
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{\bf 2.} Find the {\bf magnitude} of ${\bf r}'(t)$,
$$
\vert {\bf r}'(t) \vert=
\sqrt{ x'(t)^2+y'(t)^2+z'(t)^2} \quad,
$$
and use algebra and/or trig to simplify as much as
you can.
&
{\bf 2.} 
$$
\vert {\bf r}'(t) \vert 
=\sqrt{(2t)^2+ 2^2+ {{1} \over {t^2}}}
$$
$$
=\sqrt{{4t^4+ 4t^2+1} \over {t^2} }
=\sqrt{{(2t^2+1)^2} \over {t^2} }
={(2t^2+1) \over {t} }=2t+{{1} \over {t}} \,\, .
$$
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{\bf 3.} Integrate the expression that you got in
step 2 from $t_0$ to $t_1$.
$$
\int_{t_0}^{t_1} \vert {\bf r}'(t) \vert \, dt
$$
&
{\bf 3.} 
$$
\int_{1}^{e} \vert {\bf r}'(t) \vert \, dt=
\int_{1}^{e} [ 2t+{{1} \over {t}} ]\,  dt=
= t^2 +\ln t  \Bigl \vert_{1}^{e}=
$$
$$
e^2+\ln e - (1^2+  \ln 1)=
e^2+1-(1+0)=e^2 \quad .
$$

{\bf Ans.:} The arc length of that curve is $e^2$.
\cr}

{\bf Problem Type 13.3b}: Reparametrize the curve with
respect to arc length measured from the point when 
$t=t_0$ in the direction of increasing $t$.
$$
{\bf r} (t)= x(t) \, {\bf i}+y(t) \, {\bf j}+z(t) \, {\bf k} \quad.
$$


{\bf Example Problem 13.3b}: Reparametrize the curve with
respect to arc length measured from the point when 
$t=0$ in the direction of increasing $t$.
$$
{\bf r} (t)= 5 \sin t \, {\bf i}+3 \, {\bf j}+5 \cos t \, {\bf k} \quad.
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute ${\bf r}'(t)$, and then take its
magnitude $\vert {\bf r}'(t) \vert $.

&
{\bf 1.} 
$$
{\bf r}' (t)= (5 \sin t)' \, {\bf i}+3' \, {\bf j}+(5 \cos t)' 
\, {\bf k} \quad.
$$
$$
(5 \cos t) \, {\bf i}- (5 \sin t)\, {\bf k} \quad.
$$
So
$$
\vert {\bf r}' (t) \vert=
\sqrt{(5 \cos t)^2+(-5 \sin t)^2}=
\sqrt{25(\cos^2 t+\sin^2 t)}=5 \quad .
$$
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{\bf 2.} Integrate it from $t_0$ to $t_1$. Get an expression
in terms of $t_1$ and call it $s$. Now change $t_1$ into $t$.
Now solve for $t$ in terms of $s$.
&
{\bf 2.} 
$$
s=\int_{0}^{t_1} 5 \, dt = 5t_1 \quad .
$$
Changing the $t_1$ into $t$ we get
$$
s=5t
$$
and expressing $t$ in terms of $s$, we get
$$
t=s/5 \quad.
$$
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{\bf 3.} Go back to the original ${\bf r}(t)$ and
replace $t$ by the expression in $s$ that you found in step 2.

&
{\bf 3.} 
$$
{\bf r} (t)= 5 \sin t \, {\bf i}+3 \, {\bf j}+5 \cos t \, {\bf k} \quad.
$$
becomes
$$
5 \sin (s/5) \, {\bf i}+3 \,{\bf j}+5 \cos (s/5) \, {\bf k} \quad.
$$
This is the {\bf Ans.}.
\cr}

{\bf Problem Type 13.3c}: Find the curvature for 
$$
{\bf r}(t)=x(t) \, {\bf i}+ y(t) \, {\bf j}+z(t) \, {\bf k} \quad.
$$


{\bf Example Problem 13.3c}: Find the curvature for 
$$
{\bf r}(t)=t \, {\bf i}+ 2t \, {\bf j}+t^2 \, {\bf k} \quad.
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute ${\bf r}'(t)$ and ${\bf r}''(t)$ .
&
{\bf 1.} 
$$
{\bf r}'(t)={\bf i}+ 2 \, {\bf j}+2t \, {\bf k} \quad.
$$
$$
{\bf r}''(t)=2 \, {\bf k} \quad.
$$


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{\bf 2.} Compute the {\bf cross product}

${\bf r}'(t) \times {\bf r}''(t)$.
&
{\bf 2.} ${\bf r}'(t) \times {\bf r}''(t)$ equals
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
1 & 2 & 2t\cr
0 & 0 & 2 
} \right \vert \quad =
$$
$$
{\bf i} \left \vert \matrix{ 
2 & 2t\cr
0 & 2 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
1 & 2t\cr
0 & 2 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
1 & 2 \cr
0 & 0 
} \right \vert 
$$
$$
=4 \, {\bf i}-2 \,{\bf j}
$$
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{\bf 3.} Find the magnitude of the vector that you found in
step 2 (namely ${\bf r}'(t) \times {\bf r}''(t)$).
Also find the magnitude of ${\bf r}'(t)$, and
finally use the formula for the curvature
$$
\kappa (t)=
{{\vert {\bf r}'(t) \times {\bf r}''(t) \vert} \over
{\vert {\bf r}'(t) \vert^3}}
$$
&
{\bf 3.} 
$$
\vert {\bf r}'(t) \times {\bf r}''(t) \vert=
\sqrt{4^2+2^2+0^2}=\sqrt{20} \quad.
$$
$$
\vert {\bf r}'(t)\vert=
\vert {\bf i}+ 2 \, {\bf j}+2t \, {\bf k}\vert 
$$
$$
=\sqrt{1^2+2^2+(2t)^2}=\sqrt{5+4t^2} \quad .
$$
Finally,
$$
\kappa (t)=
{{\sqrt{20}} \over
{(\sqrt{5+4t^2})^3 } }
$$
This is the {\bf Ans.}.
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