Question

1. modeling with mathematics you are going for a hike in the woods. you hike to a waterfall that is 4 miles east of where you left your car. you then hike to a lookout point that is 2 miles north of your car. from the lookout point, you return to your car.

a. map out your route in a coordinate plane with your car at the origin. let each unit in the coordinate plane represent 1 mile. assume you travel along straight paths.
b. how far do you travel during the entire hike?
c. when you leave the waterfall, you decide to hike to an old wishing well before going to the lookout point. the wishing well is 3 miles north and 2 miles west of the lookout point. how far do you travel during the entire hike?

Explanation:

Part a

Step1: Define Coordinates

Let the car be at the origin \((0,0)\). The waterfall is 4 miles east, so its coordinates are \((4,0)\). The lookout point is 2 miles north of the car, so its coordinates are \((0,2)\).

Step2: Map the Route

Plot the points \((0,0)\) (car), \((4,0)\) (waterfall), \((0,2)\) (lookout point), and then back to \((0,0)\) (car) with straight lines.

Part b

Step1: Find Distances

• Distance from car to waterfall: \(d_{1}=\sqrt{(4 - 0)^{2}+(0 - 0)^{2}} = 4\) miles (or simply 4 miles as it's along the x - axis).
• Distance from waterfall to lookout point: Use the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\). Here, \(x_1 = 4,y_1 = 0,x_2 = 0,y_2 = 2\). So \(d_{2}=\sqrt{(0 - 4)^{2}+(2 - 0)^{2}}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\approx4.47\) miles.
• Distance from lookout point to car: \(d_{3}=\sqrt{(0 - 0)^{2}+(0 - 2)^{2}} = 2\) miles (along the y - axis).

Step2: Total Distance

Total distance \(D=d_{1}+d_{2}+d_{3}=4 + 2\sqrt{5}+2=6 + 2\sqrt{5}\approx6 + 4.47 = 10.47\) miles. But if we consider the path as car - waterfall (4), waterfall - lookout (using Pythagoras: right triangle with legs 4 and 2, hypotenuse \(\sqrt{4^{2}+2^{2}}=\sqrt{20}=2\sqrt{5}\)), lookout - car (2). Wait, actually, the original path is car to waterfall (4), waterfall to lookout (let's recalculate: from (4,0) to (0,2): \(\Delta x=- 4,\Delta y = 2\), so distance is \(\sqrt{(-4)^{2}+2^{2}}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\)), then lookout to car (2). So total distance is \(4+2\sqrt{5}+2=6 + 2\sqrt{5}\approx10.47\) miles. But maybe a simpler way: the triangle formed by car, waterfall, lookout is a right triangle with legs 4 and 2. The perimeter of the triangle is \(4 + 2+\sqrt{4^{2}+2^{2}}=6+\sqrt{20}=6 + 2\sqrt{5}\approx10.47\) miles.

Part c

Step1: Find Coordinates of Wishing Well

Lookout point is \((0,2)\). Wishing well is 3 miles north and 2 miles west of lookout point. So \(x=0 - 2=-2,y = 2+3 = 5\). So wishing well coordinates: \((-2,5)\).

Step2: Find Distances

• Car to waterfall: \(d_{1}=4\) miles (from \((0,0)\) to \((4,0)\)).
• Waterfall to wishing well: From \((4,0)\) to \((-2,5)\). \(\Delta x=-2 - 4=-6,\Delta y=5 - 0 = 5\). Distance \(d_{2}=\sqrt{(-6)^{2}+5^{2}}=\sqrt{36 + 25}=\sqrt{61}\approx7.81\) miles.
• Wishing well to lookout point: From \((-2,5)\) to \((0,2)\). \(\Delta x=0-(-2)=2,\Delta y=2 - 5=-3\). Distance \(d_{3}=\sqrt{2^{2}+(-3)^{2}}=\sqrt{4 + 9}=\sqrt{13}\approx3.61\) miles.
• Lookout point to car: \(d_{4}=2\) miles (from \((0,2)\) to \((0,0)\)).

Step3: Total Distance

Total distance \(D=d_{1}+d_{2}+d_{3}+d_{4}=4+\sqrt{61}+\sqrt{13}+2=6+\sqrt{61}+\sqrt{13}\approx6 + 7.81+3.61=17.42\) miles.

Part a Answer:

• Car: \((0,0)\), Waterfall: \((4,0)\), Lookout: \((0,2)\). Plot these points and connect with straight lines (car - waterfall - lookout - car).

Part b Answer:

\(6 + 2\sqrt{5}\) miles (or approximately \(10.47\) miles)

Part c Answer:

\(6+\sqrt{61}+\sqrt{13}\) miles (or approximately \(17.42\) miles)

Answer:

Step1: Find Coordinates of Wishing Well

Lookout point is \((0,2)\). Wishing well is 3 miles north and 2 miles west of lookout point. So \(x=0 - 2=-2,y = 2+3 = 5\). So wishing well coordinates: \((-2,5)\).

Step2: Find Distances

• Car to waterfall: \(d_{1}=4\) miles (from \((0,0)\) to \((4,0)\)).
• Waterfall to wishing well: From \((4,0)\) to \((-2,5)\). \(\Delta x=-2 - 4=-6,\Delta y=5 - 0 = 5\). Distance \(d_{2}=\sqrt{(-6)^{2}+5^{2}}=\sqrt{36 + 25}=\sqrt{61}\approx7.81\) miles.
• Wishing well to lookout point: From \((-2,5)\) to \((0,2)\). \(\Delta x=0-(-2)=2,\Delta y=2 - 5=-3\). Distance \(d_{3}=\sqrt{2^{2}+(-3)^{2}}=\sqrt{4 + 9}=\sqrt{13}\approx3.61\) miles.
• Lookout point to car: \(d_{4}=2\) miles (from \((0,2)\) to \((0,0)\)).

Step3: Total Distance

Total distance \(D=d_{1}+d_{2}+d_{3}+d_{4}=4+\sqrt{61}+\sqrt{13}+2=6+\sqrt{61}+\sqrt{13}\approx6 + 7.81+3.61=17.42\) miles.

Part a Answer:

• Car: \((0,0)\), Waterfall: \((4,0)\), Lookout: \((0,2)\). Plot these points and connect with straight lines (car - waterfall - lookout - car).

Part b Answer:

\(6 + 2\sqrt{5}\) miles (or approximately \(10.47\) miles)

Part c Answer:

\(6+\sqrt{61}+\sqrt{13}\) miles (or approximately \(17.42\) miles)