Let's solve the problem step by step: --- ## **Given:** - **A's house:** 25 km West of B's house. - **At 9am:** Both start running. - **A:** Runs East at 6 km/h. - **B:** Runs North at 8 km/h. - **Find:** a) When are A and B **closest together**? b) At what rate is the **distance between A and B changing at 10:15am**? Are they approaching one another? --- ### **Step 1: Setting up the Coordinate System** Let B's house be at the origin: \((0,0)\). A's house is at \((-25, 0)\). At time \( t \) hours after 9am: - **A's position:** \((-25 + 6t, 0)\) - **B's position:** \((0, 8t)\) --- ### **Step 2: Distance Between A and B** Let \( D(t) \) be the distance between A and B at time \( t \): \[ D(t) = \sqrt{ [(-25 + 6t) - 0]^2 + [0 - 8t]^2} = \sqrt{ (-25 + 6t)^2 + (8t)^2 } = \sqrt{ (-25 + 6t)^2 + 64t^2 } = \sqrt{ (-25 + 6t)^2 + (8t)^2 } = \sqrt{ (-25 + 6t)^2 + (8t)^2 } \] \[ D(t) = \sqrt{ (-25 + 6t)^2 + (8t)^2 } = \sqrt{ 625 - 300t + 36t^2 + 64t^2 } = \sqrt{ 625 - 300t + 100t^2 } \] --- ### **a) When are A and B closest together?** We need to **minimize \( D(t) \)** for \( t \geq 0 \) and \( t \leq 3 \) (since they run for 3 hours). Minimize \( D^2(t) = 625 - 300t + 100t^2 \). #### Take derivative and set to zero: Let \( f(t) = D^2(t) = 100t^2 - 300t + 625 \). \[ f'(t) = 200t - 300 \] Set \( f'(t) = 0 \): \[ 200t - 300 = 0 \implies t = \frac{300}{200} = 1.5 \] So, **the closest approach occurs at \( t = 1.5 \) hours after 9am**, i.e., at **10:30am**. #### Check endpoints: - \( t = 0 \): \( D(0) = 25 \) km - \( t = 3 \): \( D(3) = \sqrt{625 - 900 + 900} = \sqrt{625} = 25 \) km At \( t = 1.5 \): \[ D(1.5) = \sqrt{625 - 300(1.5) + 100(1.5)^2} = \sqrt{625 - 450 + 225} = \sqrt{400} = 20 \text{ km} \] So, **minimum distance is 20 km at 10:30am**. #### **Diagram** Draw a coordinate system: - B at (0,0) - A starts at (-25,0), moves right (East) at 6 km/h - B moves up (North) at 8 km/h --- ### **b) At what rate is the distance changing at 10:15am?** - **At 10:15am:** \( t = 1.25 \) hours Let’s differentiate \( D(t) \): \[ D(t) = \sqrt{625 - 300t + 100t^2} \] \[ \frac{dD}{dt} = \frac{1}{2}[625 - 300t + 100t^2]^{-1/2} \cdot (-300 + 200t) \] \[ = \frac{-300 + 200t}{2\sqrt{625 - 300t + 100t^2}} \] Plug in \( t = 1.25 \): \[ D(1.25) = \sqrt{625 - 300(1.25) + 100(1.25)^2} = \sqrt{625 - 375 + 156.25} = \sqrt{406.25} = 20.15 \text{ km} \] \[ \frac{dD}{dt}\bigg|_{t=1.25} = \frac{-300 + 200 \times 1.25}{2 \times 20.15} = \frac{-300 + 250}{40.3} = \frac{-50}{40.3} \approx -1.24 \text{ km/h} \] Since it's **negative**, the distance is **decreasing** at that moment (they are approaching each other). --- ## **Final Answers:** ### **a) A and B are closest together at 10:30am, at a distance of 20 km.** *(Include a diagram showing the coordinate axes, with A's and B's paths as described above.)* ### **b) At 10:15am, the distance between A and B is decreasing at a rate of approximately 1.24 km/h.** **They are approaching one another at that time.** --- If you need a diagram drawn/attached, let me know!