Asked by gabitc
ok so theres a roller coaster being built in the shape of a sinusoid, the high and low points on the track separated by 50 m horizontally and 30 m vertically, the lowest point is 3 m below the ground.
my question is: the vertical timbers are spaced every 2m starting at x=0, and ending where the track goes below the ground. write a equation that prints out the length of each timber. the programs hould also print the total length of all vertical timbers so that you will knonw how much to purchase
x=39.76 (according to my calculations) when it reaches the ground
my question is: the vertical timbers are spaced every 2m starting at x=0, and ending where the track goes below the ground. write a equation that prints out the length of each timber. the programs hould also print the total length of all vertical timbers so that you will knonw how much to purchase
x=39.76 (according to my calculations) when it reaches the ground
Answers
Answered by
oobleck
high and low points on the track separated by 30 m vertically
so, the amplitude is 30/2 = 15
high and low points on the track separated by 50 m horizontally
so, the period is 100 -- recall that sin(kx) has period 2π/k
So far, assuming we start at the top, we have y = 15cos(π/50 x)
Since the lowest point is 3m below ground, shift down by 3
y = 15cos(π/50 x) - 3
So, the kth pole (starting at k=0) has height
h(k) = 15cos(kπ/50 * 2) - 3
When does y=0? when π/50 x = π/2 -- x=25
So the sum of the heights of the first 26 poles is
25
∑ 15cos(π/50 * 2) - 3 = 390cos(π/25)-3 ≈ 383.295
k=0
So, what calculations did you make to get x=39.76? If you have an answer, it is nice to show how you got it. You may be right, but how will we know? You can check my work, but I can't check yours.
so, the amplitude is 30/2 = 15
high and low points on the track separated by 50 m horizontally
so, the period is 100 -- recall that sin(kx) has period 2π/k
So far, assuming we start at the top, we have y = 15cos(π/50 x)
Since the lowest point is 3m below ground, shift down by 3
y = 15cos(π/50 x) - 3
So, the kth pole (starting at k=0) has height
h(k) = 15cos(kπ/50 * 2) - 3
When does y=0? when π/50 x = π/2 -- x=25
So the sum of the heights of the first 26 poles is
25
∑ 15cos(π/50 * 2) - 3 = 390cos(π/25)-3 ≈ 383.295
k=0
So, what calculations did you make to get x=39.76? If you have an answer, it is nice to show how you got it. You may be right, but how will we know? You can check my work, but I can't check yours.
Answered by
what is it
dang oobleck is a pro
Answered by
oobleck
oops. I see I made a couple of mistakes.
y=0 when 15cos(π/50 x) - 3 = 0, or x=21.8
So, we just need to add the first 21.8/2 = 11 poles.
10
∑ 15cos(π/50 * 2) - 3 = 90.19
k=0
y=0 when 15cos(π/50 x) - 3 = 0, or x=21.8
So, we just need to add the first 21.8/2 = 11 poles.
10
∑ 15cos(π/50 * 2) - 3 = 90.19
k=0
Answered by
Anonymous
ty this helped a lot, i made earlier calculations to solve but i think i put down enough info to solve the final problem
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