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.

$$L(f)={\int }_a^b|f^{\prime }(t)|dt$$

Expert Answer

Finding the curve of intersection and solving the first given equation for $y$ in terms of $x$, we get:

$x^2={\displaystyle \frac{2y}{t}}$, change the first equation to parametric form by substituting $x$ for $t$, that is:

$$x=t,y={\displaystyle \frac{1}{2}}{t}^2$$

Solve second equation for $z$ in terms of $t$. we get:

$$z={\displaystyle \frac{1}{3}}(x.y)={\displaystyle \frac{1}{3}}(t.{\displaystyle \frac{1}{2}}{t}^2)={\displaystyle \frac{1}{6}}{t}^3$$

We get the coordinates $x$, $yz$ into the vector equation for the curve $r(t)$.

$$r(t)=<t,{\displaystyle \frac{1}{2}}{t}^2,{\displaystyle \frac{1}{6}}{t}^3>$$

Calculate first derivative of the vector equation $r(t)$ by components, that is,

$$r^{\prime }(t)=<1,t,{\displaystyle \frac{1}{2}}{t}^2>$$

Calculate the magnitude of $r^{\prime }(t)$.

$$|r^{\prime }(t)|=\sqrt{{\displaystyle \frac{1}{4}}{t}^4+t^2+1}$$

$$={\displaystyle \frac{1}{2}}\sqrt{t^4+4t^2+4}$$

$$={\displaystyle \frac{1}{2}}\sqrt{(t^2+2{)}^2}$$

$$={\displaystyle \frac{1}{2}}{t}^2+1$$

Solve for range of $t$ along the curve between the origin and the point $(6,18,36)$.

$$(0,0,0)\to t=0$$

$$(6,18,36)\to t=6$$

$$0\le t\le 6$$

Set the integral for the arc length from $0$ to $6$.

$$C={\int }_0^6{\displaystyle \frac{1}{2}}{t}^2+1dt$$

Evaluate the integral.

$$C=|{\displaystyle \frac{1}{6}}{t}^3+t{|}_0^6=42$$

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

$x^2={\displaystyle \frac{2y}{t}}$, change the first equation to parametric form by substituting $x$ for $t$, that is

$$x=t,y={\displaystyle \frac{1}{2}}{t}^2$$

Solve second equation for $z$ in terms of $t$. we get

$$z={\displaystyle \frac{1}{3}}(x.y)={\displaystyle \frac{1}{3}}(t.{\displaystyle \frac{1}{2}}{t}^2)={\displaystyle \frac{1}{6}}{t}^3$$

We get the coordinates $x$, $yz$ into the vector equation for the curve $r(t)$.

$$r(t)=<t,{\displaystyle \frac{1}{2}}{t}^2,{\displaystyle \frac{1}{6}}{t}^3>$$

Calculate first derivative of the vector equation $r(t)$ by components, that is,

$$r^{\prime }(t)=<1,t,{\displaystyle \frac{1}{2}}{t}^2>$$

Calculate the magnitude of $r^{\prime }(t)$.

$$|r^{\prime }(t)|=\sqrt{{\displaystyle \frac{1}{4}}{t}^4+t^2+1}$$

$$={\displaystyle \frac{1}{2}}\sqrt{t^4+4t^2+4}$$

$$={\displaystyle \frac{1}{2}}\sqrt{(t^2+2{)}^2}$$

$$={\displaystyle \frac{1}{2}}{t}^2+1$$

Solve for range of $t$ along the curve between the origin and the point $(8,24,48)$

$$(0,0,0)\to t=0$$

$$(8,24,48)\to t=8$$

$$0\le t\le 8$$

Set the integral for the arc length from $0$ to $8$

$$C={\int }_0^8{\displaystyle \frac{1}{2}}{t}^2+1dt$$

Evaluate the integral

$$C=|{\displaystyle \frac{1}{6}}{t}^3+t{|}_0^8={\displaystyle \frac{1}{6}}(8{)}^3+8=12$$

The exact length of curve $C$ from the origin to the point $(8,24,36)$ is $12$.

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