Question

a boat at anchor is bobbing up and down in the sea. the vertical distance, y, in feet, between the sea floor and the boat is given as a function of time, t, in minutes, by y = 22 + cos(2πt) ft. find the vertical velocity, v, of the boat at time t. v = ft/min

Explanation:

Step1: Recall the velocity - displacement relationship

Velocity $v$ is the derivative of displacement $y$ with respect to time $t$, i.e., $v=\frac{dy}{dt}$.

Step2: Differentiate the given function

Given $y = 22+\cos(2\pi t)$. The derivative of a constant is 0, and the derivative of $\cos(u)$ with respect to $t$ is $-\sin(u)\frac{du}{dt}$ (by the chain - rule). Here $u = 2\pi t$ and $\frac{du}{dt}=2\pi$. So $\frac{d}{dt}(22) = 0$ and $\frac{d}{dt}(\cos(2\pi t))=-\sin(2\pi t)\times(2\pi)$.

Step3: Calculate the derivative of $y$

$v=\frac{dy}{dt}=0 - 2\pi\sin(2\pi t)=- 2\pi\sin(2\pi t)$ ft/min.

Answer:

$-2\pi\sin(2\pi t)$