HELP ASAP trig question

Question

Part of a rollercoaster course precedes point A(0,35) (point F precedes point A) point F cannot go any further than 45m from point A (in the y-axis). The gradient at point A is zero. the slope cannot be greater than 65degrees. Point F must be on the x-axis (x,0)

k(x)=Mcos(Nx)+K

0 < M < 25

0 < K < 25

0 < N < 3

Find the values K and M when the gradient of F and A are 0.

Find N where point F is exactly 45 meters from point A. and thus state the equation of k(x)

Find values of N for which it reacher the criteria in the first paragraph

discuss what values of N are appropriate.

Answer 1

Let's break down the problem:

1. Understanding the Setup

1. Point A is at (0, 35).
2. Point F is on the x-axis, so its coordinates are (x, 0).
3. The vertical distance between A and F cannot exceed 45 meters.
4. The gradient (slope) at point A is zero.
5. The slope of the track (between any two points) cannot exceed 65 degrees.
6. The equation of the track is given by k(x) = Mcos(Nx) + K.
7. 0 < M < 25
8. 0 < K < 25
9. 0 < N < 3

2. Finding K and M

1. Since the gradient at point A (x=0) is zero, the derivative of k(x) at x=0 must be zero.
2. k'(x) = -MNsin(Nx)
3. k'(0) = -MNsin(0) = 0. This condition is already satisfied for any values of M and N.
4. When x = 0, k(0) = Mcos(0) + K = M + K. Since point A is at (0, 35), we have:
5. M + K = 35
6. Since 0 < M < 25 and 0 < K < 25, one possible solution is M = 15 and K = 20 (or vice versa). However, we will keep the equation M + K = 35 for now.

3. Finding N when F is 45 meters from A

1. The distance between A(0, 35) and F(x, 0) is given by the distance formula:
2. Distance = sqrt((x - 0)^2 + (0 - 35)^2) = sqrt(x^2 + 1225)
3. We are given that this distance is 45 meters:
4. 45 = sqrt(x^2 + 1225)
5. 2025 = x^2 + 1225
6. x^2 = 800
7. x = ±sqrt(800) = ±20√2 ≈ ±28.28
8. When F is 45 meters from A, k(x) = 0. Let’s consider the positive x value, x = 20√2.
9. 0 = Mcos(N * 20√2) + K
10. -K = Mcos(N * 20√2)
11. cos(N * 20√2) = -K/M
12. Since M + K = 35, we can substitute K = 35 - M:
13. cos(N * 20√2) = -(35 - M)/M = (M - 35)/M
14. If we use M = 15, then K = 20.
15. cos(N * 20√2) = (15 - 35)/15 = -20/15 = -4/3. This is impossible as the cosine function has a range of -1 to 1.
16. If we use M = 20, then K = 15.
17. cos(N * 20√2) = (20-35)/20 = -15/20 = -3/4 = -0.75
18. N * 20√2 = arccos(-0.75) ≈ 2.419 radians (or 138.6 degrees)
19. N ≈ 2.419 / (20√2) ≈ 0.085
20. Therefore, with M=20 and K=15, and N ≈ 0.085, the equation is approximately:
21. k(x) = 20cos(0.085x) + 15

4. Slope Constraint (65 degrees)

1. The slope is given by the derivative k'(x) = -MNsin(Nx).
2. We want the absolute value of the slope to be less than or equal to tan(65°).
3. |-MNsin(Nx)| ≤ tan(65°) ≈ 2.145
4. MN|sin(Nx)| ≤ 2.145

5. Appropriate Values of N

1. We found N ≈ 0.085 when the distance is 45 meters.
2. To ensure the slope condition is met, we need to consider the maximum value of |sin(Nx)|, which is 1.
3. MN ≤ 2.145
4. With M=20, this implies 20N ≤ 2.145 or N ≤ 0.10725
5. The value of N ≈ 0.085 meets this condition.

Discussion:

1. N controls the "frequency" or "wavelength" of the cosine function. Smaller values of N mean a wider, gentler curve. Larger values of N result in a tighter, more frequent oscillation.
2. The slope constraint limits how large N can be. If N is too large, the slope will exceed 65 degrees at some point.
3. The distance constraint also influences N. The larger the distance between A and F, the smaller N needs to be to maintain a smooth curve.

In conclusion, with M=20, K=15 and N=0.085, the criteria are met.

The equation is approximately k(x) = 20cos(0.085x) + 15.