Question

The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function D ( t ) = 2 sin ( π 3 t + 5 π 3 ) + 5 where t is the number of hours after midnight. Find the rate at which the depth is changing at 2 a.m. Round your answer to 4 decimal places.

Answer 1

The equation is not well formatted so I assumed it is

$D(t) = 2 \sin(\dfrac{\pi}{3}t + \dfrac{5\pi}{3}) + 5$

Answer:

1.0472 m/s

Step-by-step explanation:

If $D(t) = 2 \sin(\dfrac{\pi}{3}t + \dfrac{5\pi}{3}) + 5$, the rate at which the depth is changing is the time-derivative of D(t);

$\dfrac{d}{dt}D(t) = \dfrac{2\pi}{3}\cos(\dfrac{\pi}{3}t + \dfrac{5\pi}{3})$

At 2 am, t = 2 hours after midnight.

Substitute this in the last equation.

$\dfrac{d}{dt}D(t) = \dfrac{2\pi}{3}\cos(\dfrac{2\pi}{3} + \dfrac{5\pi}{3}) = \dfrac{2\pi}{3}(\cos \dfrac{7\pi}{3})= \dfrac{2\pi}{3}\times0.5=1.0472\text{ m/s}$

Answer 2

We want to find the rate of change of the depth function at 2 a.m. We will see that the rate of change at 2 a.m. is 0.82 ft per hour.

So we have a depth function:

$D(t) = 2*sin( (\pi/3)*t + (5*\pi/3)) + 5$

Where t is the number of hours after midnight.

We want to find the rate of change at 2 a.m., this is given by the first derivate of D(t) evaluated in t = 2   (t = 2 is equivalent to 2 hours after midnight, which is equivalent to 2 a.m.)

We will have:

$D(t) = 2*sin( (\pi/3)*t + (5*\pi/3)) + 5\\\\\frac{dD(t)}{dt} = (3/\pi)*2*cos( (\pi/3)*t + (5*\pi/3))$

Now we just evaluate this in t = 2

$\frac{dD(2)}{dt} = (3/\pi)*2*cos( (\pi/3)*2 + (5*\pi/3)) = 0.82$

It being positive means that at 2 a.m. the water level is rising at a rate of 0.82ft per hour.

If you want to learn more about rates of change, you can read:

https://brainly.com/question/18904995