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Find a vector function that represents the curve of intersection of the cylinder and the plane.

\[Cylinder\ x^2+y^2=4\]

\[Surface\ z=xy\]

The aim of this question is to find the vector function of the curve that is generated when a cylinder is intersected by a surface.

The basic concept behind this article is the Vector-Valued Function and representation of different geometrical figures in parametric equations.

A vector-valued function is defined as a mathematical function consisting of one or more variables having a range, which is a set of vectors in multi-dimensions. We can use a scalar or a vector parameter as an input for the vector-valued function, whereas its output will be a vector.

For two dimensions, the vector-valued function is:

\[r(t)=x(t)\hat{i}+y(t)\hat{j}\]

For three dimensions, the vector-valued function is:

\[r(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}\]

Or:

\[r(t)\ =\ \langle x(t),\ y(t),\ z(t) \rangle \]

Expert Answer

The Equation for Cylinder:

\[x^2+y^2=4\]

The Equation for Surface:

\[z=xy\]

When a plane surface intersects a three-dimensional cylindrical figure, the curve of intersection created will be in a three-dimensional plane in the form of a circle.

Therefore, the equation of a standard circle with Center $(0,\ 0)$ is derived by considering the position coordinates of circle centers with their constant radius $r$ as follows:

\[x^2+y^2=r^2\]

Where:

$R=$ Radius of Circle

$(x,\ y)=$ Any point on Circle

As per Cylindrical Coordinate System, the parametric equations for $x$ and $y$ are:

\[x(t)=rcos(t)\]

\[y(t)=rsin(t)\]

Where:

$t=$ Counter-clockwise angle from the x-axis in the x,y plane and having a range of:

\[0\ \le\ t\ \le\ 2\pi\]

As the Equation for Cylinder is $x^2+y^2=4$, so the radius $r$ will be:

\[x^2+y^2\ =\ {4\ =(2)}^2\]

Hence:

\[r\ =\ 2\]

By substituting the value of $r\ =\ 2$ in parametric equations for $x$ and $y$, we get:

\[x(t)\ =\ r\ cos(t)\]

\[y(t)\ =\ r\ sin(t)\]

By substituting the value of $x$ and $y$ in $z$, we get:

\[z(t)\ =\ x(t)\ \times\ y(t)\]

\[z\ =\ 2\ cos(t)\ \times\ 2\ sin(t)\]

By simplifying the equation:

\[z\ =\ 4\ sin(t)\ cos(t)\]

So the vector function will be represented as follows:

\[r(t)\ =\ \langle x(t),\ y(t),\ z(t)\rangle\]

\[r(t)\ =\ \langle\ 2\ cos(t),\ 2\ sin(t)\ \ ,\ 4\ sin(t)cos(t)\ \rangle\]

Numerical Result

The curve of intersection of cylinder and surface will be represented by a vector function as follows:

Then that represents as follows:

\[r(t)\ =\ \langle\ 2\ cos(t),\ 2\ sin(t)\ \ ,\ 4\ sin(t)cos(t)\ \rangle\]

Example

A cylinder $x^2+y^2\ =\ 36$ and surface $4y+z=21$ intersect each other and form a curve of intersection. Find its vector function.

Solution

The Equation for Cylinder:

\[x^2+y^2\ =\ 36\]

The Equation for Surface:

\[4y+z=21\]

\[z=21\ -\ 4y\]

As the Equation for Cylinder is $x^2+y^2\ =\ 36$, so the radius $r$ will be:

\[x^2+y^2\ =\ {36\ =(6)}^2\]

Hence:

\[r\ =\ 6\]

By substituting the value of $r\ =\ 6$ in parametric equations for $x$ and $y$, we get:

\[x(t)\ =\ 6\ cos(t)\]

\[y(t)\ =\ 6\ sin(t)\]

By substituting the value of $x$ and $y$ in $z$, we get:

\[z=21\ -\ 4y\]

\[z=21\ -\ 4(6\ sin(t))\]

\[z=21\ -\ 24\ sin(t)\]

So, the vector function will be:

\[r(t)\ =\ \langle\ 6\ cos(t),\ 6\ sin(t)\ \ ,\ 21\ -\ 24\ sin(t)\ \rangle\]

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