I have a trig worksheet, and I am clueless, but I want to get a good grade. Can anybody help me understand how to do this please?

Question

Throughout the day the depth of water at the end of a pier varies with the tides. High tide occurs at 4:00 am with a depth of 6 meters. Low tide occurs at 10:00 am with a depth of 2 meters.

1. model the problem by using the given trigonometric equation to show the depth (y) of the water x hours after midnight, showing all work. y=Acos(Bx+C)+D
2. a. start by sketching a graph of the situation- sketch 2 cycles (pick appropriate intervals for the x and y axes and make the horizontal axis in time, not radians. Hint: What time should x=0b?

Answer 1

The model is Y = A cos(BX + C} + D. We need to determine A, B, C, D.

At high tide cos(BX + C) = 1. At low tide cos(BX + C) =0.

We know that X = 4 am at high tide, and 10 am at low tide.

So, 4B + C = 1 and 10B + C = 0. Solve these to get

B = 15 and C =-60.

Y = 6 at high tide, so 6 = A cos(15X -60) + D and at low tide, 2 = A cos(15X -60) + D

Substitute X = 4 for high tide, and X = 10 at low tide and solve the previous two equations for A and D.

The model equation then becomes Y = 4 cos(15X - 60) + 2. Check by calculating Y when X = 4 and 10.

Answer 2

the period is 12 hours, B*12=2π cos needs to see B*12 as 2π, one period for cos

B=π/6. If 4:AM is chosen as t=0 hour, the C=0.

Important note I write level(t)=A+cos(B*(t-E))+D because the offset needs to be subtracted directly from the variable for the offset to be accurate (unless B=1)

D is where the y-axis midline is (in this case 4). A is the amplitude from midline to top, in this case 2.if t=0 is 4:am the l(t)=2*cos(π*t/6)+4

It t=0 is midnight, E positive means (t-E) moves (offsets) the cos graph E hours to the right so the equation becomes l(t)=2*cos(((π/6)*(t-4))+4

yes this could be written as l(t)=2*cos(π*t/6-2*π/3)+4

but the offset is 4 hours not 2*π/3 hours.

ask your teacher about how this is accounted for in l(t)=A*cos(B*t+C)+D.