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 $12$.