Closest Approach problem solved with derivative, Pythagorean Theorem and quadratic equation

Distance Between Solution
The distance each traveled is their speed times the elapsed time. Let t equal elapsed time. The ship traveled 40t* and the boat traveled 20t. Use Pythagorean theorem to calculate the distance between them. D(t) = $\sqrt{(20 - 40t)² + (20t)²}$ D²(t) = (20 - 40t)² + (20t)² D²(t) = 400 - 1600t +1600t² + 400t² D²(t) = 2000t² - 1600t + 400 D²(t) = 400 (5t² - 4t + 1) Take the derivative. Solve for t: 0 = 400 (10t - 4) 0 = 10t - 4 t = $\frac{2}{5}$ Apply t value to original equation. D(t) = $\sqrt{(20 - 40t)² + (20t)²}$ D² D( $\frac{2}{5}$ ) = $\sqrt{(20 - 40 2/5)² + (20 * 2/5)²}$ D( $\frac{2}{5}$ ) = $\sqrt{16 + 64}$ = $\sqrt{80}$ = 8.944 miles.
197

Distance Between
Solution

The distance each traveled is their speed
times the elapsed time. Let t equal elapsed
time. The ship traveled 40*t and the boat
traveled 20*t. Use Pythagorean theorem
to calculate the distance between them.

Drawing of map with two ships and their coordinates

Drawing of map with two ships and their coordinates

D(t) = $\sqrt{(20 - 40t)² + (20t)²}$
D²(t) = (20 - 40t)² + (20t)²
D²(t) = 400 - 1600t +1600t² + 400t²
D²(t) = 2000t² - 1600t + 400
D²(t) = 400 (5t² - 4t + 1)

Take the derivative. Solve for t:
0 = 400 (10t - 4)
0 = 10t - 4
t = $\frac{2}{5}$

Apply t value to original equation.
D(t) = $\sqrt{(20 - 40t)² + (20t)²}$
D² D( $\frac{2}{5}$ ) = $\sqrt{(20 - 40 * 2/5)² + (20 * 2/5)²}$
D( $\frac{2}{5}$ ) = $\sqrt{16 + 64}$ = $\sqrt{80}$ = 8.944 miles.

197

Distance BetweenSolution

Distance Between
Solution