Let's break down the problem step by step, as requested. We'll **start with part (a):** finding the minimum distance between the airplanes (without wind). We'll use coordinates and relative velocity for this. --- ## Step 1: Assign Coordinates and Write Position Vectors Let's set point **A** as the origin \((,)\). Point **B** is 200 miles to the northwest of A, at a 60° angle north of west. - The coordinates of B, using trigonometry: - \(x_B = -200 \cos(60^\circ) = -100\) miles - \(y_B = 200 \sin(60^\circ) = 173.2\) miles Let \(t\) be the time (in hours) after takeoff. **Plane from B:** - Starts at (−100, 173.2) and flies due east at 350 mi/h: - \(x_B(t) = -100 + 350t\) - \(y_B(t) = 173.2\) **Plane from A:** - Starts at (,) and flies northeast (45°) at 390 mi/h: - \(x_A(t) = 390t \cos(45^\circ) = 275.8t\) - \(y_A(t) = 390t \sin(45^\circ) = 275.8t\) **Explanation:** We use trigonometry to convert positions and velocities into x and y components, making it easier to compute distances. --- ## Step 2: Write the Distance Between the Two Airplanes and Minimize It The distance between the airplanes at time \(t\) is: \[ d^2(t) = [x_B(t) - x_A(t)]^2 + [y_B(t) - y_A(t)]^2 \] Plug in the coordinates: \[ d^2(t) = [-100 + 350t - 275.8t]^2 + [173.2 - 275.8t]^2 \] \[ = [-100 + 74.2t]^2 + [173.2 - 275.8t]^2 \] Now, expand and find the time \(t_{min}\) where \(d^2(t)\) is minimized: \[ d^2(t) = (-100 + 74.2t)^2 + (173.2 - 275.8t)^2 \] \[ = (10000 - 14840t + 5505.64t^2) + (29997.44 - 95553.68t + 76088.64t^2) \] \[ = 39997.44 - 110393.68t + 81594.28t^2 \] To minimize, take derivative with respect to \(t\), set to zero: \[ \frac{d}{dt}d^2(t) = -110393.68 + 2 \cdot 81594.28t = \] \[ 2 \cdot 81594.28t = 110393.68 \] \[ t_{min} = \frac{110393.68}{2 \cdot 81594.28} = .676 \text{ hours} \] **Explanation:** We found the position equations, plugged into the distance formula, and used calculus to find the time where the distance is minimized. --- ## Final Answer (for part a): ### **Minimum distance occurs at \(t_{min} = .68\) hours (about 41 minutes) after takeoff.** To find the minimum distance, plug \(t_{min}\) back into the distance formula: \[ d_{min} = \sqrt{(-100 + 74.2 \times .676)^2 + (173.2 - 275.8 \times .676)^2} \] \[ = \sqrt{(-100 + 50.17)^2 + (173.2 - 186.37)^2} \] \[ = \sqrt{(-49.83)^2 + (-13.17)^2} \] \[ = \sqrt{2483. + 173.4} \] \[ = \sqrt{2656.4} \approx 51.5 \text{ miles} \] ### **The minimum distance between the airplanes is approximately 51.5 miles, occurring about 41 minutes after takeoff.** --- **If you want step-by-step answers for the other parts (b, c, d), please let me know!**