Question

Motion on a sphere 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 given curve lies on the surface of a sphere centered at the origin if and only if the position vector (r) and its derivative (r') are orthogonal at all points. Answer: To prove this, we followed a two-step process. First, we showed that if r and r' are orthogonal, then the curve lies on the sphere's surface by demonstrating that the dot product of the position vector and its derivative is zero, which implies the curve satisfies the sphere equation. Second, we assumed that the curve lies on the surface of the sphere and differentiated the sphere equation to find that the dot product of r and r' equals zero, proving that the position vector and its derivative are orthogonal at all points of the curve.

Step-by-step solution

If r and r' are orthogonal, then the curve lies on the sphere

To prove this, we need to show that the equation for the sphere \((x^2 + y^2 + z^2 = a^2)\) is satisfied if the dot product of the position vector \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}^{\prime}(t)\) is zero. Let \(\mathbf{r}(t)=\)(x(t), y(t), z(t)) be the position vector of a point on the curve and assume that: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0\) This means that: \(x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0\) Now, we differentiate both sides of the sphere equation with respect to t: \(2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) = 0\) We can see that this equation is equivalent to the dot product of r and r', so if \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0\), then the curve must satisfy the sphere equation.

If the curve lies on the surface of the sphere, then r and r' are orthogonal

To prove this, we will assume that the curve lies on the surface of the sphere, and we will show that the dot product of r and r' equals zero. The curve lying on the surface of the sphere means that: \(x^2(t) + y^2(t) + z^2(t) = a^2\) Now, we differentiate both sides of the equation with respect to t: \(2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) = 0\) Dividing both sides by 2, we get: \(x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0\) which means that: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t) = 0\) This concludes the proof that if the curve lies on the surface of the sphere, then the position vector and its derivative are orthogonal at all points of the curve.

Key concepts

Orthogonal Vectors

In mathematics, when we refer to orthogonal vectors, we are discussing vectors that are perpendicular to one another. This is a fundamental concept in geometry and vector analysis. Orthogonality is determined by the dot product of the two vectors. If this dot product is zero, then the vectors are orthogonal. For example, given two vectors \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \), they are orthogonal if:
\[ a_1b_1 + a_2b_2 + a_3b_3 = 0 \]

• This property is crucial when dealing with curves and paths in three-dimensional space, such as the motion on a sphere.
• Orthogonal vectors imply that no change in one vector's direction affects the other, maintaining balance in their spatial orientation.

Understanding orthogonal vectors helps us analyze and interpret various physical and mathematical phenomena, such as the movement and orientation of objects in 3D space.

Dot Product

The dot product is a way of multiplying two vectors that results in a scalar—a single number. It provides a measure of how much two vectors point in the same direction. A fundamental property of the dot product relevant to motion on a sphere is its relation to orthogonal vectors.

The dot product of two vectors \( \mathbf{u} = [u_1, u_2, u_3] \) and \( \mathbf{v} = [v_1, v_2, v_3] \) is calculated as:
\[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]

• When the dot product is zero, the vectors are orthogonal, indicating they have no direct correlation in terms of direction.
• The dot product can also be used to find the angle \( \theta \) between two vectors with the formula: \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta) \).

For position vectors on a sphere, we ensure that any tangent (change in direction) does not diverge from the surface by being orthogonal to the radius vector at all times.

Differential Calculus

Differential calculus is a branch of mathematical analysis that focuses on the study of how functions change. Central to differential calculus is the concept of derivatives, which represent the instantaneous rate of change of a function relative to one of its variables. This is pivotal in understanding motion on a sphere.

In relation to vectored motion, differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \) produces the velocity vector \( \mathbf{r}'(t) \), indicating how the position changes over time.

• Knowing \( \mathbf{r}'(t) \) helps us describe motion along a curve, and understanding this helps predict future positions or reconstruct past ones.
• By applying the derivative to the sphere equation \( x^2 + y^2 + z^2 = a^2 \), we explore how movement affects the position over time while remaining on the sphere.

Differential calculus allows us to analyze changes, making it essential for solving problems where precise calculations of velocity and acceleration are key, as demonstrated by the derivatives in proving orthogonality on a sphere.