Pre cal help

Question

DEAR CONSTRUCTOR,.
I AM THE DIRECTOR OF MAGIC MOUNTAIN SKI AREA AND WE WISH TO CHOOSE A SITE FOR OUR NEW SKI LIFT. THE POSSIBLE SITES FOR THE BASE OF THE SKI LIFT MUST BE BETWEEN POINTS A AND B. THE ANDGLE OF ELEVATION FROM DA BASE OF THE FIRST SITE (A) TO THE TOP OF THE MOUNTAIN IS 33 DEGREES. tHE LAST POSSIBLE SITE (B) IS 800 FEET CLOSER TO THE BASE OF THE MOUNTAIN AND THE ANGLE OF ELEVATION IS 35 DEGREES. THE OTHER POSSIBLE SITES ARE AT ANGLES 33.5 DEGREES, 34 DEGREES, AND 34.5

AT ANY ANGLE OF ELVATION OF 33 DEGREES THE SKI LIFT WILL BE ABLE TO TRAVEL AT 5MPH. EACH ADDITIONAL DEGREE OF ELEVATION ABOVE 33 CAUSES A DECREASE IN THE SPEED OF .125MPH. 2 DEGREES OF ADDITIONAL ELEVATION WOULD CAUSE A DECREASE IN THE SPEED OF .25MPH, ETC.

we wish to build the ski lift that our skiers will have the fastest route to the top of the mountain. at wat angle should we build the lift?

Answer 1

Dear Director: To analyze this problem, let's consider the two sites A and B and the triangles they each form with the mountain. Imagine Magic Mountain as a triangle, and your ski lift goes to the top of this triangle. Call the vertical distance between the top and a horizontal line through Sites A and B a height h, and call the horizontal distance between Site B and the vertical line down from the mountain top x. From trigonometry, we have: h/x = tan(35), so h = x*tan(35) If we go out another 800 feet to Site A, we can make a second triangle with the mountain top and the vertical line down from it and the horizontal line under it. We can similarly analyze this to get: h/(x + 800) = tan(33), so h = (x + 800)*tan(33) We can combine both these expressions to get: x*tan(35) = (x + 800)*tan(33) Let's distribute tan(33) through the right hand side to give us: x*tan(35) = x*tan(33) + 800*tan(33) Collecting terms and isolating x gives us: x = (800 ft)*tan(33) = 10227 feet to the point on the horizontal ----------------- that is directly under the tan(35) - tan(33) mountain top. We then use this value to find the height h of the mountain: h = xtan(35) = 7161 feet. We then look at the lengths of the possible ski lift runs from Sites A and B: Site Angle Ski lift run length (= (7161 ft)/sin(angle)) Speed (miles/hr) A 33 13148.2 ft (1 mile/5280 ft) = 2.49 miles 5 B 35 12484.9 ft (1 mile/5280 ft) = 2.36 miles 4.75 We compare the times required to come up ski lifts from A or B: From Site A: 2.49 miles/5 mph equals about thirty minutes. From Site B: 2.36 miles/4.75 mph equals about thirty minutes. You can think of the distance that your ski lift covers as a function of the angle, between 33 and 35 degrees. You can also think of the speed as another linear function in the angle. Once you write a function for the time required to ride the ski lift as a construction of these linear functions, you can graph this constructed function versus the angle, between 33 and 35 degrees. Time required = (7161 ft/[5280 ft/mile])/[(5 - 0.125(theta-33) mph)sin(theta)] I leave it to you to graph this function for 33 <= theta <= 35 degrees, and find a minimum in the time required to ride the lift. Sincerely Yours, Joe the Constructor