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Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)/Hint 1

2026-04-15T04:55:36Z

AlanGetz: Created page with "First, solve for the steady state. Then determine the sign of <math>\frac{dS}{dt}</math> for <math>S</math> on either side of the steady state by considering very small and very large values of <math>S</math>."

First, solve for the steady state. Then determine the sign of <math>\frac{dS}{dt}</math> for <math>S</math> on either side of the steady state by considering very small and very large values of <math>S</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)/Solution 1

2026-04-15T04:53:42Z

AlanGetz: Created page with "Let <math>0<R<1</math>. Solving for the steady state <math>0=R-\frac{S}{2+S}</math>, we obtain <math>S = \frac{2R}{1-R}</math>. For <math>S</math> very close to <math>0</math>, we would have <math>R-\frac{S}{2+S}</math> close to <math>R</math>, which is positive, so <math>\frac{dS}{dt}>0</math> for <math>S<\frac{2R}{1-R}</math>. On the other hand, for <math>S</math> very large, we would have <math>\frac{S}{2+S}</math> close to <math>1</math>, so <math>R-\frac{S}{2+S}<..."

Let <math>0<R<1</math>. Solving for the steady state <math>0=R-\frac{S}{2+S}</math>, we obtain <math>S = \frac{2R}{1-R}</math>. For <math>S</math> very close to <math>0</math>, we would have <math>R-\frac{S}{2+S}</math> close to <math>R</math>, which is positive, so <math>\frac{dS}{dt}>0</math> for <math>S<\frac{2R}{1-R}</math>. On the other hand, for <math>S</math> very large, we would have <math>\frac{S}{2+S}</math> close to <math>1</math>, so <math>R-\frac{S}{2+S}</math> is negative, and therefore <math>\frac{dS}{dt}<0</math>.

The situation is described by the following phase diagram:

[[File:Q12(c) diagram, UBC math 100 December 2024 exam.png]]

File:Q12(c) diagram, UBC math 100 December 2024 exam.png

2026-04-15T04:38:02Z

AlanGetz: Uploaded a work by {{Unknown|author}} from UBC December 2024 math 100 B exam with UploadWizard

=={{int:filedesc}}==
{{Information
|description={{en|1=Q12(c) diagram, UBC math 100 December 2024 exam}}
|date=2024-12-17
|source=UBC December 2024 math 100 B exam
|author={{Unknown|author}}
|permission=
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[[Category:Mathematics]]

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (b)/Solution 1

2026-04-15T04:34:51Z

AlanGetz: Created page with "For all <math>S\geq 0</math>, <math>\frac{S}{2+S}<1</math>. Since <math>\frac{3}{2}>1</math>, we have that <math display="block"> \frac{3}{2}-\frac{S}{2+S}> \frac{3}{2} - 1 > 0, </math> so <math>\frac{dS}{dt}>0</math> for all nonnegative <math>S</math>. That is, over time the substrate builds up without bound."

For all <math>S\geq 0</math>, <math>\frac{S}{2+S}<1</math>. Since <math>\frac{3}{2}>1</math>, we have that
<math display="block">
\frac{3}{2}-\frac{S}{2+S}> \frac{3}{2} - 1 > 0,
</math>
so <math>\frac{dS}{dt}>0</math> for all nonnegative <math>S</math>. That is, over time the substrate builds up without bound.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (a)/Solution 1

2026-04-15T04:09:35Z

AlanGetz: Created page with "The phase diagram is File:Q12(a) diagram, UBC math 100 December 2024 exam.png"

The phase diagram is

[[File:Q12(a) diagram, UBC math 100 December 2024 exam.png]]

File:Q12(a) diagram, UBC math 100 December 2024 exam.png

2026-04-15T04:08:54Z

AlanGetz: Uploaded a work by {{Unknown|author}} from UBC December 2024 math 100 B exam with UploadWizard

=={{int:filedesc}}==
{{Information
|description={{en|1=Q12(a) diagram, UBC math 100 December 2024 exam}}
|date=2024-12-17
|source=UBC December 2024 math 100 B exam
|author={{Unknown|author}}
|permission=
|other versions=
}}

=={{int:license-header}}==
{{cr-ubc}}

[[Category:Mathematics]]

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (a)/Hint 1

2026-04-15T04:06:56Z

AlanGetz: Created page with "Set <math>R=0</math> and consider the differential equation <math>\frac{dS}{dt}=-\frac{S}{2+S}</math>. Find the equilibrium points by solving <math>\frac{dS}{dt}=0</math>, then determine the sign of <math>\frac{dS}{dt}</math> for the values of <math>S</math> on either side."

Set <math>R=0</math> and consider the differential equation <math>\frac{dS}{dt}=-\frac{S}{2+S}</math>. Find the equilibrium points by solving <math>\frac{dS}{dt}=0</math>, then determine the sign of <math>\frac{dS}{dt}</math> for the values of <math>S</math> on either side.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)/Statement

2026-04-15T04:00:52Z

AlanGetz: Created page with "Suppose a chemical is being converted to a product in tank. The amount of the chemical is <math>S</math>, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate <math>R</math>, then the amount of the chemical in the tank satisfies the differential equation <math display="block"> \frac{dS}{dt}=R-\frac{S}{2+S}. </math> Describe in words what happens to the amount of the chemical in the tank over a l..."

Suppose a chemical is being converted to a product in tank. The amount of the chemical is <math>S</math>, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate <math>R</math>, then the amount of the chemical in the tank satisfies the differential equation
<math display="block">
\frac{dS}{dt}=R-\frac{S}{2+S}.
</math>
Describe in words what happens to the amount of the chemical in the tank over a long period of time if <math>0 < R < 1</math>. Support your argument with a phase diagram.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (b)/Statement

2026-04-15T03:59:24Z

Suppose a chemical is being converted to a product in tank. The amount of the chemical is <math>S</math>, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate <math>R</math>, then the amount of the chemical in the tank satisfies the differential equation
<math display="block">
\frac{dS}{dt}=R-\frac{S}{2+S}.
</math>
Describe in words what happens to the amount of the chemical in the
tank over a long period of time if <math>R=\frac{3}{2}</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (a)/Statement

2026-04-15T03:55:45Z

AlanGetz: Created page with "Suppose a chemical is being converted to a product in tank. The amount of the chemical is <math>S</math>, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate <math>R</math>, then the amount of the chemical in the tank satisfies the differential equation <math display="block"> \frac{dS}{dt}=R-\frac{S}{2+S}. </math> Draw a phase diagram for the differential equation, for <math>S\geq 0</math>, whe..."

Suppose a chemical is being converted to a product in tank. The amount of the chemical is <math>S</math>, which is nonnegative and changes with respect to time. If the chemical is being added to the tank at a constant nonnegative rate <math>R</math>, then the amount of the chemical in the tank satisfies the differential equation
<math display="block">
\frac{dS}{dt}=R-\frac{S}{2+S}.
</math>
Draw a phase diagram for the differential equation, for <math>S\geq 0</math>, when <math>R=0</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 14

2026-04-15T03:52:30Z

AlanGetz:

[[Category:MER QGQ flag]][[Category:MER QGH flag]][[Category:MER QGS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 13

2026-04-15T03:52:11Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (c)

2026-04-15T03:51:52Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (b)

2026-04-15T03:51:29Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 12 (a)

2026-04-15T03:51:08Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (d)

2026-04-15T03:50:48Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)

2026-04-15T03:50:28Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (b)

2026-04-15T03:50:08Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (a)

2026-04-15T03:49:48Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 10

2026-04-15T03:49:26Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 09 (b)

2026-04-15T03:48:34Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 09 (a)

2026-04-15T03:48:13Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 08

2026-04-15T03:47:55Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 07

2026-04-15T03:47:31Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 06

2026-04-15T03:47:14Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 05

2026-04-15T03:46:55Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 04 (c)

2026-04-15T03:46:36Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 04 (b)

2026-04-15T03:46:16Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 04 (a)

2026-04-15T03:45:50Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 03 (b)

2026-04-15T03:45:26Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 03 (a)

2026-04-15T03:45:04Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 02 (c)

2026-04-15T03:44:39Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 02 (b)

2026-04-15T03:44:17Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 02 (a)

2026-04-15T03:43:52Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 01 (c)

2026-04-15T03:43:30Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 01 (b)

2026-04-15T03:42:38Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 01 (a)

2026-04-15T03:42:13Z

AlanGetz:

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (d)/Solution 1

2026-04-15T03:40:05Z

AlanGetz: Created page with "From the sketch, we see <math>y(x)</math> is concave down. That means its tangent lines lie above the curve, so linear approximations produce over approximations. Hence, the first step of Euler's method produces an over approximation. Additionally, <math>y(x)</math> is increasing, so an initial over approximation plus a linear approximation will produce an over approximation in the second step as well."

From the sketch, we see <math>y(x)</math> is concave down. That means its tangent lines lie above the curve, so linear approximations produce over approximations. Hence, the first step of Euler's method produces an over approximation. Additionally, <math>y(x)</math> is increasing, so an initial over approximation plus a linear approximation will produce an over approximation in the second step as well.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (d)/Hint 1

2026-04-15T03:36:28Z

AlanGetz: Created page with "From your sketch in part (a), does the curve look concave up or concave down? What does this imply about whether Euler's method will yield an over approximation or an under approximation?"

From your sketch in part (a), does the curve look concave up or concave down? What does this imply about whether Euler's method will yield an over approximation or an under approximation?

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (d)/Statement

2026-04-15T03:31:10Z

AlanGetz: Created page with "In this question, we’ll investigate solutions to the differential equation <math display="block"> \frac{dy}{dt}= (y-1)(3-y). </math> Is your numerical approximation from (c) using Euler’s method larger or smaller than <math>y(1)</math>? You may use your previous work, but you must explain your answer."

In this question, we’ll investigate solutions to the differential equation
<math display="block">
\frac{dy}{dt}= (y-1)(3-y).
</math>
Is your numerical approximation from (c) using Euler’s method larger or smaller than <math>y(1)</math>? You may use your previous work, but you must explain your answer.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)/Solution 1

2026-04-15T03:29:46Z

AlanGetz:

Let us denote <math>f(y) = (y-1)(3-y)</math>. At <math>t=0</math>, we have <math>y_0=y(0)=2</math> and <math>f(y_0)=f(2)=(2-1)(3-2)=1</math>. Hence, at our next time step, we have <math>y_1=1+(1/2)=5/2</math> and <math>f(y_1)=f(5/2)=(3/2)(1/2)=3/4</math>. At our final time step, we have <math>y_2=5/2+(1/2)(3/4)=23/8</math>, which is our approximation.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)/Solution 1

2026-04-15T03:28:56Z

AlanGetz: Created page with "Let us denote <math>f(y) = (y-1)(3-y)</math>. At <math>t=0</math>, we have <math>y_0=y(0)=2</math> and <math>f(y_0)=f(2)=(2-1)(3-2)=1</math>. Hence, at our next time step, we have <math>y_1=1+(1/2)=5/2</math> and <math>f(y_1)f(5/2)=(3/2)(1/2)=3/4</math>. At our final time step, we have <math>y_2=5/2+(1/2)(3/4)=23/8</math>, which is our approximation."

Let us denote <math>f(y) = (y-1)(3-y)</math>. At <math>t=0</math>, we have <math>y_0=y(0)=2</math> and <math>f(y_0)=f(2)=(2-1)(3-2)=1</math>. Hence, at our next time step, we have <math>y_1=1+(1/2)=5/2</math> and <math>f(y_1)f(5/2)=(3/2)(1/2)=3/4</math>. At our final time step, we have <math>y_2=5/2+(1/2)(3/4)=23/8</math>, which is our approximation.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)/Hint 1

2026-04-15T03:17:11Z

AlanGetz: Created page with "Since we are using Euler's method with two steps, beginning at <math>t=0</math> and ending at <math>t=1</math>, our time step will need to be <math>\Delta t = 1/2</math>."

Since we are using Euler's method with two steps, beginning at <math>t=0</math> and ending at <math>t=1</math>, our time step will need to be <math>\Delta t = 1/2</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (c)/Statement

2026-04-15T03:14:22Z

AlanGetz: Created page with "In this question, we’ll investigate solutions to the differential equation <math display="block"> \frac{dy}{dt}= (y-1)(3-y). </math> Use Euler’s method with two steps to approximate y(1)."

In this question, we’ll investigate solutions to the differential equation
<math display="block">
\frac{dy}{dt}= (y-1)(3-y).
</math>
Use Euler’s method with two steps to approximate y(1).

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (b)/Hint 1

2026-04-15T03:12:01Z

AlanGetz:

Where does it look like the curve sketched out in part (a) intersects with the line <math>x=1</math>?

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (b)/Hint 1

2026-04-15T03:11:45Z

AlanGetz: Created page with "Where does it look like the curve sketched out in part (a) intersects with the line <math>x=1</math>."

Where does it look like the curve sketched out in part (a) intersects with the line <math>x=1</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (b)/Solution 1

2026-04-15T03:05:42Z

AlanGetz: Created page with "Based on the sketch in part (a), <math>y(1)</math> is approximately <math>2.75</math>, though any answer in the interval <math>(2.5,3)</math> is acceptable."

Based on the sketch in part (a), <math>y(1)</math> is approximately <math>2.75</math>, though any answer in the interval <math>(2.5,3)</math> is acceptable.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (b)/Statement

2026-04-15T03:01:31Z

AlanGetz: Created page with "In this question, we’ll investigate solutions to the differential equation <math display="block"> \frac{dy}{dt}= (y-1)(3-y). </math> Using your sketch from (a), give an approximate value for <math>y(1)</math>."

In this question, we’ll investigate solutions to the differential equation
<math display="block">
\frac{dy}{dt}= (y-1)(3-y).
</math>
Using your sketch from (a), give an approximate value for <math>y(1)</math>.

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (a)/Statement

2026-04-15T03:00:40Z

AlanGetz:

In this question, we’ll investigate solutions to the differential equation
<math display="block">
\frac{dy}{dt}= (y-1)(3-y).
</math>
Use the slope field below to sketch the solution to the differential equation with initial value <math>y(0) = 2</math>. Your sketch should extend from <math>t=-2</math> to <math>t = 2</math>.

[[File:Q11 diagram, UBC math 100 December 2024 exam.png]]

Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 11 (a)/Solution 1

2026-04-15T02:58:42Z

AlanGetz: Created page with "The solution is given by File:Q11(a) diagram, UBC math 100 December 2024 exam.png"

The solution is given by

[[File:Q11(a) diagram, UBC math 100 December 2024 exam.png]]