I really need help constructing a function P(t) = a sin(b(t+c)) + d...?

Can someone help me out here? The question is:





A person's blood pressure varies between 80 and 120 mm Hg every 0.8 seconds (the time for one heartbeat). Given that P(0) = 90, find values of a, b, c, and d such that the function P(t) = a sin(b(t+c)) + d models the person's blood pressure P at time t.

I really need help constructing a function P(t) = a sin(b(t+c)) + d...?
Start finding the values for a, b, c, and d one at a time.





The average value for a sine wave is 0. The average value for the function here is (120+80)/2 = 100. So d = 100.





The difference between the min and max of a sin wave is 1 - -1 = 2. The difference here is 120-80 - 40. That means this function has 20 times the amplitude. So a = 20.





The wave length of a sine wave is 2 pi. The "wavelength" here is .8 seconds. That means that b = 2pi/0.8 sec = 2.5 pi / sec.





Now we just have to find c to satisfy the starting condition:





P(0) = 20 sin (2.5pi(0 + c)/sec) + 100 = 90


20 sin(2.5pi(c)/sec) = -10


sin(2.5pi(c)/sec) = -.5


2.5 pi (c)/sec = -pi/6


c = -sec/15





It makes sense that c has units of seconds, since it has to be added to t, which is also in seconds. Similarly, it makes sense that b is units of 1/sec, because it has to get rid of the units that come from the addition of t and c.





P(t) = 20 sin (2.5pi(t - (sec/15))/sec) + 100





Let's check that:


P(0) = 20 sin (2.5pi(0 - (sec/15))/sec) + 100


= 20 sin(2.5 pi (-sec/15) /sec) + 100


= 20 sin(-2.5 pi (1/15)) + 100


= 20 sin(-pi/6) + 100


= 20 (-.5) + 100


= -10 + 100


= 90





Perfect.





Note: The problem doesn't say whether the person's pressure is 90 and raising or 90 and falling at time zero. In the step where we take the arcsin of -.5 when finding a value for c, we could have used -5pi/6. That would have satisfied the initial condition given in the problem but given a function that looks different than one we found.