Problem 73 - Calculus 3

Question

Show that if the speed of a particle traveling along a curve represented by a vector-valued function is constant, then the velocity function is always perpendicular to the acceleration function.

Step-by-Step Solution

Answer

Velocity is perpendicular to acceleration if speed is constant.

Step 1: Define the Vector-Valued Function

Let the position of the particle be described by the vector-valued function \( \mathbf{r}(t) \). The velocity of the particle is given by \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \), and the acceleration is \( \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} \).

Step 2: Use Constant Speed Condition

Given that the speed of the particle is constant, the magnitude of the velocity vector \( \|\mathbf{v}(t)\| \) is constant. Thus, \( \frac{d}{dt}\|\mathbf{v}(t)\|^2 = 0 \).

Step 3: Differentiate Magnitude Squared

The magnitude squared of the velocity is \( \|\mathbf{v}(t)\|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t) \). Differentiating both sides gives: \( \frac{d}{dt}[\mathbf{v}(t) \cdot \mathbf{v}(t)] = 2\mathbf{v}(t) \cdot \frac{d\mathbf{v}}{dt} = 0 \).

Step 4: Simplify to the Perpendicularity Condition

Since \( 2\mathbf{v}(t) \cdot \mathbf{a}(t) = 0 \) leads to \( \mathbf{v}(t) \cdot \mathbf{a}(t) = 0 \), this implies that the velocity vector \( \mathbf{v}(t) \) is perpendicular to the acceleration vector \( \mathbf{a}(t) \). This is because the dot product equaling zero indicates perpendicularity.

Key Concepts

Constant Speed

When we say that a particle is moving with constant speed, we're referring to the fact that its rate of motion doesn't change as it travels along its path. For particles described by a vector-valued function

• The speed is the magnitude of the velocity vector, expressed as \( \| \mathbf{v}(t) \| \).
• If the speed is constant, \( \| \mathbf{v}(t) \| \) remains the same at all times.

To explore this concept, consider that the derivative of the square of the magnitude of the velocity (\( \frac{d}{dt}\| \mathbf{v}(t) \|^2 \) ) is zero. This happens because

• \( \| \mathbf{v}(t) \|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t) \), the dot product of velocity with itself.
• Therefore, differentiating this expression also results in a constant because the magnitudes do not change over time.

Velocity and Acceleration

Velocity and acceleration are fundamental concepts in motion.

• Velocity \( \mathbf{v}(t) \) is the rate of change of the particle's position. It's derived as the first derivative of the position vector \( \mathbf{r}(t) \).
• Acceleration \( \mathbf{a}(t) \) is the rate of change of velocity. We find it by differentiating the velocity vector.

For a particle moving with constant speed, the relationship between velocity and acceleration becomes particularly interesting.
In this scenario:

• Velocity and acceleration vectors exhibit a unique relationship as shown through their mathematical operations.
• The inner workings of these vectors reveal further insights, especially when evaluated for their perpendicularity condition.

Perpendicular Vectors

The concept of perpendicular vectors is essential when analyzing velocity and acceleration in the context of constant speed. Perpendicularity happens when:

• The dot product of two vectors equals zero: \( \mathbf{v}(t) \cdot \mathbf{a}(t) = 0 \).
• This absence of a dot product indicates the vectors are orthogonal, or in simpler terms, their paths cross at a right angle to one another.

With constant speed, this perpendicular relationship shows us:

• That the velocity vector is focused on maintaining speed along the path.
• All acceleration merely acts to redirect the path but does not alter the speed.

Understanding that these vectors are perpendicular helps to develop a deeper grasp of motion dynamics, emphasizing how motion paths reconfigure without impacting overall speed.