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Title: van der waals
Description: These notes served as our performance task last semester. This function is based on the van der Waals equation for predicting the pressure p of a gas as a function of v at a fixed temperature.

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Given problem: Consider the rational function 𝑝 =

512500v2 βˆ’449000v+19307
125vΒ²(1000vβˆ’43)


...
The function above models
the pressure p of a carbon dioxide in terms of volume v if the temperature is 500
Kelvins
...
Sketch the graph of the function
...
What is the domain of the function? (1 point)
Denominator of the function: 125v2(1000v-43)=0
125v2
1000v-43
πŸπŸπŸ“π’—Β²
𝟎
1000𝑣 βˆ’ 43 = 0
=
1000𝑣 = 43
πŸπŸπŸ“
πŸπŸπŸ“
1000𝑣
43
𝒗= 𝟎
=
1000
1000
43
𝑣=
π‘œπ‘Ÿ 0
...
What is/are the vertical asymptote(s) of the function? (1 point)
125v2(1000v-43)=0

πŸπŸπŸ“π’—Β²
𝟎
=
πŸπŸπŸ“
πŸπŸπŸ“
𝒗= 𝟎

πŸπŸŽπŸŽπŸŽπ’— βˆ’ πŸ’πŸ‘ = 𝟎
πŸπŸŽπŸŽπŸŽπ’— = πŸ’πŸ‘
πŸπŸŽπŸŽπŸŽπ’—
πŸ’πŸ‘
=
𝟏𝟎𝟎𝟎
𝟏𝟎𝟎𝟎
πŸ’πŸ‘
𝒗=
𝒐𝒓 𝟎
...
What is the horizontal asymptote of the function? (1 point)
Note: If the denominator’s degree is greater than the numerator’s degree, the
horizontal asymptote is the x-axis (v); y=0 (p)
Denominator’s degree= 3
Numerator’s degree= 2

Therefore, the horizontal
asymptote is p=0

5
...

6
...

7
...

(5 points)
INTERVAL

𝒗<0

Test Point
512500𝑣 2 βˆ’ 449000𝑣 + 19307
125𝑣 2 (1000𝑣 βˆ’ 43)
512500v 2 βˆ’ 449000v + 19307
𝑝=
125vΒ²(1000v βˆ’ 43)

-1
+
-

𝟎< 𝑣<

πŸ’πŸ‘
𝟏𝟎𝟎𝟎

or 0
...
040
+
-

πŸ’πŸ‘
𝟏𝟎𝟎𝟎
0
...
Explain the significance of the vertical asymptote(s) in practical terms
...
043 liters
...

9
...
(5 points)
As the volume increases, the pressure becomes closer to 0
...



Title: van der waals
Description: These notes served as our performance task last semester. This function is based on the van der Waals equation for predicting the pressure p of a gas as a function of v at a fixed temperature.