The Speed Of a particle moving in a plane is equal to the magnitude of its instantaneous velocity, `v=|vl= sqrt(v_x^2+v_y^2).`
(a) Show that the rate of change of the speed is `(dv)/(dt)=(v_xa_x+v_ya_y)/(sqrt(v_x^2+v_y^2))`.
(b) Show that the rate of change of speed can be expressed as `(dv)/(dt)` is equal to`a_t` the component of a that is parallelto v.

Explanation

To solve the problem step by step, we will address both parts (a) and (b) of the question regarding the rate of change of speed of a particle moving in a plane.

### Part (a)

1. **Understanding Speed and Velocity**:
The speed \( v \) of a particle is given by the magnitude of its instantaneous velocity:
\[ v = |v| = \sqrt{v_x^2 + v_y^2} \]
where \( v_x \) and \( v_y \) are the components of the velocity in the x and y directions, respectively.

2. **Differentiating Speed with Respect to Time**:
To find the rate of change of speed, we differentiate \( v \) with respect to time \( t \):
\[ \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{v_x^2 + v_y^2} \right) \]

3. **Applying the Chain Rule**:
Using the chain rule for differentiation:
\[ \frac{dv}{dt} = \frac{1}{2\sqrt{v_x^2 + v_y^2}} \cdot \frac{d}{dt}(v_x^2 + v_y^2) \]

4. **Differentiating the Inside Function**:
Now, we differentiate \( v_x^2 + v_y^2 \):
\[ \frac{d}{dt}(v_x^2 + v_y^2) = 2v_x \frac{dv_x}{dt} + 2v_y \frac{dv_y}{dt} \]
Here, \( \frac{dv_x}{dt} = a_x \) and \( \frac{dv_y}{dt} = a_y \) are the components of acceleration.

5. **Substituting Back**:
Substituting this back into our equation, we have:
\[ \frac{dv}{dt} = \frac{1}{2\sqrt{v_x^2 + v_y^2}} \cdot (2v_x a_x + 2v_y a_y) \]

6. **Simplifying**:
Simplifying gives:
\[ \frac{dv}{dt} = \frac{v_x a_x + v_y a_y}{\sqrt{v_x^2 + v_y^2}} \]

Thus, we have shown that:
\[ \frac{dv}{dt} = \frac{v_x a_x + v_y a_y}{\sqrt{v_x^2 + v_y^2}} \]

### Part (b)

1. **Understanding the Components of Acceleration**:
The acceleration vector \( \mathbf{a} \) can be expressed in terms of its components:
\[ \mathbf{a} = a_x \hat{i} + a_y \hat{j} \]

2. **Finding the Component of Acceleration Parallel to Velocity**:
The component of acceleration \( a_t \) that is parallel to the velocity \( \mathbf{v} \) can be found using the dot product:
\[ a_t = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} \]

3. **Calculating the Dot Product**:
The dot product \( \mathbf{a} \cdot \mathbf{v} \) is:
\[ \mathbf{a} \cdot \mathbf{v} = a_x v_x + a_y v_y \]

4. **Magnitude of Velocity**:
The magnitude of the velocity is:
\[ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \]

5. **Substituting into the Expression for \( a_t \)**:
Therefore, we can express \( a_t \) as:
\[ a_t = \frac{a_x v_x + a_y v_y}{\sqrt{v_x^2 + v_y^2}} \]

6. **Relating \( \frac{dv}{dt} \) and \( a_t \)**:
From part (a), we found that:
\[ \frac{dv}{dt} = \frac{v_x a_x + v_y a_y}{\sqrt{v_x^2 + v_y^2}} \]
Thus, we can conclude that:
\[ \frac{dv}{dt} = a_t \]

### Final Results
- For part (a):
\[ \frac{dv}{dt} = \frac{v_x a_x + v_y a_y}{\sqrt{v_x^2 + v_y^2}} \]
- For part (b):
\[ \frac{dv}{dt} = a_t \]