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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.

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

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 $12$.

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