Let's solve the problem step by step: ### **Given:** - \( x = \sin(t) \) - \( y = 3\cos(t) \) - \( t = \frac{\pi}{3} \) We are to **find the slope of the tangent line** at \( t = \frac{\pi}{3} \), **then find the equation of the tangent line**. --- ### **Step 1: Find \( \frac{dy}{dx} \) (Slope of the Tangent Line)** Since \( x \) and \( y \) are both given in terms of \( t \), use the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] Calculate derivatives: - \( \frac{dx}{dt} = \cos(t) \) - \( \frac{dy}{dt} = 3(-\sin(t)) = -3\sin(t) \) So, \[ \frac{dy}{dx} = \frac{-3\sin(t)}{\cos(t)} \] --- ### **Step 2: Evaluate at \( t = \frac{\pi}{3} \)** - \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) - \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) Plug these in: \[ \frac{dy}{dx} = \frac{-3 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{-\frac{3\sqrt{3}}{2}}{\frac{1}{2}} = -3\sqrt{3} \] **Slope = \( -3\sqrt{3} \)** --- ### **Step 3: Find the Point \((x, y)\) at \( t = \frac{\pi}{3} \)** - \( x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \) - \( y = 3\cos\left(\frac{\pi}{3}\right) = 3 \cdot \frac{1}{2} = \frac{3}{2} \) So the point is \( \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \). --- ### **Step 4: Equation of Tangent Line** Using point-slope form: \( y - y_1 = m(x - x_1) \) \[ y - \frac{3}{2} = -3\sqrt{3}\left(x - \frac{\sqrt{3}}{2}\right) \] --- ## **Final Answers** **Slope:** \[ \boxed{-3\sqrt{3}} \] **Equation:** \[ \boxed{y - \frac{3}{2} = -3\sqrt{3}\left(x - \frac{\sqrt{3}}{2}\right)} \] Or, you can expand it if needed: \[ y = -3\sqrt{3}x + 3\sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{3}{2} = -3\sqrt{3}x + \frac{9}{2} + \frac{3}{2} = -3\sqrt{3}x + 6 \] So, \[ \boxed{y = -3\sqrt{3}x + 6} \]

Let's carefully work through this problem step by step, providing detailed calculations and explanations. --- ## **Given:** - Parametric equations: \[ x = \sin(t) \] \[ y = 3 \cos(t) \] - At \( t = \frac{\pi}{3} \). --- ## **Objective:** Find: 1. The **slope of the tangent line** at \( t = \frac{\pi}{3} \). 2. The **equation of the tangent line** at that point. --- ## **Step 1: Find \(\frac{dy}{dx}\) in terms of \(t\)** Since both \(x\) and \(y\) are functions of \(t\), the derivative \(\frac{dy}{dx}\) can be found using **chain rule**: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] ### **Calculate \(\frac{dx}{dt}\):** \[ x = \sin(t) \implies \frac{dx}{dt} = \cos(t) \] ### **Calculate \(\frac{dy}{dt}\):** \[ y = 3 \cos(t) \implies \frac{dy}{dt} = 3 \times (-\sin(t)) = -3 \sin(t) \] ### **Therefore:** \[ \frac{dy}{dx} = \frac{-3 \sin(t)}{\cos(t)} \] --- ## **Step 2: Evaluate \(\frac{dy}{dx}\) at \( t = \frac{\pi}{3} \)** Recall the trigonometric values: \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Plug into the derivative: \[ \frac{dy}{dx} = \frac{-3 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} \] Simplify numerator: \[ -3 \times \frac{\sqrt{3}}{2} = -\frac{3 \sqrt{3}}{2} \] Now divide by \(\frac{1}{2}\): \[ \frac{-\frac{3 \sqrt{3}}{2}}{\frac{1}{2}} = -\frac{3 \sqrt{3}}{2} \times \frac{2}{1} = -3 \sqrt{3} \] ### **So, the slope of the tangent line at \( t = \frac{\pi}{3} \) is:** \[ \boxed{-3 \sqrt{3}} \] --- ## **Step 3: Find the coordinates \((x, y)\) at \( t = \frac{\pi}{3} \)** Calculate \(x\) and \(y\): \[ x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ y = 3 \cos\left(\frac{\pi}{3}\right) = 3 \times \frac{1}{2} = \frac{3}{2} \] **Point on the curve:** \[ \boxed{\left( \frac{\sqrt{3}}{2}, \frac{3}{2} \right)} \] --- ## **Step 4: Write the equation of the tangent line** Using **point-slope form**: \[ y - y_1 = m (x - x_1) \] where: - \(m = -3 \sqrt{3}\) - \( (x_1, y_1) = \left( \frac{\sqrt{3}}{2}, \frac{3}{2} \right) \) Plug in: \[ y - \frac{3}{2} = -3 \sqrt{3} \left( x - \frac{\sqrt{3}}{2} \right) \] --- ## **Step 5: Simplify the equation** Distribute: \[ y - \frac{3}{2} = -3 \sqrt{3} x + (-3 \sqrt{3}) \times \left( -\frac{\sqrt{3}}{2} \right) \] Calculate the second term: \[ -3 \sqrt{3} \times -\frac{\sqrt{3}}{2} = \left(-3 \sqrt{3}\right) \times \left(-\frac{\sqrt{3}}{2}\right) \] Multiply numerator: \[ -3 \sqrt{3} \times -\sqrt{3} = (-3) \times (-1) \times \sqrt{3} \times \sqrt{3} = 3 \times 1 \times 3 = 9 \] Divide by 2: \[ \frac{9}{2} \] So the equation becomes: \[ y - \frac{3}{2} = -3 \sqrt{3} x + \frac{9}{2} \] Add \(\frac{3}{2}\) to both sides: \[ y = -3 \sqrt{3} x + \frac{9}{2} + \frac{3}{2} = -3 \sqrt{3} x + \frac{12}{2} = -3 \sqrt{3} x + 6 \] --- ## **## **Final Result:** ### **Equation of the tangent line:** \[ \boxed{ \boxed{ y = -3 \sqrt{3} x + 6 } } \] ### **Summary:** - **Slope at \( t = \frac{\pi}{3} \):** \(\boxed{-3 \sqrt{3}}\) - **Equation of the tangent line:** \[ \boxed{ y = -3 \sqrt{3} x + 6 } \] --- Let me know if you'd like any further clarification!