help pls I don't know what to do/ the steps to take haven't learned in class but expected to know TRIGONOMETRY

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

PART 1: If the gradient of F and A are both zero, then points F and A must be critical points of the function. Since M cannot exceed 25, there is no way for either F or A to be on the midline of the graph because the amplitude of the function (the absolute value of M) would exceed 25. Therefore, this tells us that F and A are a minimum (F) and a maximum (A) of the function.

In the equation k(x)=Mcos(Nx)+K, M will be the amplitude and K will be the midline.

NOTE: since the function reaches a maximum on the y-axis, there is no vertical reflection of the graph, so M will stay positive here.

The midline is halfway between the y-values of the min and max, so it will be at (0 + 35)/2, or 17.5 . This is K

From that, M is the distance from the midline to the max or min, which is 35 - 17.5, or 17.5.

PART 2: First we will use the distance formula to find the specific location of F:

45 = sqrt[(x²-0) + (0-35²)]

so x² = 800, and x = ± 20√2

Since it says that F precedes A on the graph, then x must be the negative value.

To determine the value of N, we need the period of the function. It is always the case that the horizontal distance between a max and a min of a cosine function is half a period, so the period of the function is 2*20√2, or 40√2. The value of N is found using 2π / 40√2, which simplifies to π√2 / 40

Now, you have all values needed to write the equation of k(x).

PART 3:

Here, all you need to do is figure out the closest value that F can be to A, since you already have the farthest value from the previous part (20√2). To find the closest, we will use the restriction that the slope cannot exceed 65°. The way this is phrased could be two possible things:

1) the slope of the line connecting F to A cannot be steeper than a 65° angle. In that case, just make a right triangle with FA as the hypotenuse and solve for the x value when angle F is 65°, using tan(65) = 35/x

2) the slope of the gradient lines as you travel along the curve from F to A cannot be steeper than a 65° angle. This would require the use of the first derivative of the function and then set it equal to tan(65). This seems less likely, since it involves several variables (also, if you haven't done derivatives yet, then it can't be this option)