Q. 61 - Calculus

Question

The median rate of flow of the Lochsa River (in Idaho) at the Lowell gauge can be modeled surprisingly accurately as$f\left(t\right)=800+3.1p\left(t\right)$, where $p\left(t\right)$ is a function given by $\left(t-90\right)\left(195-t\right)$ for $t\in \left(90,195\right)$ and is zero otherwise. The time t is measured in days after the beginning of the year, and the flow is measured in cubic feet of water per second.

Part (a): Given a starting day $t_0$, use the Fundamental Theorem of Calculus to find the function $F_{t_0}\left(t\right)$ that gives the total amount of water that has flowed past the Lowell gauge since that day. Note that there are $86400$ seconds per day.

Part (b): In a non-leap year, March $31$ is day $90$ of the year and July $14$ is day $195$. Use the function you found in part (a) to compute how many cubic feet of water flowed past the gauge between those days.

Part (c): Use your function to compute how much water flows down the Lochsa River in a full year. What fraction of that amount flows by during the span of time you examined in part (b)?

Step-by-Step Solution

Answer

Part (a): The function $F_{t_0}\left(t\right)$ that gives the total amount of water that has flowed past is $F_{t_0}=86400\left[-1.03\left(t^3-{t_0}^3\right)+441.759\left(t^2-{t_0}^2\right)-53605\left(t-t_0\right)\right]$.

Part (b): $60.8$ billion cubic feet of water flowed between March $31$ and July $14$.

Part (c): The fraction of water that flows between March $31$ and July $14$ is $41.38\%$.

Part (a) Step 1. Given information.

Consider the given question,

$$\begin{gathered} f\left(t\right)=800+3.1p\left(t\right)\mathit{,} \\ p\left(t\right)=\left(t\mathrm{-}\mathrm{90}\right)\left(\mathrm{195}\mathrm{-}t\right) \end{gathered}$$

Part (a) Step 2. Find the function F t 0 t .

Substitute the function $p\left(t\right)$ in the function $f\left(t\right)$,

$$\begin{gathered} f\left(t\right)=800+3.1\left(t-90\right)\left(195-t\right) \\ =800+3.1\left(-t^2+195t+90t-17550\right) \\ =-3.1t^2+883.5t-53605 \end{gathered}$$

Integrate the function $f\left(t\right)$ from $t_0$ to t,

${\int }_{t_0}^t f\left(t\right) dt=-{\int }_{t_0}^t 3.1t^2 dt+{\int }_{t_0}^t 883.5t dt-{\int }_{t_0}^t 53605 dt$

For a function $f\left(x\right)$ which is continuous on an interval $\left[a,b\right]$, the fundamental theorem of calculus states that ${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where F is the antiderivative of the function f.

Part (a) Step 3. Use the fundamental theorem.

Using the fundamental theorem of calculus in the right hand side,

$$\begin{gathered} {\int }_{t_0}^{t}f\left(t\right) dt=-{\int }_{t_0}^{t}3.1t^{2 }dt+{\int }_{t_0}^{t}883.5t dt-{\int }_{t_0}^{t}53605 dt \\ {F}_{t_0}\left(t\right)=-3.1{{\left[\frac{t^{2+1}}{2+1}\right]}^t}_{t_0}+883.5{{\left[\frac{t^{1+1}}{2+1}\right]}^t}_{t_0}-53605{{\left[t\right]}^t}_{t_0} \\ {F}_{t_0}\left(t\right)=-\frac{3.1}{3}\left[t^3-{t_0}^3\right]+\frac{883.5}{3}\left[t^2-{t_0}^2\right]-53605\left[t-t_0\right] \\ {F}_{t_0}\left(t\right)=-1.03\left(t^3-{t_0}^3\right)+441.759\left(t^2-{t_0}^2\right)-53605\left(t-t_0\right) \end{gathered}$$

Multiply the $F_{t_0}\left(t\right)$ by $86400$,

$F_{t_0}\left(t\right)=86400\left[-1.03\left(t^3-{t_0}^3\right)+441.759\left(t^2-{t_0}^2\right)-53605\left(t-t_0\right)\right] ...... \left(i\right)$

Part (b) Step 1. Find how many feet of water flowed past the gauge during the given time.

Substitute $$\begin{gathered} t_0=90, \\ t=195 \end{gathered}$$ in equation (i),

$$\begin{gathered} F_{90}\left(195\right)=86400\left[-1.03\left({195}^3-{90}^3\right)+441.759\left({195}^2-{90}^2\right)-53605\left(195-90\right)\right] \\ \approx 60.8 billion cubic feet \end{gathered}$$

Part (c) Step 1. Find how many feet of water flowed in a full year.

For a full year, $$\begin{gathered} t_0=1, \\ t=365 \end{gathered}$$.

The function $p\left(t\right)$ is $0$ when t does not lie between $\left(90,195\right)$.

Therefore, the function $f\left(t\right)$ is $800$.

Integrate the function $f\left(t\right)=800$ from $t_0$ to t,

$$\begin{gathered} {\int }_{t_0}^{t}f\left(t\right) dt={\int }_{t_0}^{t}800 dt \\ {F}_{t_0}\left(t\right)=800{{\left[t\right]}^t}_{t_0} \\ {F}_{t_0}\left(t\right)=800\left[t-t_0\right] \end{gathered}$$

Multiply $86400$ by $F_{t_0}\left(t\right)$,

$$\begin{gathered} F_{t_0}\left(t\right)=86400\left(800\left[t-t_0\right]\right) \\ {F}_{t_0}\left(t\right)=69120000\left[t-t_0\right] \end{gathered}$$

Part (c) Step 2. Substitute t 0 = 1 , t = 365 in the function F t 0 t , followed by simplification.

Substitute $$\begin{gathered} t_0=1, \\ t=365 \end{gathered}$$ in $F_{t_0}\left(t\right)$,

$$\begin{gathered} F_{t_0}=69120000\left[365-1\right] \\ =25159680000 \end{gathered}$$

Therefore, $25.16$ billion cubic feet of water flowed in a year.

Then, to calculate the percentage,

$$\begin{gathered} =\frac{25.16}{60.8}\times 100 \\ =41.38\% \end{gathered}$$