Find A Vector Function That Represents The Curve Of Intersection

Finding a Vector Function for the Curve of Intersection

When two surfaces in three‑dimensional space meet, the set of points that belong to both surfaces forms a curve. Still, describing that curve with a vector function (also called a parametric representation) lets us compute tangents, arc length, curvature, and many other geometric quantities. Day to day, in multivariable calculus and vector analysis, the problem of “find a vector function that represents the curve of intersection” appears repeatedly—whether the surfaces are a sphere and a plane, a cylinder and a paraboloid, or more exotic implicit surfaces. Mastering this technique gives you a powerful tool for visualising and analysing the geometry of intersecting objects.

Detailed Explanation

What Is a Vector Function?

A vector function (\mathbf{r}(t)) maps a real parameter (t) (often time) to a point in (\mathbb{R}^3):

[ \mathbf{r}(t)=\langle x(t),,y(t),,z(t)\rangle . ]

Each component (x(t),y(t),z(t)) is a scalar function of (t). Practically speaking, as (t) varies over an interval (I), the tip of the vector traces out a curve in space. The curve of intersection of two surfaces (F(x,y,z)=0) and (G(x,y,z)=0) consists of all points ((x,y,z)) that satisfy both equations simultaneously.

The official docs gloss over this. That's a mistake.

[ F\bigl(x(t),y(t),z(t)\bigr)=0\quad\text{and}\quad G\bigl(x(t),y(t),z(t)\bigr)=0 ]

for every (t) in the domain Not complicated — just consistent. That alone is useful..

Why Parameterise?

Parameterising the intersection curve has several advantages:

1. Explicit computation – derivatives (\mathbf{r}'(t)) give tangent vectors directly.
2. Arc length and curvature – formulas (\displaystyle s=\int|\mathbf{r}'(t)|dt) and (\kappa=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}) become straightforward.
3. Visualization – most computer‑algebra systems plot (\mathbf{r}(t)) easily.
4. Physical interpretation – if (t) represents time, (\mathbf{r}(t)) can describe the trajectory of a particle constrained to lie on both surfaces.

The core of the problem is therefore solving a system of two equations in three unknowns and expressing the solution in terms of a single free parameter Still holds up..

General Strategy

Although each pair of surfaces may require a slightly different trick, the following workflow works in most textbook cases:

1. Identify a convenient variable to eliminate – often one of the coordinates appears linearly or quadratically in a simple way.
2. Solve one surface for that variable (or for a combination of variables).
3. Substitute into the second surface to obtain a relation between the remaining two variables.
4. Parameterise that relation using a standard curve (circle, ellipse, line, etc.).
5. Back‑substitute to obtain the third coordinate as a function of the parameter.
6. Write the vector function (\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle) and state its domain.

If the surfaces are both given implicitly, sometimes it is easier to use cylindrical or spherical coordinates because the symmetry of one surface simplifies the algebra It's one of those things that adds up. But it adds up..

Step‑by‑Step Concept Breakdown

Below is a detailed, generic procedure illustrated with a concrete example: the intersection of the sphere (x^{2}+y^{2}+z^{2}=4) and the cylinder (x^{2}+y^{2}=1). The same steps can be adapted to other pairs The details matter here. Turns out it matters..

Step 1: Write the equations

[ \begin{cases} F(x,y,z)=x^{2}+y^{2}+z^{2}-4=0\[2mm] G(x,y,z)=x^{2}+y^{2}-1=0 \end{cases} ]

Step 2: Solve the simpler equation for a combination

From (G=0) we have (x^{2}+y^{2}=1). This tells us that the projection of the curve onto the (xy)-plane is the unit circle.

Step 3: Substitute into the other equation

Replace (x^{2}+y^{2}) in the sphere equation:

[ 1+z^{2}=4;\Longrightarrow;z^{2}=3;\Longrightarrow;z=\pm\sqrt{3}. ]

Thus the intersection consists of two horizontal circles (one at (z=+\sqrt{3}), one at (z=-\sqrt{3})) each lying on the unit cylinder.

Step 4: Parameterise the circle in the (xy)-plane

A standard parametrisation of the unit circle is

[ x(t)=\cos t,\qquad y(t)=\sin t,\qquad t\in[0,2\pi). ]

Step 5: Attach the constant (z)-value

For the upper circle:

[ \mathbf{r}_{+}(t)=\langle \cos t,;\sin t,;\sqrt{3}\rangle . ]

For the lower circle:

[ \mathbf{r}_{-}(t)=\langle \cos t,;\sin t,;-\sqrt{3}\rangle . ]

Step 6: State the domain and verify

Both vector functions satisfy the original equations for every (t) in ([0,2\pi)). Their derivatives give tangent vectors (\mathbf{r}'_{\pm}(t)=\langle -\sin t,;\cos t,;0\rangle), confirming that the curves lie in planes (z=\pm\sqrt{3}) Easy to understand, harder to ignore..

Real Examples

Example 1: Plane and Paraboloid

Find the curve where the plane (z=2x+3y) meets the paraboloid (z=x^{2}+y^{2}).

1. Set the two expressions for (z) equal:

   [ x^{2}+y^{2}=2x+3y. ]

2. Rearrange to complete squares:

   [ (x^{2}-2x)+(y^{2}-3y)=0;\Longrightarrow;(x-1)^{2}+(y-\tfrac{3}{2})^{2}=1+\tfrac{9}{4}=\tfrac{13}{4}. ]

   This is a circle centred at ((1,,\tfrac{3}{2})) with radius (\sqrt{13}/2).

3. Parameterise the circle:

   [ x(t)=1+\frac{\sqrt{13}}{2}\cos t,\qquad y(t)=\frac{3}{2}+\frac{\sqrt{13}}{2}\sin t. ]

4. Obtain (z(t)) from either surface (use the plane for simplicity):

   [ z(t)=2x(t)+3y(t)=2\Bigl(1+\frac{\sqrt{13}}{2}\cos t\Bigr)+3\Bigl(\frac{3}{2}+\frac{\sqrt{13}}{2}\sin t\Bigr) =2+\sqrt{13}\cos t+\frac{9}{2}+\frac{3\sqrt{13}}{2}\sin t =\frac{13}{2}+\sqrt{13}\cos t+\frac{3\sqrt{13}}{2}\sin t. ]

5. Vector function:

   [

[ \mathbf{r}(t) = \left\langle 1+\frac{\sqrt{13}}{2}\cos t, ;\frac{3}{2}+\frac{\sqrt{13}}{2}\sin t, ;\frac{13}{2}+\sqrt{13}\cos t+\frac{3\sqrt{13}}{2}\sin t \right\rangle, \quad t \in [0, 2\pi). ]

Example 2: Two Cylinders (The Viviani Curve variant)

Find the intersection of the cylinder $x^2 + y^2 = x$ and the plane $z = y$.

1. Analyze the base projection: The equation $x^2 + y^2 = x$ can be rewritten by completing the square: [ (x - \tfrac{1}{2})^2 + y^2 = \tfrac{1}{4}. ] This is a circle in the $xy$-plane centered at $(\frac{1}{2}, 0)$ with radius $\frac{1}{2}$.

2. Parameterize the $x$ and $y$ components: Using the standard trigonometric form for a circle: [ x(t) = \frac{1}{2} + \frac{1}{2}\cos t, \qquad y(t) = \frac{1}{2}\sin t. ]

3. Determine the $z$ component: Since the curve must lie on the plane $z = y$, we simply substitute $y(t)$ into the plane equation: [ z(t) = \frac{1}{2}\sin t. ]

4. Final Vector Function: [ \mathbf{r}(t) = \left\langle \frac{1}{2}(1 + \cos t), ;\frac{1}{2}\sin t, ;\frac{1}{2}\sin t \right\rangle, \quad t \in [0, 2\pi). ]

Conclusion

Determining the intersection of two surfaces is a fundamental skill in multivariable calculus, bridging the gap between algebraic equations and geometric intuition. As demonstrated, the process generally follows a logical hierarchy: first, reducing the system of equations to a single plane (usually the $xy$-plane) to identify the "shadow" or projection of the curve; second, parameterizing that projection; and third, lifting that parameterization into three-dimensional space using the original surface equations No workaround needed..

Whether the intersection results in simple geometric shapes like circles or more complex space curves, the ability to translate implicit equations into explicit vector-valued functions $\mathbf{r}(t)$ is essential for calculating line integrals, determining surface area, or visualizing complex physical boundaries in engineering and physics.

Honestly, this part trips people up more than it should Most people skip this — try not to..

Example 3: Sphere and Paraboloid Intersection

Find the intersection of the sphere $x^2 + y^2 + z^2 = 4$ and the paraboloid $z = x^2 + y^2$.

1. Analyze the base projection: Substitute the paraboloid equation into the sphere equation: [ x^2 + y^2 + (x^2 + y^2)^2 = 4 ] Let $r^2 = x^2 + y^2$, then: [ r^2 + r^4 = 4 \implies r^4 + r^2 - 4 = 0 ] Solving this quadratic in $r^2$: [ r^2 = \frac{-1 + \sqrt{1 + 16}}{2} = \frac{-1 + \sqrt{17}}{2} ] Since $r^2 \geq 0$, we take the positive root. Thus, the projection onto the $xy$-plane is a circle of radius $\sqrt{\frac{-1 + \sqrt{17}}{2}}$.

2. Parameterize the $x$ and $y$ components: Using polar coordinates for the circle: [ x(t) = \sqrt{\frac{-1 + \sqrt{17}}{2}} \cos t, \qquad y(t) = \sqrt{\frac{-1 + \sqrt{17}}{2}} \sin t ]

3. Determine the $z$ component: From the paraboloid equation: [ z(t) = x(t)^2 + y(t)^2 = \frac{-1 + \sqrt{17}}{2} ] Note that $z$ is constant here because the intersection occurs at a fixed height.

4. Final Vector Function: [ \mathbf{r}(t) = \left\langle \sqrt{\frac{-1 + \sqrt{17}}{2}} \cos t, ;\sqrt{\frac{-1 + \sqrt{17}}{2}} \sin t, ;\frac{-1 + \sqrt{17}}{2} \right\rangle, \quad t \in [0, 2\pi). ]

General Strategy Summary

To systematically find the intersection of two surfaces:

1. Project onto a coordinate plane by eliminating one variable (typically $z$) from the system of equations. This gives the "shadow" curve in 2D.
2. Parameterize the projected curve using standard techniques (e.g., trigonometric functions for circles, lines, etc.).
3. Determine the third coordinate by substituting the parameterized expressions back into one of the original surface equations.
4. Write the vector function $\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle$ with appropriate domain restrictions.

This method ensures a structured approach to converting implicit surface intersections into explicit parametric forms, which are crucial for further analysis in vector calculus.