Question

Let C be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. Find the exact length of C from the origin to the point

Answer 1

Is there a specific point besides the origin given? "... the origin to the point[...?] "

Answer 2

I notice that it is simple to express all three variables in terms of \(x\) using the given information. \[
\quad 2y = x^2 \\
\quad y = \frac{1}{2} x^2 , \\ ~\\
\quad 3z = xy \\
\quad 3z = x(\frac{1}{2} x^2) \\
\quad 3z = \frac{1}{2} x^3 \\
\quad z = \frac{1}{6} x^3 .
\]
I write all three with a parameter \(t\): \[
\quad x(t) = t \\
\quad y(t) = \frac{1}{2} t^2 \\
\quad z(t) = \frac{1}{6} t^3 \]
\]

While you haven't given any specific point, I will use \((0,0,0)\) and \((t_0,\frac{1}{2}t_0^2, \frac{1}{6}t_0^3)\). Then, the integration with respect to \(t\) will be from \(t = 0\) to \(t = t_0\). So, I'll then use this formula: \[ \large {
\ \cdot \quad L= \int \limits_{t=a}^{t=b} \sqrt{\left(\frac{\text{dx}}{\text{dt}}\right)^2 + \left(\frac{\text{dy}}{\text{dt}}\right)^2 + \left(\frac{\text{dz}}{\text{dt}}\right)^2 } \text{dt}
} \]
We use \(a = 0\) and \(b = t_0\). Also, we may calculate the derivatives of each of the functions \(x(t)\), \(y(t)\), and \(z(t)\) and use these as well. \[
\quad \frac{\text{dx}}{\text{dt}} = 1 \\
\quad \frac{\text{dy}}{\text{dt}} = t \\
\quad \frac{\text{dz}}{\text{dt}} = \frac{1}{2} t^2
\]

Now, we insert the values into our formula and solve. \[ \large {
\quad L = \int_{t=0}^{t=t_0} \sqrt{\left(1\right)^2 + \left(t\right)^2 + \left(\frac{1}{2} t^2 \right)^2} \text{dt} \\
\quad L = \int_{t=0}^{t=t_0} \sqrt{1 + t^2 + \frac{1}{4} t^4} \text{dt}
} \]
Note that the quantity under the square root is a perfect square. \(\frac{1}{4} t^4 + t^2 + 1 = \left(\frac{1}{2} t^2 + 1\right)^2\). \[ \large {
\quad L = \int_{t=0}^{t=t_0} \sqrt{\left(\frac{1}{2}t^2 + 1 \right)^2} \text{dt} \\
\quad L = \int_{t=0}^{t=t_0} \frac{1}{2} t^2 + 1 \text{dt} \\
\quad L = \frac{1}{2} \int_{t=0}^{t=t_0} t^2 \text{dt} + \int_{t=0}^{t=t_0} 1 \text{dt} \\
\quad L = \frac{1}{2} \left. \left( \frac{1}{3} t^3 \right) \right|_{t=0}^{t=t_0} + \left. t \right|_{t=0}^{t=t_0} \\
\quad L = \frac{1}{6} t_0^3 + t_0
} \]