ENGINEERING-25
MuPAD Tutorial
Problem
The following function has exacly one INFLECTION point in the right-half of the Cartesian
Plane:
𝐴
𝑦(𝑡) =
1 + (𝐵 + 7)𝑒 −𝑡⁄𝜏
 Where, 𝐴, 𝐵, 𝑎𝑛𝑑 𝜏are all POSITIVE Constants.
For this situation find an Exact (that is, a SYMBOLIC) solution for the
CoOrdinates of the inflection point location
The Answer
(𝑡𝑖𝑛𝑓𝑙 , 𝑦𝑖𝑛𝑓𝑙 ) →
Quotient Rule
If f(x) and g(x) are differentiable functions with g(x) ≠ 0, then
d  f x 



dx  g x  
g x 
df
dg
 f x 
dx
dx
2
g x 
The HAND Solution
𝐴
𝑦(𝑡) =
1 + (𝐵 + 7)𝑒 −𝑡⁄𝜏
d  f x 

dx  g x  
g x 
df
dg
 f x 
dx
dx
2
g x 
The MuPAD NoteBook
Bruce Mayer, PE
ENGR25 • 28Nov13
ENGR25_Inflection_Point_Tutorial_1311.mn
A, B, τ, and t are all POSITIVE so use the assume command
• Either of the "assume" commands below insures positivevalue behavior
assume(A, Type::Positive); assume(B > 0);
assume(`τ` > 0); assume(t, Type::Positive)
Check that `τ` was properly accepted by MuPAD
getprop(`τ`)
The function
y := A/(1+(B+7)*E^(-t/`τ`))
When In Doubt...PLOT IT!
• for the test-plot set A=1, B=2, τ=3
The Test Function z(u)
z := 1/(1+(2+7)*E^(-u/3))
plot(z, u =-10..20, GridVisible = TRUE,
LineWidth = 0.04*unit::inch,
Width = 320*unit::mm, Height = 180*unit::mm,
AxesTitleFont = ["sans-serif", 24],
TicksLabelFont=["sans-serif", 16])
Looks like the inflection point is at about (6.6,0.5) for A=1,
B=2, τ=3
Now The First Derivative
dydt := diff(y, t)
Also Plot the first Derivative, dy/dt = m. Look for where m is
MAX or MIN
m := subs(dydt, A=1, B=2, `τ`=3)
plot(m, t =-10..20, GridVisible = TRUE,
LineWidth = 0.04*unit::inch,
Width = 320*unit::mm, Height = 180*unit::mm,
AxesTitleFont = ["sans-serif", 24],
TicksLabelFont=["sans-serif", 16])
The Second Derivative
d2ydt2 := diff(dydt, t)
Set the 2nd derivative to ZERO to find the location of the
Inflection Point
ti := solve(d2ydt2,t)
tiS := Simplify(ti)
Double Check solution using the simplified version of d2y/dt2
d2ydt2S := Simplify(d2ydt2)
tsolS := solve(d2ydt2S,[t])
Checks OK. Now extract solution from SET NOTATION to
back-sub into y(x)
tiback := solvelib::getElement(tiS)
Then the y-value of the inflection point
yi := subs(y, t = tiback)
Simplify yi
yiS := Simplify(yi)
Now that is REALLY Simple
Finally the inflection Point Location.
[tinfl,yinfl] := [tiback, yi]
Check the test Case
[titest,yitest] := [subs(tiback, B =2,
`τ`=3), subs(yiS, A = 1)]
float([titest,yitest])