Q. 62 - Calculus

Question

Ian is climbing every day, using a camp at the base of a snowfield. His only supply of water is a trickle that comes out of the the snowfield. The trickle dries at night, because the temperature drops and the snow stops melting.

Since he has run out of books, to entertain himself he uses measurements of the water in his cooking pot to model the flow as $w\left(t\right)=15\left(1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right)$, where t is the time in hours after midnight and $w\left(t\right)$ is the rate at which snow is melting into his pot, in gallons per hour.

Part (a): About how long does it take Ian to fill his $2$-quart pot at $4$ in the afternoon?

Part (b): What is the total amount of water that flows out of the snowfield in a single day?

Part (c): Write an expression for the total amount of water that flows from the snowfield between any starting time $t_0$ and an ending time t.

Part (d): If Ian stashes his pot under the trickle at $5$ in the morning, how long must he wait until he comes back to a full pot? 

Step-by-Step Solution

Answer

Part (a): The pot will be full by $4:01$ PM.

Part (b): The amount of water that flows out of snowfield in a single day is $360$ gallons.

Part (c): The expression for the total amount of water that flows from snowfield between $t_0$ and t is$v=15\left[t-t_0\right]+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_0-16\right)}{12}\right)\right]$.

Part (d): After $3.34$ hours from $5$ AM in the morning, the pot is filled. Thus, the pot is full by $8:34$ AM.

Part (a) Step 1. Given information.

Consider the given question,

$w\left(t\right)=15\left(1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right)$

Part (a) Step 2. Determine how long it will take Ian to fill his pot.

The function $w\left(t\right)$ indicates the rate at which the snow is melting. Therefore, the function $w\left(t\right)$ can be written as $\frac{dv}{dt}$, where v is the quantity of pot.

$\frac{dv}{dt}=15\left(1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right)$

Integrate both sides of the equation form $t_1$ to $t_2$ with respect to t.

$$\begin{gathered} {\int }_{t_1}^{t_2}\frac{dv}{dt} dt={\int }_{t_1}^{t_2}15\left(1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right) dt \\ {\int }_{t_1}^{t_2}\frac{dv}{dt} dt={\int }_{t_1}^{t_2}15 dt+15{\int }_{t_1}^{t_2}\cos \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right) dt ...... \left(i\right) \end{gathered}$$

For a function $f\left(x\right)$ which is continuous on a 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 of calculus.

Using the fundamental theorem of calculus in equation (i),

$$\begin{gathered} v=15{{\left[t\right]}^{t_2}}_{t_1}+15{{\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}}\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{12}}\right)}\right]}^{t_2}}_{t_1} \\ v=15\left[t_2-t_1\right]+15\left(\frac{12}{\mathrm{\pi }}\right)\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_1-16\right)}{12}\right)\right] ...... \left(ii\right) \end{gathered}$$

There are $0.25$ gallon in $1$ quart.

To calculate the no. of gallons,

$$\begin{gathered} =2\times 0.25 \\ =0.5 \end{gathered}$$

The time $4$in the afternoon corresponds to the time $t=16$ as time t indicates the no. of hours after midnight.

Part (a) Step 4. Substitute v = 0 . 5 , t 1 = 16 in equation (ii), followed by simplification.

Substitute $$\begin{gathered} v=0.5, \\ {t}_1=16 \end{gathered}$$ in equation (ii),

$$\begin{gathered} 0.5=15\left[t_2-16\right]+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left(16-16\right)}{12}\right)\right] \\ 0.5=15t_2-240+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)-\sin \left(0\right)\right] \\ 15t_2+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)\right]=240.5 \\ 15t_2+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)\right]-240.5=0 \end{gathered}$$

Consider the function as $f\left(t\right)=15t_2+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)\right]-240.5$.

Part (a) Step 5. Substitute t 2 = 10 in the function f t .

Substituting $t_2=10$ in the function $f\left(t\right)$,

$$\begin{gathered} f\left(10\right)=15\left(10\right)+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left(10-16\right)}{12}\right)\right]-240.5 \\ =150+\frac{180}{\mathrm{\pi }}\sin \left(\frac{-6\mathrm{\pi }}{12}\right)-240.5 \\ \approx -147.79 \end{gathered}$$

Make the table value of the function for the different values of $t_2$,

Part (a) Step 6. Plot the graph.

On plotting the graph,

From the graph, it is observed that the graph of the function passes through the point $t_2=16.01$.

Therefore, the solution of the function $f\left(t\right)$ is $t_2=16.01$.

Subtract $t_1,t_2$ to calculate the no. of hours,

$$\begin{gathered} t_2-t_1=16.017-16 \\ =0.017 \end{gathered}$$

Multiply $0.017$ by data-custom-editor="chemistry" $60$ to calculate the no. of minutes,

$0.017\times 60=1.02$.

Thus, the pot will be full by $4:01 PM$.

Part (b) Step 1. Find the total amount of water that flows out of the snowfield in a single day.

The function for the amount of water that flows,

$v=15\left[t_2-t_1\right]+15\left(\frac{12}{\mathrm{\pi }}\right)\left[\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_2-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_1-16\right)}{12}\right)\right] ...... \left(iii\right)$

Here, t indicates the no. of hours after midnight.

For a single day, $$\begin{gathered} t_1=0, \\ {t}_2=24 \end{gathered}$$. Then substituting it in equation (iii),

$$\begin{gathered} v=15\left[24-0\right]+15\left(\frac{12}{\mathrm{\pi }}\right)\left[\sin \left(\frac{\mathrm{\pi }\left(24-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left(0-16\right)}{12}\right)\right] \\ v=360 \end{gathered}$$

Thus, the amount of water that flows out of snowfield in a single day is $360$ gallons.

Part (c) Step 1. Write an expression for the total amount of water that flows from the snowfield.

The function $w\left(t\right)$ indicates the rate at which the snow is melting. Therefore, the function can be written as $\frac{dv}{dt}$, where v is the quantity of pot.

$\frac{dv}{dt}=15\left[1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right]$

Integrate both sides from $t_0$ to t,

$$\begin{gathered} {\int }_{t_0}^t\frac{dv}{dt} dt={\int }_{t_0}^{t}15\left[1+\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right)\right] dt \\ {\int }_{t_0}^t\frac{dv}{dt} dt={\int }_{t_0}^{t}15 dt+15{\int }_{t_0}^t\cos \left(\frac{2\mathrm{\pi }\left(\mathrm{t}-16\right)}{24}\right) dt ...... \left(iv\right) \end{gathered}$$

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 (c) Step 2. Use the fundamental theorem of calculus.

Using the fundamental theorem of calculus in equation (iv),

$$\begin{gathered} v=15{{\left[t\right]}^t}_{t_0}+15{{\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}}\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{12}}\right)}\right]}^t}_{t_0} \\ v=15\left[t-t_0\right]+15\left(\frac{12}{\mathrm{\pi }}\right)\left[\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_0-16\right)}{12}\right)\right] \\ v=15\left[t-t_0\right]+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left({\mathrm{t}}_0-16\right)}{12}\right)\right] ...... \left(v\right) \end{gathered}$$

Part (d) Step 1. Find how long he will have to wait until he comes back to a full pot.

Time time $5$ AM in morning corresponds to the time $t=5$ as time t indicates the no. of hours after midnight.

For full pot, the value of $v=1$.

Substitute $$\begin{gathered} v=1, \\ {t}_0=5 \end{gathered}$$ in equation (v),

$$\begin{gathered} 1=15\left[t-5\right]+\frac{180}{\mathrm{\pi }}\left[\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-\sin \left(\frac{\mathrm{\pi }\left(5-16\right)}{12}\right)\right] \\ 1=15t-75+\frac{180}{\mathrm{\pi }}\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-\frac{180}{\mathrm{\pi }}\sin \left(\frac{-11\mathrm{\pi }}{12}\right) \\ 0=15t+\frac{180}{\mathrm{\pi }}\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-73.122 \end{gathered}$$

Consider the function $f\left(t\right)=15t+\frac{180}{\mathrm{\pi }}\sin \left(\frac{\mathrm{\pi }\left(\mathrm{t}-16\right)}{12}\right)-73.122 ...... \left(vi\right)$

Part (d) Step 2. Substitute t = - 2 in equation (vi), followed by simplification.

Substitute $t=-2$in equation (vi),

$$\begin{gathered} f\left(-2\right)=15\left(-2\right)+\frac{180}{\mathrm{\pi }}\sin \left(\frac{\mathrm{\pi }\left(-2-16\right)}{12}\right)-73.122 \\ =-45.82 \end{gathered}$$

Make the table of values of the function for different values of t,

Part (a) Step 3. Plot the graph.

On plotting the graph,

From the graph, it is observed that the graph of the function passes through the point $\left(8.34,0\right)$.

Therefore, the solution of the equation (vi) is $8.34$.

To calculate the no. of hours,

$8.34-5=3.34$

After $3.34$ hours from $5$ AM in the morning, the pot is filled. Therefore, the pot is full by $8:34$AM.