Let's solve each problem step-by-step: --- ### **13. Graph a Polar function given by \( r = 2 \sin 3\theta \)** #### **Step 1: Recognize the Type of Polar Graph** - The equation \( r = a \sin(n\theta) \) is a "rose curve". - Here, \( a = 2 \) and \( n = 3 \). #### **Step 2: Key Features** - Since \( n = 3 \) (odd), the rose curve will have **3 petals**. - The maximum value of \( r \) is 2 (when \( \sin 3\theta = 1 \)). - The petals are spaced evenly around the origin. #### **Step 3: Petal Angles** - Petals occur where \( \sin 3\theta = 1 \Rightarrow 3\theta = \frac{\pi}{2} + k\pi \) - \( \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} \) #### **Step 4: Sketch the Graph** - Draw 3 petals, each reaching out to a length of 2, at \( \theta = 30^\circ, 150^\circ, 270^\circ \). --- ### **14. Find the derivative \(\frac{dy}{dx}\) of the parametric set:** \[ x(t) = 3t^2 + t - 11 \] \[ y(t) = \sqrt{2t} + 1 \] #### **Step 1: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)** - \(\frac{dx}{dt} = \frac{d}{dt}(3t^2 + t - 11) = 6t + 1\) - \(\frac{dy}{dt} = \frac{d}{dt}(\sqrt{2t} + 1) = \frac{1}{2}(2t)^{-1/2} \cdot 2 = (2t)^{-1/2} = \frac{1}{\sqrt{2t}}\) #### **Step 2: Express \(\frac{dy}{dx}\)** \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{\sqrt{2t}}}{6t + 1} \] \[ \boxed{\frac{dy}{dx} = \frac{1}{(6t + 1)\sqrt{2t}}} \] --- ### **Final Answers** **13.** _Graph:_ A rose curve with 3 petals, each of length 2, centered at \( \theta = 30^\circ, 150^\circ, 270^\circ \). **14.** \[ \boxed{\frac{dy}{dx} = \frac{1}{(6t + 1)\sqrt{2t}}} \] If you need the polar graph drawn, let me know!

Certainly! Here's a step-by-step guide to sketching the graph of the polar function \( r = 2 \sin 3\theta \): --- ### **Graphing \( r = 2 \sin 3\theta \) (Rose Curve with 3 Petals)** #### **Step 1: Understand the equation** - It is a rose curve with: - \( a = 2 \) - \( n = 3 \) (odd) - For \( r = a \sin n\theta \), if \( n \) is odd: - Number of petals = \( n \) - The petals are symmetric and evenly spaced. #### **Step 2: Key points** - Maximum \( r \) = 2 (when \( \sin 3\theta = 1 \)) - Petal tips occur at angles where \( \sin 3\theta = 1 \): - \( 3\theta = \frac{\pi}{2} + 2k\pi \) - For \( k=0 \): - \( 3\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{6} = 30^\circ \) - For \( k=1 \): - \( 3\theta = \frac{5\pi}{2} \Rightarrow \theta = \frac{5\pi}{6} = 150^\circ \) - For \( k=2 \): - \( 3\theta = \frac{9\pi}{2} \Rightarrow \theta = \frac{3\pi}{2} = 270^\circ \) These are the angles where petals reach their maximum length. #### **Step 3: Plot key points** - **At \( \theta = 30^\circ \):** \( r=2 \) - **At \( \theta = 150^\circ \):** \( r=2 \) - **At \( \theta = 270^\circ \):** \( r=2 \) Between these angles, \( r \) varies smoothly, reaching zero at intermediate points. #### **Step 4: Draw petals** - From the origin, draw three petal loops: - Petal 1 centered at \( 30^\circ \), extending out to \( r=2 \) - Petal 2 centered at \( 150^\circ \), extending out to \( r=2 \) - Petal 3 centered at \( 270^\circ \), extending out to \( r=2 \) - The petals will be symmetric and have a petal shape, with the loops passing through the origin at angles where \( \sin 3\theta = 0 \): - \( 3\theta = 0, \pi, 2\pi, \ldots \) - These zeros occur at: - \( \theta=0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ \) #### **Step 5: Final sketch notes** - Draw smooth loops for each petal. - The entire graph is symmetric about the origin. - The rose will have 3 petals, each reaching out to length 2. --- ### **Summary:** - The rose curve has 3 petals. - Petals are centered at \( 30^\circ, 150^\circ, 270^\circ \). - Each petal extends up to \( r=2 \). --- If you'd like, I can generate a visual sketch for you! However, since I can't draw directly here, you can follow these steps on graph paper or use graphing software like Desmos: **Desmos polar graph input:** ```plaintext r = 2 * sin(3 * θ) ``` Set the mode to polar, and you'll see the rose curve with 3 petals. --- Let me know if you'd like further assistance!