Accelerated Precalculus November 8, 2018 Happy Thursday!

One-Minute Question Write the equation of the curve on the board.

Homework?? Questions?

A Deer Problem To avoid a hunter a deer runs in a sinusoidal path that crosses a stream. At time = 2 sec., the deer is 30 feet to the north of the stream and at time = 20 sec., the deer is 10 feet to the south of the stream. If these are maximum distances from the stream that runs east-west, write an equation of the deer’s path.

Extensions . Where is the deer at t = 0 seconds? . What are the first 3 times at which the deer will cross the stream?

Answers An equation is: y = 20cos((π/18)(x – 2)) + 10 . At t = 0 seconds, the deer is 28.79’ north of the stream. . At t = 13 seconds, he is 3.16’ north of the stream.

Answers To find where he crosses the stream algebraically, let 20cos((π/18)(x – 2)) + 10 = 0 So 20cos((π/18)(x – 2)) = -10 cos((π/18)(x – 2)) = -1/2 arccos(cos((π/18)(x – 2)) = arccos(-1/2) (π/18)(x – 2) = ±2π/3 + 2πk x – 2 = ±12 + 36k **** Why is 36 right? X = 14 + 36k or x = -10 + 36k so… X = {14, 26, 50} so at t = 14, t = 26 and t = 50, the deer crosses the stream.

Another Example A water wheel 14 feet in diameter is rotating counterclockwise. You start a stopwatch and observe a point P on the rim of the wheel. At t = 2 seconds, P is at its highest, 13 feet above the water. At t = 7 seconds, P is at its maximum depth below the water.

Questions: . Write an equation of the motion. . At what time does the wheel first emerge from the water? . Where is P at time = 6 seconds?

Questions: . Y = 7cos (π/5(x – 2)) + 6 . The wheel first emerges from the water at t = 7.861 seconds. . At time = 6 seconds, P is.3369’ above the water.