For a person at rest, the function p(t) = -20 cos(300˚t) + 100 models blood pressure, in millimeters of mercury ( mm Hg), at time t seconds. What is the rate of change in blood pressure at 3 s?
To find the rate of change of blood pressure at 3 seconds, we need to calculate the derivative of the given function with respect to time (t) and then substitute the value of t as 3 seconds.
Step 1: Find the derivative of the function p(t) = -20 cos(300˚t) + 100.
The derivative of the function can be found using the chain rule, which states that the derivative of f(g(x)) is f'(g(x)) * g'(x).
Differentiating each term separately:
d/dt(-20cos(300˚t)) = -20 * d/dt(cos(300˚t))
d/dt(100) = 0 [since it is a constant]
Now, let's evaluate the derivative of the cosine function:
d/dt(cos(300˚t))
To differentiate the cosine function, we multiply it by the derivative of the inner function (300˚t):
d/dt(cos(u)) = -sin(u) * du/dt
Using the chain rule, we have:
d/dt(cos(300˚t)) = -sin(300˚t) * d/dt(300˚t)
The derivative of 300˚t with respect to t is simply 300˚.
So, d/dt(cos(300˚t)) = -sin(300˚t) * 300˚ = -300˚sin(300˚t)
Step 2: Substitute the value of t as 3 seconds into the derivative expression.
We have:
d/dt(p(t)) = -20 * d/dt(cos(300˚t)) + 0
Substituting t as 3 seconds:
d/dt(p(t)) = -20 * (-300˚sin(300˚(3)))
= -20 * (-300˚sin(900˚))
= 6000˚sin(900˚)
Therefore, the rate of change in blood pressure at 3 seconds is 6000˚sin(900˚) mm Hg/s.