Transcript Differential Equations, Let`s Get This Party Started!

Let’s Get this Party Started…
You’re Invited
 Differential Equations Party
 Hosted by Beth Tsai ([email protected]), Wake Technical
Community College
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Where: AMATYC Conference
When: November 2016
Why: Introduce Differential Equations in a fun way
How: party games and participation required
The gift of prior math knowledge necessary, refreshments will
be provided
Every good party has a cake. Let’s bake one in this
session. Each baker knows that before we can put
“icing” on this cake/session, we must allow the
cake to cool sufficiently.
According to Newton’s Law of Cooling, we know
that the rate at which our cake cools is
proportional to the difference between the
temperature of the cake and its surrounding
medium… hmmm…
With the Cake in the oven, it’s time for the
Opening Dance
words
DEQ
rate
Solution
function
predictions
Gifts we bring
 Prior Knowledge
 Algebraic Equations
3𝑥 + 1 = 7
3𝑥 = 6
𝑥=2
 New Math
 Differential Equations
𝑑𝑦
=𝑥
𝑑𝑥
1 2
𝑦 = 𝑥 +𝐶
2
Games we play
Games we play...follow the leader
DEQ:
𝑑𝑦
𝑑𝑥
=𝑥
 Analyze:

𝑑𝑦
𝑑𝑥
+, 𝑤ℎ𝑒𝑛 𝑥 +
= −, 𝑤ℎ𝑒𝑛 𝑥 −
0, 𝑤ℎ𝑒𝑛 𝑥 = 0
 Illustrate:
y'
(x,m)
x
Games we play...follow the leader
DEQ:
𝑑𝑦
𝑑𝑥
=𝑥
 Create a slope field of the
solution
slopes are constant for x
which occur along vertical
lines in the plane
Red, m +
Green, m=0
Blue, m -
 illustration
y
x
Games we play…follow the leader
Graph of DEQ
Graph of solution field
y'
y
(x,m)
x
x-
x=0
x+
x
m-
m=0
m+
Games we play…follow the leader
DEQ:
𝑑𝑦
𝑑𝑥
=𝑥
 Solve the differential equation
Preview of Separation of
Variables
𝑑𝑦 = 𝑥𝑑𝑥
𝑑𝑦 =
 illustration
y
𝑥𝑑𝑥
1 2
𝑦 = 𝑥 +𝐶
2
x
Games we play…follow the leader
DEQ:
𝑑𝑦
𝑑𝑥
=𝑥
 Check your solution
1
2
Solution: 𝑦 = 𝑥 2 + 𝐶
1
2
 𝑦 = 𝑥2 + 𝐶
𝑦′ =
1
2
𝑦′ = 𝑥
2𝑥 + 0
Games we play…follow the leader
Team A:
𝑑𝑦
𝑑𝑥
=𝑥
versus
 Analyze
 Plot the differential equation
 Create a slope field of the
solution
 Solve the differential equation
 Check your solution
Team B:
𝑑𝑦
𝑑𝑥
=𝑦
 Analyze
 Plot the differential equation
 Create a slope field of the
solution
 Solve the differential equation
 Check your solution
Games we play…follow the leader
Team A:
𝑑𝑦
𝑑𝑥
=𝑥
Team B:
y'
𝑑𝑦
𝑑𝑥
=𝑦
y'
x
y
Games we play…follow the leader
Team A:
𝑑𝑦
𝑑𝑥
=𝑥
Team B:
y
𝑑𝑦
𝑑𝑥
=𝑦
y
x
x
Games we play…follow the leader
Team A:
𝑑𝑦
𝑑𝑥
=𝑥
Team B:
 Solution
 Solution
1 2
𝑦 = 𝑥 +𝐶
2
𝑑𝑦
𝑑𝑥
=𝑦
𝑦 = 𝐶𝑒 𝑥
y
y
x
x
This isn’t your Grandma’s Calc I
More fun and games we play…slippery slopes
Match the differential equation to its slope field
Differential Equation
𝑑𝑦
=𝑦−3
𝑑𝑥
𝑑𝑦
=3
𝑑𝑥
𝑑𝑦
= 3𝑥𝑦
𝑑𝑥
𝑑𝑦
=𝑥−3
𝑑𝑥
𝑑𝑦
= 3𝑥 − 𝑦
𝑑𝑥
𝑑𝑦
= 3cos(𝑦)
𝑑𝑥
𝑑𝑦
= 3sin(𝑦)
𝑑𝑥
Slope Field
More fun and games we play…slippery slopes
Match the differential equation to its slope field
Differential Equation
Slope Field
𝑑𝑦
=𝑦−3
𝑑𝑥
y
𝑑𝑦
=3
𝑑𝑥
𝑑𝑦
= 3𝑥𝑦
𝑑𝑥
𝑑𝑦
=𝑥−3
𝑑𝑥
𝑑𝑦
= 3𝑥 − 𝑦
𝑑𝑥
x
𝑑𝑦
= 3cos(𝑦)
𝑑𝑥
𝑑𝑦
= 3sin(𝑦)
𝑑𝑥
y
y
x
y
x
x
Slippery slopes…
Post Game Analysis…
𝑑𝑦
= 3𝑥 − 𝑦
𝑑𝑥
𝑑𝑦
= 3cos(𝑦)
𝑑𝑥
y
y
x
𝑑𝑦
= 3sin(𝑦)
𝑑𝑥
y
x
x
More Games we play…
Pin the Graph to its Differential Equation
𝑑𝑦
= 0.5𝑦(20 − 𝑦)
𝑑𝑥
𝑑𝑦
= −32 + .0001(64𝑦 2 )
𝑑𝑥
𝑑𝑦
= 0.5𝑦
𝑑𝑥
𝑑𝑦
= −0.5(𝑦 − 70)
𝑑𝑥
More Games we play…
Pin the Graph to its Differential Equation
y'
y'
y
𝑑𝑦
= 0.5𝑦(20 − 𝑦)
𝑑𝑥
y
𝑑𝑦
= −32 + .0001(64𝑦 2 )
𝑑𝑥
y'
y'
𝑑𝑦
= 0.5𝑦
𝑑𝑥
y
𝑑𝑦
= −0.5(𝑦 − 70)
𝑑𝑥
y
More Games we play…
Spreading a rumor…whisper…time
for cake?
A sneak peek at the sequel…
y
y'
y
𝑑𝑦
= 0.005𝑦 20 − 𝑦 ,
𝑑𝑥
𝑦 0 =1
x
separation of
variables
20𝑒 𝑥/10
𝑦 = 𝑥/10
𝑒
+ 19
Time for cake
According to Newton’s Law of Cooling,
we know that the rate at which our cake
cools is proportional to the difference
between the temperature of the cake and
its surrounding medium. The cake was
taken out of a 350 degree oven into a
room that is a constant temperature of 70
degrees. Let the constant of
proportionality be -0.5. It has been 20
minutes since the party started, is it time
to ice the cake?
𝑑𝑦
= −0.5 𝑦 − 70 , 𝑦 0 = 350
𝑑𝑥
Icing on the cake…
Graph of DEQ
𝑑𝑦
= −0.5(𝑦 − 70)
𝑑𝑥
Analysis
y'
y
 Graph is linear with negative
slope
 If object starts at a temperature
above 70, the slopes are negative
(decaying to room temperature
 If object starts at a temperature
below 70, slopes are positive
(increasing to room
temperature)
 If an object is at 70 degrees,
there is no change in the
temperature (equilibrium)
Icing on the cake…
Graph of Slope field
y
Analysis
 Confirms DEQ
 Confirms equilibrium solution
at room temperature of 70
degrees
x
Icing on the cake…
𝑑𝑦
= −0.5 𝑦 − 70 , 𝑦 0 = 350
𝑑𝑥
 Separation of Variables
𝑑𝑦
= −0.5𝑑𝑥
𝑦 − 70
𝑑𝑦
= −0.5𝑑𝑥
𝑦 − 70
𝑙𝑛 𝑦 − 70 =−.05𝑥 + 𝐶
𝑒 𝑙𝑛 𝑦−70 = 𝑒 −.05𝑥+𝐶
Symbolic Solution
Continued
𝑦 − 70 = 𝐶𝑒 −.05𝑥
As Temp is above 70,
𝑦 − 70 = 𝐶𝑒 −.05𝑥
y = 𝐶𝑒 −.05𝑥 +70
Solving for C,
y = 280𝑒 −.05𝑥 +70
Is it time for cake?
Symbolic Solution
 y = 280𝑒 −.05𝑥 + 70
Make a prediction
 It has been twenty minutes, is
it time for cake?
 Find temperature after 20
minutes.
 y = 280𝑒 −.05(20) + 70 ≈
70.01 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
Last Dance
words
DEQ
rate
Solution
function
predictions
Last Dance
words
DEQ
rate
Solution
function
According to Newton’s Law of Cooling, we know that the rate at which our cake cools
is proportional to the difference between the temperature of the cake and its
surrounding medium. The cake was taken out of a 350 degree oven into a room that is
a constant temperature of 70 degrees. Let the constant of proportionality be 0.5. It
has been 20 minutes since the party started, is it time to ice the cake?
𝑑𝑦
𝑑𝑥
y'
= −0.5 𝑦 − 70 , 𝑦 0 = 350
where x is time in minutes and y is
temperature in degrees Farenheit
y = 280𝑒 −.05𝑥 + 70
y
y
By separation of variables
predictions
Find temperature after 20 minutes.
y = 280𝑒 −.05(20) + 70 ≈ 70.01 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
x
Time for Presents
Let us eat cake…
Enjoy the rest of the Conference/Party
Thanks for attending!!!
Beth Tsai: [email protected]