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Sagot :
Sure! Let's tackle the problem step by step.
### Part a: How long does it take the tank to fill?
Given:
- The total volume the tank can hold is \(V_{\text{full }} = \pi \left( 20 + \frac{38}{3} \sqrt{2} \right) \approx 119.11 \text{ cubic feet}\).
- The actual height of the tank is \(h_{\text{full }} = \sqrt{8} + 5 \approx 7.83 \text{ feet}\).
To determine the time it takes for the tank to fill, we need to understand the rate at which water is being added to the tank. For this, let's assume water flows into the tank at a constant rate. From the context, assume this rate is such that it aligns with the volume calculations used earlier.
Without specific information about the flow rate, we could assume a constant flow that fills the tank in a reasonable amount of time. Since time, \(t\), in minutes, might not be directly given, let's deduce it in the following general manner:
1. Flow rate assumption: Suppose a certain known flow rate \(R \text{ cubic feet per minute}\).
2. Time to fill the tank: Given the volume \(V_{\text{full }} = 119.11\) cubic feet,
[tex]\[ t_{\text{full}} = \frac{V_{\text{full}}}{R} \][/tex]
For now, we are concerned with how the tank volume and height of water vary over time.
### Part b: Sketch of graphs \(V\) vs \(t\) and \(h\) vs \(t\)
Let's create qualitative sketches of the graphs. Here's what we can infer about each relationship:
1. Graph of \(V\) vs \(t\):
- \(V\) increases from 0 to \(119.11\) cubic feet as \(t\) increases.
- If filling is at a constant rate, the graph will be a straight line from \((0, 0)\) to \((t_{\text{full}}, 119.11)\).
2. Graph of \(h\) vs \(t\):
- \(h\) increases from 0 to \(7.83\) feet as \(t\) increases.
- The rate of change of \(h\) is not constant because the tank likely has a varying cross-sectional area. The increase in \(h\) might slow down as it progressively fills the spherical base and speeds up going into the chimney, resulting in a concave or convex curve depending on specific tank geometry especially as it transitions from the hemisphere to the cylindrical part.
#### Sketch:
##### Graph of \(V\) vs \(t\)
- Horizontal axis (t): Time in minutes
- Vertical axis (V): Volume in cubic feet
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = [tex]$t$[/tex] (minutes),
ylabel = {[tex]$V$[/tex] (cubic feet)},
]
% The straight line from (0,0) to (t_full, 119.11)
\addplot [
domain=0:10,
samples=2,
color=blue,
]
{11.911*x}; % Assuming t_full = 10 minutes for simplicity
\addlegendentry{[tex]$V(t)$[/tex]}
\end{axis}
\end{tikzpicture}
\][/tex]
##### Graph of \(h\) vs \(t\)
- Horizontal axis (t): Time in minutes
- Vertical axis (h): Depth of water in feet
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = [tex]$t$[/tex] (minutes),
ylabel = {[tex]$h$[/tex] (feet)},
]
\addplot [
domain=0:10,
samples=100,
color=blue,
]
{7.83 (1 - exp(-0.3x))}; % Hypothetical concave shape
\addlegendentry{[tex]$h(t)$[/tex]}
\end{axis}
\end{tikzpicture}
\][/tex]
In this sketch, the concave-up shape signifies the initial slower rise in height (filling spherical part) and faster rise near the end (filling the cylindrical chimney).
Points to label:
- At \(t=0\): \(V(0) = 0, h(0) = 0\)
- At [tex]\(t=t_{\text{full}}\)[/tex]: [tex]\(V(t_{\text{full}}) \approx 119.11, h(t_{\text{full}}) \approx 7.83\)[/tex]
### Part a: How long does it take the tank to fill?
Given:
- The total volume the tank can hold is \(V_{\text{full }} = \pi \left( 20 + \frac{38}{3} \sqrt{2} \right) \approx 119.11 \text{ cubic feet}\).
- The actual height of the tank is \(h_{\text{full }} = \sqrt{8} + 5 \approx 7.83 \text{ feet}\).
To determine the time it takes for the tank to fill, we need to understand the rate at which water is being added to the tank. For this, let's assume water flows into the tank at a constant rate. From the context, assume this rate is such that it aligns with the volume calculations used earlier.
Without specific information about the flow rate, we could assume a constant flow that fills the tank in a reasonable amount of time. Since time, \(t\), in minutes, might not be directly given, let's deduce it in the following general manner:
1. Flow rate assumption: Suppose a certain known flow rate \(R \text{ cubic feet per minute}\).
2. Time to fill the tank: Given the volume \(V_{\text{full }} = 119.11\) cubic feet,
[tex]\[ t_{\text{full}} = \frac{V_{\text{full}}}{R} \][/tex]
For now, we are concerned with how the tank volume and height of water vary over time.
### Part b: Sketch of graphs \(V\) vs \(t\) and \(h\) vs \(t\)
Let's create qualitative sketches of the graphs. Here's what we can infer about each relationship:
1. Graph of \(V\) vs \(t\):
- \(V\) increases from 0 to \(119.11\) cubic feet as \(t\) increases.
- If filling is at a constant rate, the graph will be a straight line from \((0, 0)\) to \((t_{\text{full}}, 119.11)\).
2. Graph of \(h\) vs \(t\):
- \(h\) increases from 0 to \(7.83\) feet as \(t\) increases.
- The rate of change of \(h\) is not constant because the tank likely has a varying cross-sectional area. The increase in \(h\) might slow down as it progressively fills the spherical base and speeds up going into the chimney, resulting in a concave or convex curve depending on specific tank geometry especially as it transitions from the hemisphere to the cylindrical part.
#### Sketch:
##### Graph of \(V\) vs \(t\)
- Horizontal axis (t): Time in minutes
- Vertical axis (V): Volume in cubic feet
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = [tex]$t$[/tex] (minutes),
ylabel = {[tex]$V$[/tex] (cubic feet)},
]
% The straight line from (0,0) to (t_full, 119.11)
\addplot [
domain=0:10,
samples=2,
color=blue,
]
{11.911*x}; % Assuming t_full = 10 minutes for simplicity
\addlegendentry{[tex]$V(t)$[/tex]}
\end{axis}
\end{tikzpicture}
\][/tex]
##### Graph of \(h\) vs \(t\)
- Horizontal axis (t): Time in minutes
- Vertical axis (h): Depth of water in feet
[tex]\[ \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = [tex]$t$[/tex] (minutes),
ylabel = {[tex]$h$[/tex] (feet)},
]
\addplot [
domain=0:10,
samples=100,
color=blue,
]
{7.83 (1 - exp(-0.3x))}; % Hypothetical concave shape
\addlegendentry{[tex]$h(t)$[/tex]}
\end{axis}
\end{tikzpicture}
\][/tex]
In this sketch, the concave-up shape signifies the initial slower rise in height (filling spherical part) and faster rise near the end (filling the cylindrical chimney).
Points to label:
- At \(t=0\): \(V(0) = 0, h(0) = 0\)
- At [tex]\(t=t_{\text{full}}\)[/tex]: [tex]\(V(t_{\text{full}}) \approx 119.11, h(t_{\text{full}}) \approx 7.83\)[/tex]
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