Question

Prove that \(\mathbf{r}\) describes a curve that lies on the surface of a sphere centered at the origin \(\left(x^{2}+y^{2}+z^{2}=a^{2}\right.\) with \(a \geq 0\) ) if and only if \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points of the curve.

Short answer

Question: Prove that a curve with parametric equations \(\mathbf{r}(t)\) lies on the surface of a sphere centered at the origin if and only if \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points of the curve. Answer: To prove that a curve \(\mathbf{r}\) lies on the surface of a sphere centered at the origin, we need to show that the equation \(x^2 + y^2 + z^2 = a^2\) holds true for the given curve. By taking the dot product of \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) and integrating with respect to \(t\), we arrive at the equation of the sphere, \(x^2(t) + y^2(t) + z^2(t) = a^2\). This is possible if and only if \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points of the curve.

Step-by-step solution

Computing the dot product of \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\)

Let \(\mathbf{r} = \begin{pmatrix} x(t) \\ y(t) \\ z(t) \end{pmatrix}\) be a parametric representation of the curve. The derivatives of \(x(t)\), \(y(t)\), and \(z(t)\) with respect to the parameter \(t\) will give us the components of the tangent vector \(\mathbf{r}^{\prime}\). Hence, \(\mathbf{r}^{\prime} = \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix}\). Now, calculating the dot product of \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\): $$\mathbf{r} \cdot \mathbf{r}^\prime = x(t)x'(t) + y(t)y'(t) + z(t)z'(t)$$ Since \(\mathbf{r}\) and \(\mathbf{r}^\prime\) are orthogonal, their dot product is zero: $$x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0$$

Integrating both sides with respect to \(t\)

Now we integrate both sides with respect to \(t\): $$\int (x(t)x'(t) + y(t)y'(t) + z(t)z'(t)) dt = \int 0 dt$$ Integration gives us: $$\frac{x^2(t)}{2} + \frac{y^2(t)}{2} + \frac{z^2(t)}{2} = C \; (\text{where } C \text{ is the constant of integration})$$

Determining the equation of the sphere

Finally, multiply the entire equation by 2: $$x^2(t) + y^2(t) + z^2(t) = 2C$$ We recognize that \(2C\) is, in fact, the radius squared of the sphere. Let \(a^2 = 2C\): $$x^2(t) + y^2(t) + z^2(t) = a^2$$

Conclusion

Thus, the curve with parametric equations \(\mathbf{r}(t)\) lies on the surface of a sphere centered at the origin if and only if \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points of the curve. We have proved this by reaching the equation of the sphere using the given conditions.

Key concepts

Parametric Equations

Parametric equations are a versatile tool in mathematics, particularly useful for describing the paths of objects in motion or the shape of complex curves and surfaces. They represent functions of one or more variables, called parameters.
For instance, a parametric equation may define a curve in three-dimensional space as

• \(x(t) = \text{function of } t\)
• \(y(t) = \text{function of } t\)
• \(z(t) = \text{function of } t\)

The variable \(t\) in this context is known as a parameter. As \(t\) varies, \(x(t), y(t), \) and \(z(t)\) trace out a path in space. This technique allows us to model and explore curves that may not be easy to describe using just traditional coordinates.
In the context of the exercise provided, parametric equations define a vector \( \mathbf{r} = \begin{pmatrix} x(t) \cr y(t) \cr z(t) \end{pmatrix} \). The motion or position of \(\mathbf{r}\) is depicted by the functions \(x(t), y(t),\) and \(z(t)\). These equations are essential for revealing the nature of the paths, such as whether they trace along the surface of a sphere.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra that results in a scalar. It measures the extent to which two vectors align with each other. In three-dimensional space, for vectors \(\mathbf{a} = \begin{pmatrix} a_1 \cr a_2 \cr a_3 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} b_1 \cr b_2 \cr b_3 \end{pmatrix}\), the dot product is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]If the dot product is zero, the vectors are orthogonal, or perpendicular, to each other. This concept was crucial in the original exercise, where the vectors \(\mathbf{r}\) and its derivative \(\mathbf{r}^{\prime}\) are proven to be orthogonal to ensure the path lies on a sphere.
Orthogonality through the dot product highlights an essential property: when a position vector and its rate of change (derivative) are orthogonal, it indicates that the motion is restricted in such a way that keeps the path on the desired geometric surface, like a sphere. Therefore, understanding the dot product is key in analyzing and proving the behavior of vector-based parametric paths.

Curve on a Sphere

A sphere is a perfectly symmetrical shape in three-dimensional space, defined mathematically by the equation \[ x^2 + y^2 + z^2 = r^2 \]where \(r\) is the sphere's radius. A curve on a sphere implies that every point along the curve satisfies this equation for a given \(r\). In the given exercise, the goal was to verify under what conditions a curve, described geometrically by a parametric vector \(\mathbf{r}(t)\), indeed lies on such a sphere.
To ensure a curve remains on the sphere, the relationship between the curve’s position and its change in position (i.e., its derivative) becomes vital. When the vector \(\mathbf{r}\), representing a point on the curve, is orthogonal to \(\mathbf{r}^{\prime}\), the tangent vector, every stage of this movement restricts the path to the sphere’s surface.
This geometrical constraint means that while the point changes its position, it remains equidistant from the center (the origin in this case), preserving the equation of the sphere—\(x^2(t) + y^2(t) + z^2(t) = a^2\), where \(a\) is the chosen radius of the sphere. Hence, understanding how curves interact with spheres in a vector context helps in examining several physical and mathematical phenomena.