Question

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

Short answer

The exact length from the origin to the point \((6,18,36)\;is\;42\) .

Step-by-step solution

Step 1: Find the curve of intersection and change first equation into parametric

Let us find the curve of intersection and solve the first given equation for y in terms of x, we get,

\(x{}^2 = 2y \to y = \frac{1}{2}{x^2}\)

Next change the first equation into parametric form by replacing x with t, that is,\(\begin{aligned}{l}x = t\\y = \frac{1}{2}{t^2}\end{aligned}\)

Step 2: Solve the second equation and plug in values of x, y and z coordinates into a vector equation for the curve.

Solve the second equation for z in terms of t, we get,

\(\begin{aligned}{l}z = \frac{1}{3}(x \cdot y)\\ = \frac{1}{3}\left( {t \cdot \frac{1}{2}{t^2}} \right)\\ = \frac{1}{6}{t^3}\end{aligned}\)

Plug the x, y and z coordinates into a vector equation for the curve, r(t), we get,

\(r(t) = \left\langle {t,\frac{1}{2}{t^2},\frac{1}{6}{t^3}} \right\rangle \)

Step 3: Calculate the derivative of vector equation and its the magnitude

Calculate the first derivative of the vector equation r(t) component wise that is,

\(r'(t) = \left\langle {1,t,\frac{1}{2}{t^2}} \right\rangle \)

Calculate the magnitude of the obtained equation :

\(\begin{aligned}{l}|r'(t)| = \sqrt {\frac{1}{4}{t^4} + {t^2} + 1} \\ = \frac{1}{2}\sqrt {{t^4} + 4{t^2} + 4} \\ = \frac{1}{2}\sqrt {{{({t^2} + 2)}^2}} = \frac{1}{2}{t^2} + 1\end{aligned}\)

Step 4: Solve for range of t along the curve between the origin and the point (6,18,36)

\(\begin{aligned}{l}(0,0,0) \to t = 0\\(6,18,36) \to t = 6\\\therefore 0 \le t \le 6\end{aligned}\)

Step 5: Set up integral for the arc length

Set up an integral for the arc length from 0 to 6.

\(\begin{aligned}{l}C = \int\limits_0^6 {\left( {\frac{1}{2}{t^2} + 1} \right)dt} \\C = \left( {\frac{1}{6}{t^3} + t} \right)_0^6 = 42\end{aligned}\)