This article aims to find the length of the curve $ C $ from origin to point $ (6,18,36) $. This article uses the concept of finding the length of arc length. The length of the curve defined by $f$ can be defined as the limit of the sum of lengths of linear segments for the regular partition $(a,b)$ as the number of segments approaches infinity.
\[L(f) = \int _{a} ^{b} |f'(t)| dt \]
Expert Answer
Finding the curve of intersection and solving the first given equation for $ y $ in terms of $ x $, we get:
$x^{2} = \dfrac{2y}{t}$, change the first equation to parametric form by substituting $ x $ for $ t $, that is:
\[x= t, y = \dfrac{1}{2} t^{2}\]
Solve second equation for $ z $ in terms of $t$. we get:
\[z= \dfrac{1}{3}(x.y) = \dfrac{1}{3}(t. \dfrac{1}{2}t^{2}) = \dfrac{1}{6}t^{3}\]
We get the coordinates $x$, $yz$ into the vector equation for the curve $r(t)$.
\[r(t) = <t, \dfrac{1}{2}t^{2} , \dfrac{1}{6}t^{3}>\]
Calculate first derivative of the vector equation $r(t)$ by components, that is,
\[r'(t) = <1,t, \dfrac{1}{2}t^{2}>\]
Calculate the magnitude of $r'(t)$.
\[|r'(t) | = \sqrt {\dfrac{1}{4}t^{4} + t^{2}+1 }\]
\[= \dfrac{1}{2} \sqrt{t^{4}+4t^{2}+4} \]
\[= \dfrac{1}{2} \sqrt{(t^{2}+2)^{2}}\]
\[= \dfrac{1}{2} t^{2}+1 \]
Solve for range of $t$ along the curve between the origin and the point $(6,18,36)$.
\[(0,0,0)\rightarrow t = 0\]
\[(6,18,36)\rightarrow t = 6\]
\[0\leq t\leq 6\]
Set the integral for the arc length from $0$ to $6$.
\[C = \int_{0}^{6} \dfrac{1}{2} t^{2}+1 dt\]
Evaluate the integral.
\[C = |\dfrac{1}{6} t^{3} +t |_{0}^{6} = 42\]
The exact length of curve $C$ from the origin to the point $ (6,18,36)$ is $42$.
Numerical Result
The exact length of curve $C$ from the origin to the point $ (6,18,36)$ is $42$.
Example
Let $C$ be the intersection of the curve of the parabolic cylinder $x^{2} = 2y$ and surface $3z= xy $. Find exact length of $C$ from the origin to the point $(8,24,48)$.
Solution
$x^{2} = \dfrac{2y}{t}$, change the first equation to parametric form by substituting $ x $ for $ t $, that is
\[x= t, y = \dfrac{1}{2} t^{2}\]
Solve second equation for $ z $ in terms of $t$. we get
\[z= \dfrac{1}{3}(x.y) = \dfrac{1}{3}(t. \dfrac{1}{2}t^{2}) = \dfrac{1}{6}t^{3}\]
We get the coordinates $x$, $yz$ into the vector equation for the curve $r(t)$.
\[r(t) = <t, \dfrac{1}{2}t^{2} , \dfrac{1}{6}t^{3}>\]
Calculate first derivative of the vector equation $r(t)$ by components, that is,
\[r'(t) = <1,t, \dfrac{1}{2}t^{2}>\]
Calculate the magnitude of $r'(t)$.
\[|r'(t) | = \sqrt {\dfrac{1}{4}t^{4} + t^{2}+1 }\]
\[= \dfrac{1}{2} \sqrt{t^{4}+4t^{2}+4} \]
\[= \dfrac{1}{2} \sqrt{(t^{2}+2)^{2}}\]
\[= \dfrac{1}{2} t^{2}+1 \]
Solve for range of $t$ along the curve between the origin and the point $(8,24,48)$
\[(0,0,0)\rightarrow t = 0\]
\[(8,24,48)\rightarrow t = 8\]
\[0\leq t\leq 8\]
Set the integral for the arc length from $0$ to $8$
\[C = \int_{0}^{8} \dfrac{1}{2} t^{2}+1 dt\]
Evaluate the integral
\[C = |\dfrac{1}{6} t^{3} +t |_{0}^{8} = \dfrac{1}{6}(8)^{3}+8 = 12\]
The exact length of curve $C$ from the origin to the point $ (8,24,36)$ is $12$.