# Let C be the curve intersection of the parabolic cylinder x^2=2y and the surface 3z=xy. Find the exact length of C from the origin to the point (6,18,36).

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.

$L(f) = \int _{a} ^{b} |f'(t)| dt$

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 $12$.