Asked by Anonymous
The given function H(d) = ad^3 + bd^2 + cd + e represents a roller coaster, where H(d) represents the height above the ground, and d represents the horizontal distance the roller coaster has travelled.
Make it so the following parameters are true:
1) The roller coaster must have a local maximum at the point when d = (1)/2
2) The roller coaster must have a local minimum when d = 2 , h = 0.5
3) The roller coaster must have a point of inflection at the point when d = 1.25
4) The roller coaster starts at a height of 1 m above the ground.
I've tried each part separately, and I can get it to work but as soon as I put the parts together it seems to fall apart, I'm confused
Make it so the following parameters are true:
1) The roller coaster must have a local maximum at the point when d = (1)/2
2) The roller coaster must have a local minimum when d = 2 , h = 0.5
3) The roller coaster must have a point of inflection at the point when d = 1.25
4) The roller coaster starts at a height of 1 m above the ground.
I've tried each part separately, and I can get it to work but as soon as I put the parts together it seems to fall apart, I'm confused
Answers
Answered by
oobleck
so, just what did you do? Too bad you don't show anything we can work with.
H(d) = ad^3 + bd^2 + cd + e
H'(d) = 3ad^2 + 2bd + c
H"(d) = 6ad + 2b
So, we need
H"(1/2) < 0
H"(2) > 0
H(2) = 1/2
H"(5/4) = 0
H(0) = 1
Putting all that into equations, we need to solve
6a(5/4) + 2b = 0
e = 1
3a*2^2 + 2b*2 + c = 0
3a(1/2)^2 + 2b(1/2) + c = 0
or,
15a+4b = 0
12a+4b+c = 0
3a+4b+4c = 0
Solving those, we get
H(d) = 4d^3 - 15d^2 + 12d + 1
H'(d) = 6(2x^2-5x+2)
H"(d) = 6(4x-5)
H has a local maximum at (1/2, 15/4)
H has a local minimum at (2,-3)
H has a point inflection at (5/4, 3/8)
H(0) = 1
H(d) = ad^3 + bd^2 + cd + e
H'(d) = 3ad^2 + 2bd + c
H"(d) = 6ad + 2b
So, we need
H"(1/2) < 0
H"(2) > 0
H(2) = 1/2
H"(5/4) = 0
H(0) = 1
Putting all that into equations, we need to solve
6a(5/4) + 2b = 0
e = 1
3a*2^2 + 2b*2 + c = 0
3a(1/2)^2 + 2b(1/2) + c = 0
or,
15a+4b = 0
12a+4b+c = 0
3a+4b+4c = 0
Solving those, we get
H(d) = 4d^3 - 15d^2 + 12d + 1
H'(d) = 6(2x^2-5x+2)
H"(d) = 6(4x-5)
H has a local maximum at (1/2, 15/4)
H has a local minimum at (2,-3)
H has a point inflection at (5/4, 3/8)
H(0) = 1
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