a graphing problem!!

summerhill

2009-03-15 5:03 pm

The function represents a roller coaster's route is
y= 100- (1/4)x^2

But at x=10, the euquation is changed to concave up parabola with eqation y=a(x-40)^2 in order to avoid an unpleasant meeting with the ground.

thus: y=100-(1/4)x^2 -----0≤x＜10
and y=a(x-40)^2 ------x＞10

*Find the value of a in the second piece of the function which makes the roller coaster work ( must be continuous and differentiable )
* On the graph state starting point, meeting point of the two functions and the point where the roller coaster safely returns to the ground.

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in order to make the graph continuous and differentiable, there shouldn't be any corners and breaks ...etc, but how can you connect the two?

回答 (1)

用戶10012

2009-03-15 10:22 pm

✔ 最佳答案

For y = 100 - x^2/4 0 < x < 10.
when x = 0, y = 100, so the roller coaster starts at (0, 100).
Meeting point of the 2 functions is when x = 10, so y = 100 - (10^2)/4 = 75, so the meeting point is (10,75).
Put (10, 75) into the second function we get
75 = a(10 - 40)^2 = 900a, so a = 75/900 = 1/12.
For y = 100 - x^2/4, dy/dx = -x/2. At x = 10, dy/dx = -5.
For y =(x-40)^2/12, dy/dx = (x - 40)/6. At x = 10, dy/dx = -30/6 = -5.
So the graph is continuous and differentiable at point ( 10, 75).
For y = (x-40)^/12, when y = 0, x = 40. So the landing point is (40,0).

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