Question

Prove that if \(\mathbf{v}(t)\) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t) .\) Prove the converse that if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t),\) then \(|\mathbf{v}(t)|\) is constant. [Hint: Consider the derivative of \(\mathbf{v}^{2}\).] This is a very handy result. It explains why, in two- dimensional polars, \(d \hat{\mathbf{r}} / d t\) has to be in the direction of \(\hat{\boldsymbol{\phi}}\) and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

Short answer

If \(|\mathbf{v}(t)|\) is constant, \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\), and vice versa.

Step-by-step solution

Establish the expression for the magnitude of the vector

Start by considering a time-dependent vector \(\mathbf{v}(t)\) with constant magnitude. The magnitude squared is given by \(\mathbf{v}(t) \cdot \mathbf{v}(t)\). Since the magnitude is constant, the derivative with respect to time of the magnitude squared must be zero.

Differentiate magnitude squared with respect to time

We have \(\frac{d}{dt} (\mathbf{v}(t) \cdot \mathbf{v}(t)) = 2 \mathbf{v}(t) \cdot \dot{\mathbf{v}}(t)\). Since the magnitude is constant, this derivative is zero: \(\mathbf{v}(t) \cdot \dot{\mathbf{v}}(t) = 0\). This equation shows that \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\).

Prove converse - If \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\), then magnitude is constant

Assume that \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\), meaning \(\mathbf{v}(t) \cdot \dot{\mathbf{v}}(t) = 0\). Consider the time derivative of the magnitude squared: \(\frac{d}{dt} (\mathbf{v}(t) \cdot \mathbf{v}(t)) = 2 \mathbf{v}(t) \cdot \dot{\mathbf{v}}(t) = 0 \). Since the derivative is zero, \(\mathbf{v}(t) \cdot \mathbf{v}(t)\) is constant, indicating that the magnitude of \(\mathbf{v}(t)\) is constant.

Conclusion of the proof

Both parts of the proof are now established: if the magnitude of \(\mathbf{v}(t)\) is constant, \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\), and conversely, if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\), then the magnitude is constant. Thus, the required properties are mutually inclusive.

Key concepts

Vector calculus

Vector calculus is an important field of mathematics that extends ordinary calculus to vector-valued functions. It is particularly useful in physics and engineering, where we often deal with quantities that have both magnitudes and directions, like force and velocity.

The basics of vector calculus involve operations like addition, scalar multiplication, and dot product. Dot product, in particular, is crucial when analyzing problems concerning orthogonal vectors. Two vectors are orthogonal if their dot product is zero. This concept is leveraged in the exercise solution.

In dealing with time-dependent vectors in vector calculus, you often perform differentiation with respect to time. This helps in understanding how these vectors change over time, which is essential for dynamics and kinematics.

Time-dependent vectors

Time-dependent vectors refer to vectors whose magnitude and/or direction can change over time. These vectors are commonly found in motion contexts, such as the velocity of a moving object. Movement in physics is often described using time-dependent vectors to capture the dynamic nature of real-world scenarios.

Understanding time-dependent vectors is key in analyzing how an object moves. For example, velocity \(\mathbf{v}(t)\) and acceleration vectors depict how an object's state changes as time progresses. When studying these vectors, calculus is employed to find derivatives, which provide information about rates of change.

In the context of the original exercise, we see a time-dependent vector \(\mathbf{v}(t)\), which we differentiate to understand its changes over time, revealing its relationship with its derivative \(\dot{\mathbf{v}}(t)\).

Derivative of vector magnitude

The derivative of a vector's magnitude with respect to time provides insight into how the size of the vector changes as time passes. This is especially important when considering vectors with constant magnitude but changing directions.

To determine the rate of change of a vector's magnitude, we look at the derivative of the square of the vector's magnitude. For a vector \(\mathbf{v}(t)\) with constant magnitude, the magnitude squared is \(\mathbf{v}(t) \cdot \mathbf{v}(t)\). Differentiating this gives: \(\frac{d}{dt}(\mathbf{v}(t) \cdot \mathbf{v}(t)) = 2 \mathbf{v}(t) \cdot \dot{\mathbf{v}}(t)\).

From the exercise, since the magnitude is constant, the derivative of this expression is zero, confirming that \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t)\).

Velocity and acceleration

Velocity and acceleration are fundamental concepts in physics. They are both vector quantities. Velocity \(\mathbf{v}(t)\) indicates the speed and direction an object is moving, while acceleration \(\dot{\mathbf{v}}(t)\) describes how the velocity changes over time.

When a vector like velocity has constant magnitude, its corresponding acceleration vector must be orthogonal to it. This is because any change in direction, without changing speed, corresponds to a perpendicular acceleration. This is highlighted in the exercise's derivation.

Understanding that acceleration is often perpendicular to velocity in systems with constant speed is important in many physical contexts, such as objects moving in a circle or in the presence of a magnetic field.

Constant magnitude vectors

Vectors with constant magnitude have a fixed length, although their direction can change. A common example is uniform circular motion, where the speed remains constant, but direction continuously changes at every point along the path.

The exercise emphasizes that the derivative of the vector magnitude remains zero for constant magnitude vectors. This means that only the direction component of the vector changes.

• Constant magnitude in velocity implies motion is at a constant speed.
• Any change in the vector involves a perpendicular component, which is why acceleration (\(\dot{\mathbf{v}}(t)\)) is orthogonal to the vector (\(\mathbf{v}(t)\)).

This principle is useful for understanding circular motion and magnetic forces on charged particles, among other phenomena.