Freshmen Quotient Rule and More: Problem...
36
Freshmen Quotient Rule and More: Problem 1941 Presentation of DiMuro’s Solution Chuck F. Rocca Jr. Western Connecticut State University C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 1/1
Transcript of Freshmen Quotient Rule and More: Problem...
Freshmen Quotient Rule and More: Problem 1941Presentation of DiMuro’s Solution
Chuck F. Rocca Jr.
Western Connecticut State University
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 1 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 2 / 1
Problem 1941
Find all pairs of nonzero real valued functions (u(x), v(x)) that satisfyboth of the following equations:(u
v
)′=
u′
v ′and u′v ′ = uv .[1]
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 3 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 4 / 1
Review of Calculus I
Theorem (Quotient Rule [3])
If f (x) and g(x) are differentiable functions of x and g(x) 6= 0 for all x ,then
d
dx
f
g=
g dfdx − f dg
dx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 5 / 1
Review of Calculus I
Proof.
Assume f and g are as stated in the hypothesis, then
d
dx
f
g= lim
h→x
f (h)g(h) −
f (x)g(x)
h − x
= limh→x
1
h − x
f (h)g(x)− f (x)g(h)
g(x)g(h)
= limh→x
1
h − x
(f (h)− f (x))g(x)− f (x)(g(h)− g(x))
g(x)g(h)
= limh→x
1
g(x)g(h)
[(f (h)− f (x)
h − x
)g(x)− f (x)
(g(h)− g(x)
h − x
)]=
g dfdx − f dg
dx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 6 / 1
Review of Calculus I
Proof.
Assume f and g are as stated in the hypothesis, then
d
dx
f
g= lim
h→x
f (h)g(h) −
f (x)g(x)
h − x
= limh→x
1
h − x
f (h)g(x)− f (x)g(h)
g(x)g(h)
= limh→x
1
h − x
(f (h)− f (x))g(x)− f (x)(g(h)− g(x))
g(x)g(h)
= limh→x
1
g(x)g(h)
[(f (h)− f (x)
h − x
)g(x)− f (x)
(g(h)− g(x)
h − x
)]=
g dfdx − f dg
dx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 6 / 1
Review of Calculus I
Proof.
Assume f and g are as stated in the hypothesis, then
d
dx
f
g= lim
h→x
f (h)g(h) −
f (x)g(x)
h − x
= limh→x
1
h − x
f (h)g(x)− f (x)g(h)
g(x)g(h)
= limh→x
1
h − x
(f (h)− f (x))g(x)− f (x)(g(h)− g(x))
g(x)g(h)
= limh→x
1
g(x)g(h)
[(f (h)− f (x)
h − x
)g(x)− f (x)
(g(h)− g(x)
h − x
)]=
g dfdx − f dg
dx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 6 / 1
Review of Calculus I
Proof.
Assume f and g are as stated in the hypothesis, then
d
dx
f
g= lim
h→x
f (h)g(h) −
f (x)g(x)
h − x
= limh→x
1
h − x
f (h)g(x)− f (x)g(h)
g(x)g(h)
= limh→x
1
h − x
(f (h)− f (x))g(x)− f (x)(g(h)− g(x))
g(x)g(h)
= limh→x
1
g(x)g(h)
[(f (h)− f (x)
h − x
)g(x)− f (x)
(g(h)− g(x)
h − x
)]
=g df
dx − f dgdx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 6 / 1
Review of Calculus I
Proof.
Assume f and g are as stated in the hypothesis, then
d
dx
f
g= lim
h→x
f (h)g(h) −
f (x)g(x)
h − x
= limh→x
1
h − x
f (h)g(x)− f (x)g(h)
g(x)g(h)
= limh→x
1
h − x
(f (h)− f (x))g(x)− f (x)(g(h)− g(x))
g(x)g(h)
= limh→x
1
g(x)g(h)
[(f (h)− f (x)
h − x
)g(x)− f (x)
(g(h)− g(x)
h − x
)]=
g dfdx − f dg
dx
g2
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 6 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 7 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0
u′v−v ′uv2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0u′v−v ′u
v2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0u′v−v ′u
v2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0u′v−v ′u
v2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0u′v−v ′u
v2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Analysis and Solution of the Problem
(uv
)′=
u′
v ′and u′v ′ = uv .
v 6= 0u′v−v ′u
v2 = u′
v ′
u = u′v ′
v
u′v−v ′uv2 =
u′v−v ′(
u′v′v
)v2 = u′v2−u′(v ′)2
v3 = u′
v ′
v ′
v −(v ′
v
)3= 1
X 3 − X + 1 = 0, where X = v ′
v
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 8 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 9 / 1
A Little History of Math
Nicolo Tartaglia (1500-1557) [5]
Girolamo Cardano (1501-1576) [4]
Solution for the depressed cubic t2 + pt + q = 0:
3
√−q
2+
√q2
4+
p3
27+
3
√−q
2−√
q2
4+
p3
27[2]
Solution for our cubic X 3 − X + 1 = 0:
3
√−1
2+
√1
4− 1
27+
3
√−1
2−√
1
4− 1
27≈ −1.324717957244 . . .
We’ll just call it c for constant.
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 10 / 1
A Little History of Math
Nicolo Tartaglia (1500-1557) [5]
Girolamo Cardano (1501-1576) [4]
Solution for the depressed cubic t2 + pt + q = 0:
3
√−q
2+
√q2
4+
p3
27+
3
√−q
2−√
q2
4+
p3
27[2]
Solution for our cubic X 3 − X + 1 = 0:
3
√−1
2+
√1
4− 1
27+
3
√−1
2−√
1
4− 1
27≈ −1.324717957244 . . .
We’ll just call it c for constant.
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 10 / 1
A Little History of Math
Nicolo Tartaglia (1500-1557) [5]
Girolamo Cardano (1501-1576) [4]
Solution for the depressed cubic t2 + pt + q = 0:
3
√−q
2+
√q2
4+
p3
27+
3
√−q
2−√
q2
4+
p3
27[2]
Solution for our cubic X 3 − X + 1 = 0:
3
√−1
2+
√1
4− 1
27+
3
√−1
2−√
1
4− 1
27≈ −1.324717957244 . . .
We’ll just call it c for constant.
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 10 / 1
A Little History of Math
Nicolo Tartaglia (1500-1557) [5]
Girolamo Cardano (1501-1576) [4]
Solution for the depressed cubic t2 + pt + q = 0:
3
√−q
2+
√q2
4+
p3
27+
3
√−q
2−√
q2
4+
p3
27[2]
Solution for our cubic X 3 − X + 1 = 0:
3
√−1
2+
√1
4− 1
27+
3
√−1
2−√
1
4− 1
27≈ −1.324717957244 . . .
We’ll just call it c for constant.
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 10 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 11 / 1
Some Differential Equations
v ′
v = c or dvdx = cv
v = Aecx so that dvdx = cAecx = cv
v ′
v= c =⇒
∫v ′
vdv =
∫c dx
=⇒ ln(v) = cx + k
=⇒ v = Aecx (A = ek)
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 12 / 1
Some Differential Equations
v ′
v = c or dvdx = cv
v = Aecx so that dvdx = cAecx = cv
v ′
v= c =⇒
∫v ′
vdv =
∫c dx
=⇒ ln(v) = cx + k
=⇒ v = Aecx (A = ek)
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 12 / 1
Some Differential Equations
v ′
v = c or dvdx = cv
v = Aecx so that dvdx = cAecx = cv
v ′
v= c =⇒
∫v ′
vdv =
∫c dx
=⇒ ln(v) = cx + k
=⇒ v = Aecx (A = ek)
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 12 / 1
Outline
Introduction to the Problem
Review of Calculus I
Analysis and Solution of the Problem
A Little History of Math
Some Differential Equations
The Final Solution
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 13 / 1
The Final Solution
u′v ′ = uv
u′
u = vv ′
u′
u = 1c
u = Bex/c
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 14 / 1
The Final Solution
u′v ′ = uvu′
u = vv ′
u′
u = 1c
u = Bex/c
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 14 / 1
The Final Solution
u′v ′ = uvu′
u = vv ′
u′
u = 1c
u = Bex/c
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 14 / 1
The Final Solution
u′v ′ = uvu′
u = vv ′
u′
u = 1c
u = Bex/c
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 14 / 1
The Final Solution
(uv
)′=
v 1c u − u cv
v2
=1c u(1− c2)
v
=1c u(c − c3)
cv
=u′
v ′
u′v ′ =1
cu cv = uv
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 15 / 1
The Final Solution
(uv
)′=
v 1c u − u cv
v2
=1c u(1− c2)
v
=1c u(c − c3)
cv
=u′
v ′
u′v ′ =1
cu cv = uv
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 15 / 1
The Final Solution
(uv
)′=
v 1c u − u cv
v2
=1c u(1− c2)
v
=1c u(c − c3)
cv
=u′
v ′
u′v ′ =1
cu cv = uv
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 15 / 1
The Final Solution
(uv
)′=
v 1c u − u cv
v2
=1c u(1− c2)
v
=1c u(c − c3)
cv
=u′
v ′
u′v ′ =1
cu cv = uv
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 15 / 1
References:Joseph DiMuro, Solution to Freshmen Quotient Rule (Problem 1941), Mathematics
Magazine vol. 38 (2015), no. 3, Mathematical Association of America, Washington D.C.,pp. 237-238.
William Dunham, Journey Through Genius, Penguin Books, New York, NY, 1991, p. 145.
Ron Larson and Bruce Edwards, Calculus: Early Transcendental Functions, 6e, CengageLearning, Boston, MA 2015, p. 141.
J. J. O’Connor and E.F. Robertson, Girolamo Cardano, MacTutor History of Mathematics,June 1998, http://www-history.mcs.st-andrews.ac.uk/Biographies/Cardan.html[Accessed 8/27/2015]
, Nicolo Tartaglia, MacTutor History of Mathematics, September 2005,http://www-history.mcs.st-andrews.ac.uk/Biographies/Tartaglia.html [Accessed8/27/2015]
C.F. Rocca Jr. (WCSU) Freshmen Quotient Rule and More: Problem 1941 16 / 1
