Water Flow
The rate of water flow into a reservoir is modeled by
r(t) = 2\cos(t) + 3
cubic meters
per hour.
V(t)
is the volume of water after
t
hours:
1. FindV(π)
if V(0) = 5
.
2. Explain what V'(\pi/2)
represents.
3. Determine when the volume is increasing most rapidly.
1. Find
V(π)
if
V(0) = 5
.
We need to find the antiderivative, solve for
C
, and then find
V(\pi)
.
\begin{align*}
r(t) &= 2\cos(t) + 3 \\
R(t) &= 2\sin(t) + 3t + C \\
V(t) &= R(t) \\
V(0) &= 5 \\
&= 2\sin(0) + 3(0) + C \\
&= C \\
C &= 5 \\
V(\pi) &= 2\sin(\pi) + 3\pi + 5 \\
&= 3\pi + 5
\end{align*}
2. Explain what
V'(\pi/2)
represents.
V(t)
is the volume of water in the reservoir after t
hours.
V'(t)
is the rate the volume is changing after t
hours.
V'(\pi/2)
is the rate that water is entering the reservoir after \pi/2
hours. 3. Determine when the volume is increasing most rapidly.
The volume is increasing most rapidly when its rate of change is at its highest.
V'(t)
is at its maximum.
We must find and test the critical points of V'(t)
, which are where its slope is 0
or undefined. V'(t) = r(t) = 2\cos(t) + 3
V''(t) = -2\sin(t)
The slope of V'(t)
is 0
when t
is a multiple of \pi
.
t=\pi k, k \in \mathbb{Z}
(k
is an integer)
Where V'(t)
is concave down, it can have local maxima.
V'(t)
is concave down when V'''(t)
is negative. V''(t) = -2\sin(t)
V'''(t) = -2\cos(t)
V'''(t)
is negative where \cos(t)
is positive.
-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}, \quad \frac{3\pi}{2} \lt t \lt \frac{5\pi}{2}, \quad \text{etc}
V'(t)
has local maxima where t=2\pi k, k \in \mathbb{Z}
.
When t
is a multiple of 2\pi
, V'(t)=2(1) + 3=5
.
Since all of the local maxima are equal, they are all the global maxima.
Therefore, the volume is increasing most rapidly when t
is a multiple of
2\pi
. Below is a graph of
V(t)
and its derivative. We can check our answer to part three by seeing where
V'(t)
is at its maximum and
V(t)
is increasing
most rapidly.