Find a vector function, $\mathbf{r}(t)$, that represents the curve of intersection of the two surfaces: The cylinder $x^2 + y^2 = 36$ and the surface $z = xy$

$\mathbf{r}(t) = \boxed{\langle \dots, \dots, \dots \rangle}$

Neetesh Kumar | December 14, 2024

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This is the solution to Math 1C
Assignment: 13.1 Question Number 10
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Step-by-step solution:

Step 1: Parametrize the cylinder $x^2 + y^2 = 36$

The equation of the cylinder $x^2 + y^2 = 36$ represents a circle in the $xy$-plane with radius $6$ centered at the origin.

A standard parametrization of this circle is:
$x = 6\cos(t), \quad y = 6\sin(t),$
where $t$ is the parameter, and $t \in [0, 2\pi)$.

Step 2: Use $z = xy$ to find $z$

The surface $z = xy$ relates $z$ to the $x$ and $y$ coordinates.

Substitute the parametric expressions for $x$ and $y$ into $z = xy$:
$z = \left(6\cos(t)\right)\left(6\sin(t)\right) = 36\cos(t)\sin(t).$
Using the trigonometric identity $\sin(2t) = 2\sin(t)\cos(t)$, we can rewrite $z$ as:
$z = 18\sin(2t).$

Step 3: Write the vector function

Combine the parametric expressions for $x$, $y$, and $z$ into a single vector function:
$\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle 6\cos(t), 6\sin(t), 18\sin(2t) \rangle.$

Final Answer:

$\mathbf{r}(t) = \boxed{\langle 6\cos(t), 6\sin(t), 18\sin(2t) \rangle}$

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