---### Exercise 3: Orthogonality of Velocity and Acceleration Vectors**Problem Statement:**Show that if a particle moves with constant speed, then the velocity and acceleration vectors are orthogonal.**Explanation:**To solve this problem, follow the steps below:1. **Constant Speed:** - A particle moving with constant speed means the magnitude of its velocity vector `|v|` is constant. 2. **Velocity Vector:** - Denote the velocity vector by **v**. Since its speed is constant, the derivative of the speed (which is the magnitude of the velocity) with respect to time is zero: \[ \frac{d}{dt} (\| \mathbf{v} \|) = 0 \] 3. **Dot Product of the Velocity with Itself:** - Write the speed squared in terms of the dot product of the velocity vector with itself: \[ \|\mathbf{v}\|^2 = \mathbf{v} \cdot \mathbf{v} \] 4. **Differentiate the Dot Product:** - Differentiate both sides of the equation with respect to time: \[ \frac{d}{dt} (\mathbf{v} \cdot \mathbf{v}) = \frac{d}{dt} (\|\mathbf{v}\|^2) \] Since the right-hand side is zero (because the speed is constant): \[ 2 \mathbf{v} \cdot \frac{d\mathbf{v}}{dt} = 0 \] Simplifying this gives: \[ \mathbf{v} \cdot \mathbf{a} = 0 \] where $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ is the acceleration vector.5. **Conclusion:** - The result shows that the dot product of the velocity and acceleration vectors is zero: $\mathbf{v} \cdot \mathbf{a} = 0$ This implies that the velocity and acceleration vectors are orthogonal.Therefore, if a particle moves with constant speed, the velocity and acceleration vectors must be orthogonal (perpendicular) to each other.---

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3. Show that if a particle moves with constant speed, then the velocity and acceleration vectors are orthog- onal.

Calculus: Early Transcendentals

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Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

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### Exercise 3: Orthogonality of Velocity and Acceleration Vectors

**Problem Statement:**
Show that if a particle moves with constant speed, then the velocity and acceleration vectors are orthogonal.

**Explanation:**
To solve this problem, follow the steps below:

1. **Constant Speed:**
- A particle moving with constant speed means the magnitude of its velocity vector `|v|` is constant.

2. **Velocity Vector:**
- Denote the velocity vector by **v**. Since its speed is constant, the derivative of the speed (which is the magnitude of the velocity) with respect to time is zero:
\[
\frac{d}{dt} (\| \mathbf{v} \|) = 0
\]

3. **Dot Product of the Velocity with Itself:**
- Write the speed squared in terms of the dot product of the velocity vector with itself:
\[
\|\mathbf{v}\|^2 = \mathbf{v} \cdot \mathbf{v}
\]

4. **Differentiate the Dot Product:**
- Differentiate both sides of the equation with respect to time:
\[
\frac{d}{dt} (\mathbf{v} \cdot \mathbf{v}) = \frac{d}{dt} (\|\mathbf{v}\|^2)
\]
Since the right-hand side is zero (because the speed is constant):
\[
2 \mathbf{v} \cdot \frac{d\mathbf{v}}{dt} = 0
\]
Simplifying this gives:
\[
\mathbf{v} \cdot \mathbf{a} = 0
\]
where $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ is the acceleration vector.

5. **Conclusion:**
- The result shows that the dot product of the velocity and acceleration vectors is zero:
$\mathbf{v} \cdot \mathbf{a} = 0$
This implies that the velocity and acceleration vectors are orthogonal.

Therefore, if a particle moves with constant speed, the velocity and acceleration vectors must be orthogonal (perpendicular) to each other.

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