Let's break down the problem and provide a step-by-step solution, including the plot of the polar curve. ### **Given:** - Polar equation: \( r = e^{\theta/2} \) - Interval: \( 0 \leq \theta \leq \frac{\pi}{2} \) --- ## **Step 1: Understanding the Equation** - In polar coordinates, \( r \) is the distance from the origin, and \( \theta \) is the angle from the positive x-axis. - Here, \( r \) increases exponentially as \( \theta \) increases. --- ## **Step 2: Calculate Key Points** Let’s calculate a few values to understand the curve: | \( \theta \) | \( r = e^{\theta/2} \) | |------------------------|----------------------------| | \( 0 \) | \( e^{0/2} = 1 \) | | \( \frac{\pi}{6} \) | \( e^{\pi/12} \approx 1.284 \) | | \( \frac{\pi}{4} \) | \( e^{\pi/8} \approx 1.454 \) | | \( \frac{\pi}{2} \) | \( e^{\pi/4} \approx 2.193 \) | --- ## **Step 3: Plotting the Curve** To plot this, you can use polar graph paper, or (in Python/Matplotlib): ```python import numpy as np import matplotlib.pyplot as plt theta = np.linspace(0, np.pi/2, 200) r = np.exp(theta / 2) plt.figure(figsize=(6,6)) ax = plt.subplot(111, polar=True) ax.plot(theta, r) ax.set_title(r"$r = e^{\theta/2}$, $0 \leq \theta \leq \frac{\pi}{2}$") plt.show() ``` --- ## **Step 4: Explanation** - The curve starts at \( r = 1 \) when \( \theta = 0 \). - As \( \theta \) increases, \( r \) increases exponentially. - At \( \theta = \frac{\pi}{2} \), \( r = e^{\pi/4} \approx 2.193 \). - The curve will be a smooth outward spiral in the first quadrant. --- ## **Step 5: Final Answer (Description and Plot)** **Final Answer:** The polar curve \( r = e^{\theta/2} \) for \( 0 \leq \theta \leq \frac{\pi}{2} \) starts at \( (1, 0^\circ) \) and spirals outward in the first quadrant, ending at approximately \( (2.193, 90^\circ) \). **Plot:**  *(If you run the provided Python code, you will get this plot!)* --- If you need the plot in another form or more explanation, let me know!

Let's provide a detailed step-by-step explanation with calculations for plotting the polar curve given by: \[ r = e^{\frac{\theta}{2}}, \quad \text{for } 0 \leq \theta \leq \frac{\pi}{2} \] --- ### **Step 1: Understand the given polar equation** - The formula relates the radius \( r \) to the angle \( \theta \). - Since the exponential function is involved, the radius increases exponentially as \( \theta \) increases. - The interval for \( \theta \) is from 0 to \( \frac{\pi}{2} \). --- ### **Step 2: Calculate key points** Let's compute \( r \) at important values of \( \theta \): **At \( \theta = 0 \):** \[ r = e^{0/2} = e^{0} = 1 \] **At \( \theta = \frac{\pi}{6} \) (30°):** \[ r = e^{(\pi/6)/2} = e^{\pi/12} \] Numerically: \[ \pi \approx 3.1416 \] \[ \pi/12 \approx 0.2618 \] \[ r \approx e^{0.2618} \approx 1.299 \] **At \( \theta = \frac{\pi}{4} \) (45°):** \[ r = e^{(\pi/4)/2} = e^{\pi/8} \] Calculate: \[ \pi/8 \approx 0.3927 \] \[ r \approx e^{0.3927} \approx 1.481 \] **At \( \theta = \frac{\pi}{2} \) (90°):** \[ r = e^{(\pi/2)/2} = e^{\pi/4} \] Calculate: \[ \pi/4 \approx 0.7854 \] \[ r \approx e^{0.7854} \approx 2.193 \] --- ### **Step 3: Summarize key points** | \( \theta \) | \( r = e^{\theta/2} \) | Approximate \( r \) | |--------------|------------------------|---------------------| | 0 | \( e^{0} \) | 1.000 | | \( \pi/6 \) | \( e^{\pi/12} \) | 1.299 | | \( \pi/4 \) | \( e^{\pi/8} \) | 1.481 | | \( \pi/2 \) | \( e^{\pi/4} \) | 2.193 | --- ### **Step 4: Plotting the curve** To visualize: - Use polar coordinate plotting tools or software. - For example, in Python with `matplotlib`, use the following code: ```python import numpy as np import matplotlib.pyplot as plt # Generate theta values from 0 to pi/2 theta = np.linspace(0, np.pi/2, 200) # Calculate corresponding r values r = np.exp(theta / 2) # Plot in polar coordinates plt.figure(figsize=(6,6)) ax = plt.subplot(111, polar=True) ax.plot(theta, r) ax.set_title(r'$r = e^{\theta/2}$ for $0 \leq \theta \leq \pi/2$') plt.show() ``` This will produce a smooth spiral starting at \( r=1 \) and expanding outward to about \( r=2.193 \). --- ### **Step 5: Interpretation of the plot** - The curve begins at the point: \[ \text{At } \theta=0, \quad r=1 \quad \Rightarrow \quad \text{Point } (r, \theta) = (1, 0^\circ) \] - As \( \theta \) approaches \( \frac{\pi}{2} \), \( r \) approaches approximately 2.193, so the point moves outward in the first quadrant. - The curve is a **spiral** that outwardly expands as \( \theta \) increases. --- ### **Final Answer:** - **Equation:** \( r = e^{\theta/2} \) - **Interval:** \( 0 \leq \theta \leq \frac{\pi}{2} \) - **Key points:** - \( (r, \theta) = (1, 0^\circ) \) - \( (1.299, 30^\circ) \) - \( (1.481, 45^\circ) \) - \( (2.193, 90^\circ) \) - **Description:** The curve is a smooth exponential spiral starting at radius 1 when \( \theta=0 \) and expanding outward to radius approximately 2.193 at \( \theta=\pi/2 \), covering the first quadrant. --- If you'd like, I can help generate the plot or analyze further properties!