We denote the derivative of a function
f
{\displaystyle f}
at a number
a
{\displaystyle a}
as
f
′
(
a
)
{\displaystyle f'(a)\,\!}
.
The derivative of a function
f
{\displaystyle f}
at a number
a
{\displaystyle a}
a is given by the following limit (if it exists):
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}}
An analagous equation can be defined by letting
x
=
(
a
+
h
)
{\displaystyle x=(a+h)}
. Then
h
=
(
x
−
a
)
{\displaystyle h=(x-a)}
, which shows that when
x
{\displaystyle x}
approaches
a
{\displaystyle a}
,
h
{\displaystyle h}
approaches
0
{\displaystyle 0}
:
f
′
(
a
)
=
lim
x
→
a
f
(
x
)
−
f
(
a
)
x
−
a
{\displaystyle f'(a)=\lim _{x\rightarrow a}{\frac {f(x)-f(a)}{x-a}}}
Given a function
y
=
f
(
x
)
{\displaystyle y=f(x)\,\!}
, the derivative
f
′
(
a
)
{\displaystyle f'(a)\,\!}
can be understood as the slope of the tangent line to
f
(
x
)
{\displaystyle f(x)}
at
x
=
a
{\displaystyle x=a}
:
Find the equation of the tangent line to
y
=
x
2
{\displaystyle y=x^{2}}
at
x
=
1
{\displaystyle x=1}
.
To find the slope of the tangent, we let
y
=
f
(
x
)
{\displaystyle y=f(x)}
and use our first definition:
f
′
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
⇒
lim
h
→
0
(
a
+
h
)
2
−
(
a
)
2
h
⇒
lim
h
→
0
a
2
+
2
a
h
+
h
2
−
a
2
h
⇒
lim
h
→
0
h
(
2
a
+
h
)
h
⇒
lim
h
→
0
(
2
a
+
h
)
{\displaystyle f'(a)=\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}(a+h)^{2}-(a)^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}a^{2}+2ah+h^{2}-a^{2}}{h}}\Rightarrow \lim _{h\rightarrow 0}{\frac {\color {Blue}h(2a+h)}{h}}\Rightarrow \lim _{h\rightarrow 0}{\color {Blue}(2a+h)}}
It can be seen that as
h
{\displaystyle h}
approaches
0
{\displaystyle 0}
, we are left with
f
′
(
a
)
=
2
a
{\displaystyle f'(a)={\color {Blue}2a}\,\!}
. If we plug in
1
{\displaystyle 1}
for
a
{\displaystyle a}
:
f
′
(
1
)
=
2
(
1
)
⇒
2
{\displaystyle f'({\color {Red}1})=2({\color {Red}1})\Rightarrow {\color {Red}2}}
So our preliminary equation for the tangent line is
y
=
2
x
+
b
{\displaystyle y={\color {Red}2}x+b}
. By plugging in our tangent point
(
1
,
1
)
{\displaystyle (1,1)}
to find
b
{\displaystyle b}
, we can arrive at our final equation:
1
=
2
(
1
)
+
b
⇒
−
1
=
b
{\displaystyle {\color {Red}1}=2({\color {Red}1})+b\Rightarrow -1=b}
So our final equation is
y
=
2
x
−
1
{\displaystyle y=2x-1\,\!}
.
The derivative of a function
f
(
x
)
{\displaystyle f(x)}
at a number
a
{\displaystyle a}
can be understood as the instantaneous rate of change of
f
(
x
)
{\displaystyle f(x)}
when
x
=
a
{\displaystyle x=a}
.
So far we have only examined the derivative of a function
f
{\displaystyle f}
at a certain number
a
{\displaystyle a}
. If we move from the constant
a
{\displaystyle a}
to the variable
x
{\displaystyle x}
, we can calculate the derivative of the function as a whole, and come up with an equation that represents the derivative of the function
f
{\displaystyle f}
at any arbitrary
x
{\displaystyle x}
value. For clarification, the derivative of
f
{\displaystyle f}
at
a
{\displaystyle a}
is a number , whereas the derivative of
f
{\displaystyle f}
is a function .
Likewise to the derivative of
f
{\displaystyle f}
at
a
{\displaystyle a}
, the derivative of the function
f
(
x
)
{\displaystyle f(x)}
is denoted
f
′
(
x
)
{\displaystyle f'(x)\,\!}
.
The derivative of the function
f
{\displaystyle f}
is defined by the following limit:
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}
Also,
f
′
(
x
)
=
lim
h
→
x
f
(
x
)
−
f
(
h
)
x
−
h
{\displaystyle f'(x)=\lim _{h\rightarrow x}{\frac {f(x)-f(h)}{x-h}}}
or
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
−
h
)
2
h
{\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x-h)}{2h}}}
Consider the sequences:
y
=
x
2
;
y
+
Δ
y
=
(
x
+
h
)
2
=
x
2
+
2
x
h
+
h
2
{\displaystyle y=x^{2};\ y+\Delta y=(x+h)^{2}=x^{2}+2xh+h^{2}}
y
=
x
3
;
y
+
Δ
y
=
(
x
+
h
)
3
=
x
3
+
3
x
2
h
+
3
x
h
2
+
h
3
{\displaystyle y=x^{3};\ y+\Delta y=(x+h)^{3}=x^{3}+3x^{2}h+3xh^{2}+h^{3}}
y
=
x
4
;
y
+
Δ
y
=
(
x
+
h
)
4
=
x
4
+
4
x
3
h
+
6
x
2
h
2
+
4
x
h
3
+
h
4
{\displaystyle y=x^{4};\ y+\Delta y=(x+h)^{4}=x^{4}+4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}}
y
=
x
5
;
y
+
Δ
y
=
(
x
+
h
)
5
=
x
5
+
5
x
4
h
+
10
x
3
h
2
+
10
x
2
h
3
+
5
x
h
4
+
h
5
{\displaystyle y=x^{5};\ y+\Delta y=(x+h)^{5}=x^{5}+5x^{4}h+10x^{3}h^{2}+10x^{2}h^{3}+5xh^{4}+h^{5}}
y
=
x
6
;
y
+
Δ
y
=
(
x
+
h
)
6
=
x
6
+
6
x
5
h
+
{\displaystyle y=x^{6};\ y+\Delta y=(x+h)^{6}=x^{6}+6x^{5}h\ +}
many terms containing
h
2
{\displaystyle h^{2}}
y
=
x
n
;
y
+
Δ
y
=
(
x
+
Δ
x
)
n
=
x
n
+
n
x
n
−
1
Δ
x
+
{\displaystyle y=x^{n};\ y+\Delta y=(x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x\ +}
many terms containing
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Δ
y
=
n
x
n
−
1
Δ
x
+
{\displaystyle \Delta y=nx^{n-1}\Delta x\ +}
many terms containing
(
Δ
x
)
2
{\displaystyle (\Delta x)^{2}}
Therefore:
Δ
y
Δ
x
=
n
x
n
−
1
+
{\displaystyle {\frac {\Delta y}{\Delta x}}=nx^{n-1}\ +}
many terms with each and every term containing
Δ
x
{\displaystyle \Delta x}
d
y
d
x
=
lim
Δ
x
→
0
Δ
y
Δ
x
{\displaystyle {\frac {dy}{dx}}=\lim _{\Delta x\rightarrow 0}{\frac {\Delta y}{\Delta x}}}
d
(
x
n
)
d
x
=
n
x
n
−
1
{\displaystyle {\frac {d(x^{n})}{dx}}=nx^{n-1}}
read as "derivative with respect to
x
{\displaystyle x}
of
x
{\displaystyle x}
to the power
n
{\displaystyle n}
."
Later it will be shown that this is valid for all real
n
{\displaystyle n}
, positive or negative, integer or fraction.
Graph of cubic function illustrating use of associated quadratic. When
x
==
7.5
,
{\displaystyle x==7.5,}
there is exactly 1 value of
x
{\displaystyle x}
that gives
f
(
x
)
=
f
(
7.5
)
.
{\displaystyle f(x)=f(7.5).}
When
x
==
3
,
{\displaystyle x==3,}
there are 3 values of
x
{\displaystyle x}
that give
f
(
x
)
=
f
(
3
)
.
{\displaystyle f(x)=f(3).}
When
x
==
p
,
{\displaystyle x==p,}
there are exactly 2 values of
x
{\displaystyle x}
that give
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
Point
(
p
,
f
(
p
)
)
{\displaystyle (p,f(p))}
is a stationary point.
In the diagram there is a stationary point at
x
=
p
.
{\displaystyle x=p.}
When
x
=
p
,
{\displaystyle x=p,}
there are exactly 2 values of
x
{\displaystyle x}
that produce
f
(
x
)
=
f
(
p
)
.
{\displaystyle f(x)=f(p).}
Aim of this section is to derive the condition that produces exactly 2 values of
x
.
{\displaystyle x.}
See Cubic function as
product
of linear function and quadratic.
The associated quadratic when
x
=
p
:
{\displaystyle x=p:}
A
=
a
{\displaystyle A=a}
B
=
A
p
+
b
=
a
p
+
b
{\displaystyle B=Ap+b=ap+b}
C
=
B
p
+
c
=
a
p
p
+
b
p
+
c
.
{\displaystyle C=Bp+c=app+bp+c.}
Divide
g
(
x
)
=
a
x
2
+
(
a
p
+
b
)
x
+
(
a
p
2
+
b
p
+
c
)
{\displaystyle g(x)=ax^{2}+(ap+b)x+(ap^{2}+bp+c)}
by
(
x
−
p
)
.
{\displaystyle (x-p).}
This division gives a quotient of
a
x
+
(
2
a
p
+
b
)
{\displaystyle ax+(2ap+b)}
and remainder of
h
(
p
)
=
3
a
p
2
+
2
b
p
+
c
.
{\displaystyle h(p)=3ap^{2}+2bp+c.}
Factor
(
x
−
p
)
{\displaystyle (x-p)}
divides
g
(
x
)
{\displaystyle g(x)}
exactly. Therefore, remainder
h
(
p
)
=
0
,
{\displaystyle h(p)=0,}
the desired condition.
When
x
=
p
,
h
(
x
)
=
3
a
x
2
+
2
b
x
+
c
,
{\displaystyle x=p,\ h(x)=3ax^{2}+2bx+c,}
the derivative of
f
(
x
)
.
{\displaystyle f(x).}
Let
y
=
u
{\displaystyle y=u}
⋅
v
{\displaystyle v}
where
u
=
U
(
x
)
;
v
=
V
(
x
)
{\displaystyle u=U(x);\ v=V(x)}
y
+
Δ
y
=
(
u
+
Δ
u
)
{\displaystyle y+\Delta y=(u+\Delta u)}
⋅
(
v
+
Δ
v
)
=
u
{\displaystyle (v+\Delta v)=u}
⋅
v
+
u
{\displaystyle v+u}
⋅
Δ
v
+
v
{\displaystyle \Delta v+v}
⋅
Δ
u
+
Δ
u
{\displaystyle \Delta u+\Delta u}
⋅
Δ
v
{\displaystyle \Delta v}
Δ
y
=
u
{\displaystyle \Delta y=u}
⋅
Δ
v
+
v
{\displaystyle \Delta v+v}
⋅
Δ
u
+
Δ
u
{\displaystyle \Delta u+\Delta u}
⋅
Δ
v
{\displaystyle \Delta v}
Δ
y
Δ
x
=
u
{\displaystyle {\frac {\Delta y}{\Delta x}}=u}
⋅
Δ
v
Δ
x
+
v
{\displaystyle {\frac {\Delta v}{\Delta x}}+v}
⋅
Δ
u
Δ
x
+
(
Δ
u
)
(
Δ
v
)
Δ
x
{\displaystyle {\frac {\Delta u}{\Delta x}}+{\frac {(\Delta u)(\Delta v)}{\Delta x}}}
d
y
d
x
=
u
{\displaystyle {\frac {dy}{dx}}=u}
⋅
d
v
d
x
+
v
{\displaystyle {\frac {dv}{dx}}+v}
⋅
d
u
d
x
{\displaystyle {\frac {du}{dx}}}
Let
y
=
x
−
n
=
1
x
n
{\displaystyle y=x^{-n}={\frac {1}{x^{n}}}}
. Calculate
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
.
y
x
n
=
1
{\displaystyle yx^{n}=1}
Differentiate both sides.
y
n
x
n
−
1
+
x
n
d
y
d
x
=
0
{\displaystyle ynx^{n-1}+x^{n}{\frac {dy}{dx}}=0}
d
y
d
x
=
−
y
n
x
n
−
1
x
n
=
−
n
x
n
−
1
x
n
x
n
=
−
n
x
n
−
1
x
−
2
n
=
−
n
x
−
n
−
1
{\displaystyle {\frac {dy}{dx}}=-{\frac {ynx^{n-1}}{x^{n}}}=-{\frac {nx^{n-1}}{x^{n}x^{n}}}=-nx^{n-1}x^{-2n}=-nx^{-n-1}}
This shows that
d
(
x
n
)
d
x
{\displaystyle {\frac {d(x^{n})}{dx}}}
as above is valid for negative
n
{\displaystyle n}
.
Used where
y
=
U
(
x
)
V
(
x
)
{\displaystyle y={\frac {U(x)}{V(x)}}}
y
=
u
v
{\displaystyle y={\frac {u}{v}}}
y
v
=
u
{\displaystyle yv=u}
y
v
′
+
v
y
′
=
u
′
{\displaystyle yv'+vy'=u'}
v
y
′
=
u
′
−
y
v
′
=
u
′
−
u
v
′
v
{\displaystyle vy'=u'-yv'=u'-{\frac {uv'}{v}}}
y
′
=
u
′
v
−
u
v
′
v
2
=
v
u
′
v
2
−
u
v
′
v
2
=
v
u
′
−
u
v
′
v
2
{\displaystyle y'={\frac {u'}{v}}-{\frac {uv'}{v^{2}}}={\frac {vu'}{v^{2}}}-{\frac {uv'}{v^{2}}}={\frac {vu'-uv'}{v^{2}}}}
y
=
sin
(
x
)
{\displaystyle y=\sin(x)}
y
+
Δ
y
=
sin
(
x
+
Δ
x
)
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
{\displaystyle y+\Delta y=\sin(x+\Delta x)=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)}
Δ
y
=
sin
(
x
)
cos
(
Δ
x
)
+
cos
(
x
)
sin
(
Δ
x
)
−
sin
(
x
)
{\displaystyle \Delta y=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)-\sin(x)}
Δ
y
Δ
x
=
sin
(
x
)
{\displaystyle {\frac {\Delta y}{\Delta x}}=\sin(x)}
⋅
cos
(
Δ
x
)
−
1
Δ
x
+
cos
(
x
)
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)}
⋅
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
The value
cos
(
Δ
x
)
−
1
Δ
x
{\displaystyle {\frac {\cos(\Delta x)-1}{\Delta x}}}
:
>>> # python code
>>> [ ( math . cos ( Δx ) - 1 ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ - 0.049958347219742905 , - 0.004999958333473664 , - 0.0004999999583255033 ,
- 4.999999969612645e-05 , - 5.000000413701855e-06 , - 5.000444502911705e-07 ,
- 4.9960036108132044e-08 , 0.0 , 0.0 ]
>>>
lim
Δ
x
→
0
cos
(
Δ
x
)
−
1
Δ
x
=
lim
Δ
x
→
0
−
sin
(
Δ
x
)
1
=
0
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)-1}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {-\sin(\Delta x)}{1}}=0}
by L'Hôpital's rule.
The value
sin
(
Δ
x
)
Δ
x
{\displaystyle {\frac {\sin(\Delta x)}{\Delta x}}}
:
>>> # python code
>>> [ math . sin ( Δx ) / Δx for Δx in ( .1 , .01 , .001 , .0001 , .0000_1 , .0000_01 , .0000_001 , .0000_0001 , .0000_0000_1 ) ]
[ 0.9983341664682815 , 0.9999833334166665 , 0.9999998333333416 ,
0.9999999983333334 , 0.9999999999833332 , 0.9999999999998334 ,
0.9999999999999983 , 1.0 , 1.0 ]
>>>
lim
Δ
x
→
0
sin
(
Δ
x
)
Δ
x
=
lim
Δ
x
→
0
cos
(
Δ
x
)
1
=
1
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {\sin(\Delta x)}{\Delta x}}=\lim _{\Delta x\rightarrow 0}{\frac {\cos(\Delta x)}{1}}=1}
by L'Hôpital's rule.
d
y
d
x
=
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\sin(x)}
⋅
0
+
cos
(
x
)
{\displaystyle 0+\cos(x)}
⋅
1
=
cos
(
x
)
{\displaystyle 1=\cos(x)}
y
=
cos
(
x
)
=
1
−
sin
2
(
x
)
{\displaystyle y=\cos(x)={\sqrt {1-\sin ^{2}(x)}}}
y
2
=
1
−
sin
(
x
)
{\displaystyle y^{2}=1-\sin(x)}
⋅
sin
(
x
)
{\displaystyle \sin(x)}
Differentiate both sides:
2
y
{\displaystyle 2y}
⋅
d
y
d
x
=
−
(
2
sin
(
x
)
cos
(
x
)
)
{\displaystyle {\frac {dy}{dx}}=-(2\sin(x)\cos(x))}
d
y
d
x
=
−
sin
(
x
)
cos
(
x
)
cos
(
x
)
=
−
sin
(
x
)
.
{\displaystyle {\frac {dy}{dx}}=-{\frac {\sin(x)\cos(x)}{\cos(x)}}=-\sin(x).}
y
=
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle y=\tan(x)={\frac {\sin(x)}{\cos(x)}}}
y
cos
(
x
)
=
sin
(
x
)
{\displaystyle y\cos(x)=\sin(x)}
Differentiate both sides:
y
(
−
sin
(
x
)
)
+
cos
(
x
)
{\displaystyle y(-\sin(x))+\cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)}
cos
(
x
)
{\displaystyle \cos(x)}
⋅
d
y
d
x
=
cos
(
x
)
+
y
sin
(
x
)
{\displaystyle {\frac {dy}{dx}}=\cos(x)+y\sin(x)}
d
y
d
x
=
cos
(
x
)
+
tan
(
x
)
sin
(
x
)
cos
(
x
)
=
1
+
tan
2
(
x
)
=
sec
2
(
x
)
{\displaystyle {\frac {dy}{dx}}={\frac {\cos(x)+\tan(x)\sin(x)}{\cos(x)}}=1+\tan ^{2}(x)=\sec ^{2}(x)}
y
=
a
x
{\displaystyle y=a^{x}}
y
+
Δ
y
=
a
x
+
Δ
x
=
a
x
a
Δ
x
{\displaystyle y+\Delta y=a^{x+\Delta x}=a^{x}a^{\Delta x}}
Δ
y
=
a
x
a
Δ
x
−
a
x
=
a
x
(
a
Δ
x
−
1
)
{\displaystyle \Delta y=a^{x}a^{\Delta x}-a^{x}=a^{x}(a^{\Delta x}-1)}
Δ
y
Δ
x
=
a
x
⋅
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {\Delta y}{\Delta x}}=a^{x}\cdot {\frac {a^{\Delta x}-1}{\Delta x}}}
Consider the value
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
specifically
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
. L'Hôpital's rule cannot be used here because
d
(
a
x
)
d
x
{\displaystyle {\frac {d(a^{x})}{dx}}}
is what we are trying to find.
a
1
2
=
a
{\displaystyle a^{\frac {1}{2}}={\sqrt {a}}}
(
a
1
2
)
1
2
=
a
{\displaystyle (a^{\frac {1}{2}})^{\frac {1}{2}}={\sqrt {\sqrt {a}}}}
a
1
2
3
=
a
{\displaystyle a^{\frac {1}{2^{3}}}={\sqrt {\sqrt {\sqrt {a}}}}}
a
1
2
n
=
a
{\displaystyle a^{\frac {1}{2^{n}}}={\sqrt {a}}}
taken
n
{\displaystyle n}
times.
We will look at the expression
a
Δ
x
−
1
Δ
x
{\displaystyle {\frac {a^{\Delta x}-1}{\Delta x}}}
for different values of
a
{\displaystyle a}
with
Δ
x
=
1
2
70
=
8.5
e
−
22
{\displaystyle \Delta x={\frac {1}{2^{70}}}=8.5e-22}
(approx) in which case the expression becomes
(
a
1
/
2
70
−
1
)
∗
(
2
70
)
{\displaystyle (a^{1/2^{70}}-1)*(2^{70})}
where
a
1
/
2
70
{\displaystyle a^{1/2^{70}}}
is
a
70
{\displaystyle {\sqrt[{70}]{a}}}
or
a
{\displaystyle {\sqrt {a}}}
taken
70
{\displaystyle 70}
times. This approach is used here because function sqrt()
can be written so
that it does not depend on logarithmic or exponential operations.
>>> # python code
>>> N = Decimal ( 2 )
>>> v2 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v2
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
>>>
>>> N = Decimal ( 8 )
>>> v8 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v8
Decimal ( '2.07944154167983592825352768227031325913255072732801782513664' )
>>>
>>> N = Decimal ( 32 )
>>> v32 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v32
Decimal ( '3.46573590279972654709124760144583715956091114572543812435968' )
>>>
>>> N = Decimal ( 128 )
>>> v128 = ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 ) ; v128
Decimal ( '4.85203026391961716593059535875094644210510807293198187102208' )
>>>
Compare the values v8, v32, v128
with v2
:
>>> v8 / v2 ; v32 / v2 ; v128 / v2
Decimal ( '3.00000000000000000000176135549769209744528640235368520610520' )
Decimal ( '5.00000000000000000000587118499230699148996545343403093128046' )
Decimal ( '7.00000000000000000001232948848384468209997247971055570994695' )
>>>
We know that
8
=
2
3
;
32
=
2
5
;
128
=
2
7
{\displaystyle 8=2^{3};\ 32=2^{5};\ 128=2^{7}}
. The values v2, v8, v32, v128
are behaving like logarithms.
In fact
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}}
is the natural logarithm of
a
{\displaystyle a}
written as
ln
(
a
)
.
{\displaystyle \ln(a).}
Figure 5: Graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
for a =
2
,
8
,
32
,
128
{\displaystyle 2,\ 8,\ 32,\ 128}
.
Figure 5 contains graphs of
a
x
−
1
x
{\displaystyle {\frac {a^{x}-1}{x}}}
for
a
=
2
,
8
,
32
,
128
{\displaystyle a=2,\ 8,\ 32,\ 128}
with
graph of
e
x
−
1
x
{\displaystyle {\frac {e^{x}-1}{x}}}
included for reference.
All values of
x
{\displaystyle x}
are valid for all curves except where
x
=
0.
{\displaystyle x=0.}
The correct value of
ln
(
2
)
{\displaystyle \ln(2)}
is:
>>> Decimal ( 2 ) . ln ()
Decimal ( '0.693147180559945309417232121458176568075500134360255254120680' )
>>>
Our calculation produced:
Decimal ( '0.69314718055994530941743560122437474084363865015406919942144' )
accurate to 21 places of decimals, not bad for one line of simple python code using high-school math.
This method for calculation of
ln
(
a
)
{\displaystyle \ln(a)}
supposes that function sqrt()
is available.
Programming language python interprets expression a**b
as
a
b
.
{\displaystyle a^{b}.}
Therefore, in python,
ln
(
2
)
{\displaystyle \ln(2)}
can be calculated in accordance with the basic definition above:
# python code
>>> getcontext () . prec
101 # Precision of 101.
>>> dx = Decimal ( '1E-50' )
>>> ( 2 ** dx - 1 ) / dx
Decimal ( '0.69314718055994530941723212145817656807550013436026' ) # ln(2)
>>> ( 2 ** ( - dx ) - 1 ) / ( - dx )
Decimal ( '0.693147180559945309417232121458176568075500134360253' ) # ln(2)
>>>
When
y
=
a
x
,
d
y
d
x
=
a
x
⋅
ln
(
a
)
.
{\displaystyle y=a^{x},{\frac {dy}{dx}}=a^{x}\cdot \ln(a).}
The base of natural logarithms is the value of
a
{\displaystyle a}
that gives
lim
Δ
x
→
0
a
Δ
x
−
1
Δ
x
=
1.
{\displaystyle \lim _{\Delta x\rightarrow 0}{\frac {a^{\Delta x}-1}{\Delta x}}=1.}
This value of
a
{\displaystyle a}
, usually called
e
,
=
2.718281828459045235360287471352662497757247093699959574.....
{\displaystyle e,=2.718281828459045235360287471352662497757247093699959574.....}
>>> # python code
>>> N = e = Decimal ( 1 ) . exp (); N
Decimal ( '2.71828182845904523536028747135266249775724709369995957496697' )
>>> ( [ n for n in ( N ,) for p in range ( 0 , 70 ) for n in ( n . sqrt (),) ][ - 1 ] - 1 ) * ( 2 ** 70 )
Decimal ( '1.0000_0000_0000_0000_0000_0423_51647362715016770651530003719651328' )
>>> # ln(e) = 1. Our calculation of ln(e) is accurate to 21 places of decimals.
When
a
=
e
,
d
(
e
x
)
d
x
=
e
x
{\displaystyle a=e,{\frac {d(e^{x})}{dx}}=e^{x}}
⋅
ln
(
e
)
=
e
x
{\displaystyle \ln(e)=e^{x}}
⋅
1
=
e
x
.
{\displaystyle 1=e^{x}.}
y
=
l
n
(
x
)
{\displaystyle y=ln(x)}
x
=
e
y
{\displaystyle x=e^{y}}
d
x
d
y
=
e
y
{\displaystyle {\frac {dx}{dy}}=e^{y}}
d
y
d
x
=
1
e
y
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {1}{e^{y}}}={\frac {1}{x}}}
y
=
ln
(
a
x
)
{\displaystyle y=\ln(ax)}
d
y
d
x
=
d
d
x
ln
(
a
x
)
=
d
d
x
(
ln
(
a
)
+
ln
(
x
)
)
=
d
d
x
ln
(
a
)
+
d
d
x
ln
(
x
)
=
1
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(ax)={\frac {d}{dx}}(\ln(a)+\ln(x))={\frac {d}{dx}}\ln(a)+{\frac {d}{dx}}\ln(x)={\frac {1}{x}}}
y
=
ln
(
x
a
)
{\displaystyle y=\ln(x^{a})}
d
y
d
x
=
d
d
x
ln
(
x
a
)
=
d
d
x
(
a
⋅
ln
(
x
)
)
=
a
⋅
d
d
x
ln
(
x
)
=
a
x
{\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}\ln(x^{a})={\frac {d}{dx}}(a\cdot \ln(x))=a\cdot {\frac {d}{dx}}\ln(x)={\frac {a}{x}}}
y
=
x
m
n
{\displaystyle y=x^{\frac {m}{n}}}
Calculate
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
y
n
=
x
m
{\displaystyle y^{n}=x^{m}}
n
⋅
ln
(
y
)
=
m
⋅
ln
(
x
)
{\displaystyle n\cdot \ln(y)=m\cdot \ln(x)}
n
⋅
1
y
⋅
d
y
d
x
=
m
⋅
1
x
{\displaystyle n\cdot {\frac {1}{y}}\cdot {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}}
d
y
d
x
=
m
⋅
1
x
⋅
y
n
{\displaystyle {\frac {dy}{dx}}=m\cdot {\frac {1}{x}}\cdot {\frac {y}{n}}}
=
m
n
⋅
x
m
n
x
{\displaystyle ={\frac {m}{n}}\cdot {\frac {x^{\frac {m}{n}}}{x}}}
=
m
n
⋅
x
m
n
−
1
{\displaystyle ={\frac {m}{n}}\cdot x^{{\frac {m}{n}}-1}}
Careful manipulation of logarithms converts exponents into simple constants.
Used where
y
=
G
(
H
(
I
(
x
)
)
)
{\displaystyle y=G(H(I(x)))}
y
=
cos
(
2
x
)
{\displaystyle y=\cos(2x)}
Let
y
=
cos
(
u
)
{\displaystyle y=\cos(u)}
where
u
=
2
x
{\displaystyle u=2x}
.
d
y
d
x
=
d
y
d
u
⋅
d
u
d
x
=
−
sin
(
u
)
⋅
2
=
−
2
sin
(
2
x
)
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dx}}\\=&-\sin(u)\cdot 2\\=&-2\sin(2x)\\\end{aligned}}}
y
=
sin
2
(
ln
(
5
x
)
)
{\displaystyle y=\sin ^{2}(\ln(5x))}
Let
y
=
u
2
{\displaystyle y=u^{2}}
where
u
=
sin
(
v
)
;
v
=
ln
(
w
)
;
w
=
5
x
{\displaystyle u=\sin(v);\ v=\ln(w);\ w=5x}
.
d
y
d
x
=
d
y
d
u
⋅
d
u
d
v
⋅
d
v
d
w
⋅
d
w
d
x
=
2
u
⋅
cos
(
v
)
⋅
1
w
⋅
5
=
2
sin
(
v
)
⋅
cos
(
ln
(
w
)
)
⋅
1
5
x
⋅
5
=
2
sin
(
ln
(
w
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
=
2
sin
(
ln
(
5
x
)
)
⋅
cos
(
ln
(
5
x
)
)
⋅
1
x
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}=&{\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dw}}\cdot {\frac {dw}{dx}}\\=&2u\cdot \cos(v)\cdot {\frac {1}{w}}\cdot 5\\=&2\sin(v)\cdot \cos(\ln(w))\cdot {\frac {1}{5x}}\cdot 5\\=&2\sin(\ln(w))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\=&2\sin(\ln(5x))\cdot \cos(\ln(5x))\cdot {\frac {1}{x}}\\\end{aligned}}}
The first derivative of
f
(
x
)
=
y
′
{\displaystyle f(x)=y'}
or
f
′
(
x
)
.
{\displaystyle f'(x).}
As shown above,
f
′
(
x
)
{\displaystyle f'(x)}
at any point
x
1
{\displaystyle x_{1}}
gives the slope of
f
(
x
)
{\displaystyle f(x)}
at point
x
1
.
{\displaystyle x_{1}.}
f
′
(
x
1
)
{\displaystyle f'(x_{1})}
is the slope of
f
(
x
)
{\displaystyle f(x)}
when
x
=
x
1
.
{\displaystyle x=x_{1}.}
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle f(x)=x^{2}-x-2}
and
f
′
(
x
)
=
2
x
−
1.
{\displaystyle f'(x)=2x-1.}
When
x
=
0.5
,
{\displaystyle x=0.5,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
x
=
0
,
{\displaystyle x=0,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
−
1.
{\displaystyle f(x)=-1.}
When
x
=
1
,
{\displaystyle x=1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
1.
{\displaystyle f(x)=1.}
Point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is absolute minimum.
In the example to the right,
y
=
f
(
x
)
=
x
2
−
x
−
2
{\displaystyle y=f(x)=x^{2}-x-2}
and
y
′
=
f
′
(
x
)
=
2
x
−
1.
{\displaystyle y'=f'(x)=2x-1.}
Of special interest is the point at which
f
′
(
x
)
{\displaystyle f'(x)}
or slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
0.5
{\displaystyle f'(x)=0,\ x=0.5}
and
f
(
x
)
=
0.5
2
−
0.5
−
2
=
−
2.25.
{\displaystyle f(x)=0.5^{2}-0.5-2=-2.25.}
The point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is called a critical point or stationary point of
f
(
x
)
.
{\displaystyle f(x).}
Because
y
′
{\displaystyle y'}
has exactly one solution for
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
has exactly one critical point.
The value of
y
{\displaystyle y}
at point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is less than
y
{\displaystyle y}
at both
(
0
,
−
2
)
,
(
1
,
−
2
)
.
{\displaystyle (0,-2),\ (1,-2).}
Therefore the critical point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is a minimum of
f
(
x
)
.
{\displaystyle f(x).}
In this curve
y
=
x
2
−
x
−
2
,
{\displaystyle y=x^{2}-x-2,}
the point
(
0.5
,
−
2.25
)
{\displaystyle (0.5,-2.25)}
is both local minimum and absolute minimum.
Figure 1: Diagram illustrating relationship between
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
and
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle f'(x)={\frac {6x^{2}+6x-12}{8}}.}
When
x
=
−
2
{\displaystyle x=-2}
or
x
=
1
,
{\displaystyle x=1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
x
=
−
3
,
{\displaystyle x=-3,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
When
x
=
−
1
,
{\displaystyle x=-1,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
−
1.5.
{\displaystyle f(x)=-1.5.}
When
x
=
2
,
{\displaystyle x=2,}
both
f
′
(
x
)
{\displaystyle f'(x)}
and slope of
f
(
x
)
=
3.
{\displaystyle f(x)=3.}
Point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is local maximum. Point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is local minimum.
In the example to the right,
y
=
f
(
x
)
=
2
x
3
+
3
x
2
−
12
x
−
8
8
{\displaystyle y=f(x)={\frac {2x^{3}+3x^{2}-12x-8}{8}}}
and
y
′
=
f
′
(
x
)
=
6
x
2
+
6
x
−
12
8
.
{\displaystyle y'=f'(x)={\frac {6x^{2}+6x-12}{8}}.}
Of special interest are the points at which
f
′
(
x
)
{\displaystyle f'(x)}
or slope of
f
(
x
)
=
0.
{\displaystyle f(x)=0.}
When
f
′
(
x
)
=
0
,
x
=
−
2
{\displaystyle f'(x)=0,\ x=-2}
and
f
(
x
)
=
1.5
{\displaystyle f(x)=1.5}
or
When
f
′
(
x
)
=
0
,
x
=
1
{\displaystyle f'(x)=0,\ x=1}
and
f
(
x
)
=
−
1
7
8
.
{\displaystyle f(x)=-1{\frac {7}{8}}.}
The points
(
−
2
,
1.5
)
,
(
1
,
−
1
7
8
)
{\displaystyle (-2,1.5),\ (1,-1{\frac {7}{8}})}
are critical or stationary points of
f
(
x
)
.
{\displaystyle f(x).}
Because
y
′
{\displaystyle y'}
has exactly two real solutions for
y
′
=
0
,
f
(
x
)
{\displaystyle y'=0,\ f(x)}
has exactly two critical points.
Slope of
f
(
x
)
{\displaystyle f(x)}
to the left of
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is positive and
adjacent slope of
f
(
x
)
{\displaystyle f(x)}
to the right of
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is negative.
Therefore point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is local maximum. Point
(
−
2
,
1.5
)
{\displaystyle (-2,1.5)}
is not absolute maximum.
Adjacent slope of
f
(
x
)
{\displaystyle f(x)}
to the left of
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is negative and
slope of
f
(
x
)
{\displaystyle f(x)}
to the right of
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is positive.
Therefore point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is local minimum. Point
(
1
,
−
1
7
8
)
{\displaystyle (1,-1{\frac {7}{8}})}
is not absolute minimum.
Figure 1(c): Plan of county road between Town A and Town B to be constructed so that cost is minimum.
Town B is 40 miles East and 50 miles North of Town A. The county is going to construct a road from Town A to Town B.
Adjacent to Town A the cost to build a road is $500k/mile.
Adjacent to Town B the cost to build a road is $200k/mile.
The dividing line runs East-West 30 miles North of Town A.
Calculate the position of point C so that the cost of the road from Town A to Town B is minimum.
Let point
C
=
(
x
,
30
)
.
{\displaystyle C=(x,30).}
Then distance from Town A to point
C
=
x
2
+
30
2
=
x
2
+
900
.
{\displaystyle C={\sqrt {x^{2}+30^{2}}}={\sqrt {x^{2}+900}}.}
Distance from Town B to point
C
=
(
40
−
x
)
2
+
20
2
{\displaystyle C={\sqrt {(40-x)^{2}+20^{2}}}}
=
1600
−
80
x
+
x
2
+
400
{\displaystyle ={\sqrt {1600-80x+x^{2}+400}}}
=
2000
−
80
x
+
x
2
.
{\displaystyle ={\sqrt {2000-80x+x^{2}}}.}
x
=
10.52322823517
…
{\displaystyle x=10.52322823517\dots }
Figure 1(e): Sheet of cardboard to be cut and folded to make box of maximum possible volume. Cut on purple lines, fold on red lines. Design of box includes top.
A piece of cardboard of length
4
f
t
{\displaystyle 4\ ft}
and width
3
f
t
{\displaystyle 3\ ft}
will be used to make a box with a top. Some waste will be cut out of the piece of cardboard
and the remaining cardboard will be folded to make a box so that the volume of the box is maximum.
What is the height of the box?
Figure 1: Graph of ellipse showing semi-chord through center. When length of
t
{\displaystyle t}
is maximum, length of major axis
=
2
t
.
{\displaystyle =2t.}
When length of
t
{\displaystyle t}
is minimum, length of minor axis
=
2
t
.
{\displaystyle =2t.}
An ellipse with center at origin has equation:
A
x
2
+
B
x
y
+
C
y
2
+
F
=
0
…
(
1
)
{\displaystyle Ax^{2}+Bxy+Cy^{2}+F=0\ \dots \ (1)}
Given values
A
,
B
,
C
,
F
{\displaystyle A,B,C,F}
calculate:
In Figure 1
t
{\displaystyle t}
is any line from origin to ellipse and
θ
{\displaystyle \theta }
is angle between
X
{\displaystyle X}
axis and
t
.
{\displaystyle t.}
Aim of this section is to calculate
θ
{\displaystyle \theta }
so that length of
t
{\displaystyle t}
is maximum,
in which case length of major axis =
2
t
.
{\displaystyle 2t.}
Let
c
=
cos
(
θ
)
{\displaystyle c=\cos(\theta )}
and
s
=
sin
(
θ
)
.
{\displaystyle s=\sin(\theta ).}
Then
x
=
t
cos
(
θ
)
=
t
c
,
{\displaystyle x=t\cos(\theta )=tc,}
and
y
=
t
sin
(
θ
)
=
t
s
.
{\displaystyle y=t\sin(\theta )=ts.}
Substitute these values in
(
1
)
:
{\displaystyle (1):}
A
t
t
c
c
+
B
t
t
c
s
+
C
t
t
s
s
+
F
=
0
…
(
2
)
{\displaystyle Attcc+Bttcs+Cttss+F=0\ \dots \ (2)}
Calculate
t
′
=
d
t
d
θ
{\displaystyle t'={\frac {dt}{d\theta }}}
A
(
t
t
(
−
2
c
s
)
+
c
c
2
t
t
′
)
+
B
(
t
t
(
c
c
−
s
s
)
+
c
s
2
t
t
′
)
+
C
(
t
t
(
2
s
c
)
+
s
s
2
t
t
′
)
=
0
{\displaystyle A(tt(-2cs)+cc2tt')+B(tt(cc-ss)+cs2tt')+C(tt(2sc)+ss2tt')=0}
A
t
t
(
−
2
c
s
)
+
A
c
c
2
t
t
′
+
B
t
t
(
c
c
−
s
s
)
+
B
c
s
2
t
t
′
+
C
t
t
(
2
s
c
)
+
C
s
s
2
t
t
′
=
0
{\displaystyle Att(-2cs)+Acc2tt'+Btt(cc-ss)+Bcs2tt'+Ctt(2sc)+Css2tt'=0}
−
2
A
t
t
c
s
+
A
c
c
2
t
t
′
+
B
t
t
c
c
−
B
t
t
s
s
+
B
c
s
2
t
t
′
+
2
C
t
t
s
c
+
2
C
s
s
t
t
′
=
0
{\displaystyle -2Attcs+Acc2tt'+Bttcc-Bttss+Bcs2tt'+2Cttsc+2Csstt'=0}
+
A
c
c
2
t
t
′
+
2
B
c
s
t
t
′
+
2
C
s
s
t
t
′
−
2
A
t
t
c
s
+
B
t
t
c
c
−
B
t
t
s
s
+
2
C
t
t
s
c
=
0
{\displaystyle +Acc2tt'+2Bcstt'+2Csstt'-2Attcs+Bttcc-Bttss+2Cttsc=0}
+
t
′
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
{\displaystyle +t'(Acc2t+2Bcst+2Csst)=+2Attcs-Bttcc+Bttss-2Cttsc}
t
′
=
+
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
(
A
c
c
2
t
+
2
B
c
s
t
+
2
C
s
s
t
)
{\displaystyle t'={\frac {+2Attcs-Bttcc+Bttss-2Cttsc}{(Acc2t+2Bcst+2Csst)}}}
For maximum or minimum
t
:
{\displaystyle t:}
2
A
t
t
c
s
−
B
t
t
c
c
+
B
t
t
s
s
−
2
C
t
t
s
c
=
0
{\displaystyle 2Attcs-Bttcc+Bttss-2Cttsc=0}
2
A
c
s
−
B
c
c
+
B
s
s
−
2
C
s
c
=
0
{\displaystyle 2Acs-Bcc+Bss-2Csc=0}
2
A
c
s
−
2
C
s
c
=
B
c
c
−
B
s
s
{\displaystyle 2Acs-2Csc=Bcc-Bss}
Square both sides, substitute
1
−
s
s
{\displaystyle 1-ss}
for
c
c
{\displaystyle cc}
and result is:
a
S
2
+
b
S
+
c
=
0
…
(
3
)
{\displaystyle aS^{2}+bS+c=0\ \dots \ (3)}
where:
S
=
sin
2
(
θ
)
{\displaystyle S=\sin ^{2}(\theta )}
a
=
(
+
4
A
A
−
8
A
C
+
4
B
B
+
4
C
C
)
{\displaystyle a=(+4AA-8AC+4BB+4CC)}
b
=
−
a
{\displaystyle b=-a}
c
=
B
B
{\displaystyle c=BB}
Let equation of ellipse be:
55
x
2
−
24
x
y
+
48
y
2
−
2496
=
0
{\displaystyle 55x^{2}-24xy+48y^{2}-2496=0}
# python code
>>> A , B , C = 55 , - 24 , 48
>>> a = ( + 4 * A * A - 8 * A * C + 4 * B * B + 4 * C * C ); a
2500
>>> b = - a ; b
- 2500
>>> c = B * B ; c
576
>>>
>>> a , b , c = [ v / 4 for v in ( a , b , c )] ; a , b , c
( 625.0 , - 625.0 , 144.0 )
>>> S = .36
>>> a * S * S + b * S + c
0.0
>>> S = .64
>>> a * S * S + b * S + c
0.0
>>>
The solutions of quadratic equation
(
3
)
{\displaystyle (3)}
are
.36
{\displaystyle .36}
or
.64
{\displaystyle .64}
.
Therefore
sin
(
θ
)
=
±
0.6
{\displaystyle \sin(\theta )=\pm 0.6}
or
sin
(
θ
)
=
±
0.8
{\displaystyle \sin(\theta )=\pm 0.8}
.
From
(
2
)
{\displaystyle (2)}
above:
t
=
−
F
A
c
c
+
B
c
s
+
C
s
s
{\displaystyle t={\sqrt {\frac {-F}{Acc+Bcs+Css}}}}
# python code
A , B , C , F = 55 , - 24 , 48 , - 2496
t1 = ( 0.6 , 0.8 )
dict1 = dict ()
for v1 in ( t1 , t1 [:: - 1 ]) :
c1 , s1 = v1
for c in ( c1 , - c1 ) :
for s in ( s1 , - s1 ) :
t = ( - F / ( A * c * c + B * c * s + C * s * s )) ** .5
if t in dict1 : dict1 [ t ] += (( c , s ),)
else : dict1 [ t ] = (( c , s ),)
L1 = [ ( v , dict1 [ v ]) for v in sorted ([ v for v in dict1 ]) ]
for v in L1 : print ( v )
(6.244997998398398, ((0.8, -0.6), (-0.8, 0.6)))
(6.34287855135306, ((0.6, -0.8), (-0.6, 0.8)))
(7.806247497997998, ((0.8, 0.6), (-0.8, -0.6)))
(7.999999999999999, ((0.6, 0.8), (-0.6, -0.8)))
Minimum value of
t
=
6.244997998398398.
{\displaystyle t=6.244997998398398.}
Length of minor axis
=
2
∗
6.244997998398398
{\displaystyle =2*6.244997998398398}
Maximum value of
t
=
7.999999999999999.
{\displaystyle t=7.999999999999999.}
Length of major axis
=
8
∗
2
{\displaystyle =8*2}
Figure 3: Curves and values associated with car jack. When
x
=
1
,
d
y
d
x
=
0.1
…
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1\dots }
When
θ
=
45
∘
,
d
y
d
x
=
1
{\displaystyle \theta =45^{\circ },\ {\frac {dy}{dx}}=1}
When
x
=
9
,
d
y
d
x
=
2.06
…
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06\dots }
At what rate is point
A
{\displaystyle A}
moving upwards:
(a) when
x
=
9
{\displaystyle x=9}
?
(b) when
x
=
1
{\displaystyle x=1}
?
(c) when
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
?
We have to calculate
d
y
d
t
{\displaystyle {\frac {dy}{dt}}}
when
d
x
d
t
{\displaystyle {\frac {dx}{dt}}}
is given.
d
y
d
t
=
d
y
d
x
⋅
d
x
d
t
{\displaystyle {\frac {dy}{dt}}={\frac {dy}{dx}}\cdot {\frac {dx}{dt}}}
x
2
+
y
2
=
10
2
{\displaystyle x^{2}+y^{2}=10^{2}}
(equation of circle)
y
=
100
−
x
2
{\displaystyle y={\sqrt {100-x^{2}}}}
y
2
=
100
−
x
2
{\displaystyle y^{2}=100-x^{2}}
2
y
⋅
d
y
d
x
=
−
2
x
{\displaystyle 2y\cdot {\frac {dy}{dx}}=-2x}
d
y
d
x
=
−
2
x
2
y
=
−
x
100
−
x
2
{\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}={\frac {-x}{\sqrt {100-x^{2}}}}}
For convenience we'll use the negative value of the square root and say that
d
y
d
x
=
x
100
−
x
2
.
{\displaystyle {\frac {dy}{dx}}={\frac {x}{\sqrt {100-x^{2}}}}.}
Relative to line
B
C
:
{\displaystyle BC:}
When
x
=
9
,
d
y
d
x
=
2.06
,
d
y
d
t
=
2.06
⋅
5
=
10.3
{\displaystyle x=9,\ {\frac {dy}{dx}}=2.06,\ {\frac {dy}{dt}}=2.06\cdot 5=10.3}
inches
/
{\displaystyle /}
minute.
When
x
=
1
,
d
y
d
x
=
0.1
,
d
y
d
t
=
0.1
⋅
5
=
0.5
{\displaystyle x=1,\ {\frac {dy}{dx}}=0.1,\ {\frac {dy}{dt}}=0.1\cdot 5=0.5}
inches
/
{\displaystyle /}
minute.
When
θ
=
45
∘
,
x
=
10
⋅
cos
45
∘
=
7.071
{\displaystyle \theta =45^{\circ },x=10\cdot \cos 45^{\circ }=7.071}
and
d
y
d
t
=
1
⋅
5
=
5
{\displaystyle {\frac {dy}{dt}}=1\cdot 5=5}
inches
/
{\displaystyle /}
minute.
This example highlights the mechanical advantage of this simple but effective tool. When the top of the jack is low, it moves quickly.
As the jack takes more and more weight, the top of the jack moves more slowly.
Figure 4: Graph of
a
{\displaystyle a}
and
d
a
d
x
{\displaystyle {\frac {da}{dx}}}
. Area of
Δ
A
B
C
{\displaystyle \Delta ABC}
is maximum when
b
=
h
=
50
=
10
cos
(
45
∘
)
{\displaystyle b=h={\sqrt {50}}=10\cos(45^{\circ })}
At what rate is the area of
Δ
A
B
C
{\displaystyle \Delta ABC}
changing when
(i)
x
=
9
{\displaystyle x=9}
?
(ii)
x
=
1
{\displaystyle x=1}
?
(iii)
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
?
d
a
d
t
=
d
a
d
x
⋅
d
x
d
t
{\displaystyle {\frac {da}{dt}}={\frac {da}{dx}}\cdot {\frac {dx}{dt}}}
where
a
{\displaystyle a}
is area of
Δ
A
B
C
{\displaystyle \Delta ABC}
and
d
x
d
t
=
5
{\displaystyle {\frac {dx}{dt}}=5}
inches
/
{\displaystyle /}
minute.
a
=
b
⋅
h
2
=
x
⋅
y
2
=
x
100
−
x
2
2
{\displaystyle a={\frac {b\cdot h}{2}}={\frac {x\cdot y}{2}}={\frac {x{\sqrt {100-x^{2}}}}{2}}}
Calculate
d
a
d
x
:
{\displaystyle {\frac {da}{dx}}:}
2
a
=
x
100
−
x
2
{\displaystyle 2a=x{\sqrt {100-x^{2}}}}
4
a
2
=
x
2
(
100
−
x
2
)
=
100
x
2
−
x
4
{\displaystyle 4a^{2}=x^{2}(100-x^{2})=100x^{2}-x^{4}}
8
a
⋅
d
a
d
x
=
200
x
−
4
x
3
{\displaystyle 8a\cdot {\frac {da}{dx}}=200x-4x^{3}}
d
a
d
x
=
200
x
−
4
x
3
8
a
=
50
x
−
x
3
2
a
=
50
x
−
x
3
x
100
−
x
2
{\displaystyle {\frac {da}{dx}}={\frac {200x-4x^{3}}{8a}}={\frac {50x-x^{3}}{2a}}={\frac {50x-x^{3}}{x{\sqrt {100-x^{2}}}}}}
=
50
−
x
2
100
−
x
2
{\displaystyle ={\frac {50-x^{2}}{\sqrt {100-x^{2}}}}}
When
x
=
9
,
d
a
d
x
=
7.11
…
{\displaystyle x=9,\ {\frac {da}{dx}}=7.11\dots }
and area of
Δ
A
B
C
{\displaystyle \Delta ABC}
is increasing at rate of
(
5
⋅
7.11
…
)
{\displaystyle (5\cdot 7.11\dots )}
square inches/minute.
When
x
=
1
,
d
a
d
x
=
−
4.92
…
{\displaystyle x=1,\ {\frac {da}{dx}}=-4.92\dots }
and area of
Δ
A
B
C
{\displaystyle \Delta ABC}
is decreasing at rate of
(
5
⋅
4.92
…
)
{\displaystyle (5\cdot 4.92\dots )}
square inches/minute.
When
θ
=
45
∘
,
h
=
b
=
x
=
y
=
10
cos
(
45
∘
)
=
10
⋅
2
2
=
5
2
,
{\displaystyle \theta =45^{\circ },\ h=b=x=y=10\cos(45^{\circ })=10\cdot {\frac {\sqrt {2}}{2}}=5{\sqrt {2}},\ }
d
a
d
x
=
0
,
a
{\displaystyle {\frac {da}{dx}}=0,\ a}
is a maximum of
25
{\displaystyle 25}
square inches and
d
a
d
t
=
0.
{\displaystyle {\frac {da}{dt}}=0.}
{\displaystyle }
Figure 5: Image of analog clock showing minute and hour hands at
3
{\displaystyle 3}
o'clock.
An old fashion analog clock with American style face (12 hours) keeps accurate time.
The length of the minute hand is
4
{\displaystyle 4}
inches and the length of the hour hand is
3
{\displaystyle 3}
inches.
At what rate is the tip of the minute hand approaching the tip of the hour hand at 3 o'clock?
Let
a
{\displaystyle a}
be distance between the two tips.
The task is to calculate
d
a
d
t
.
{\displaystyle {\frac {da}{dt}}.}
d
a
d
t
=
d
a
d
θ
⋅
d
θ
d
t
{\displaystyle {\frac {da}{dt}}={\frac {da}{d\theta }}\cdot {\frac {d\theta }{dt}}}
where
θ
{\displaystyle \theta }
is
angle subtended by side
a
{\displaystyle a}
at center of clock.
Calculating
d
θ
d
t
:
{\displaystyle {\frac {d\theta }{dt}}:}
Angular velocity of minute hand
=
2
π
1
{\displaystyle ={\frac {2\pi }{1}}}
radians/hour.
Angular velocity of hour hand
=
2
π
12
=
π
6
{\displaystyle ={\frac {2\pi }{12}}={\frac {\pi }{6}}}
radians/hour.
d
θ
d
t
=
{\displaystyle {\frac {d\theta }{dt}}=}
angular velocity of minute hand relative to hour hand
=
2
π
−
π
6
=
11
π
6
{\displaystyle =2\pi -{\frac {\pi }{6}}={\frac {11\pi }{6}}}
radians/hour.
Calculating
d
a
d
θ
:
{\displaystyle {\frac {da}{d\theta }}:}
a
2
=
3
2
+
4
2
−
2
⋅
3
⋅
4
⋅
cos
θ
=
25
−
24
cos
θ
.
{\displaystyle a^{2}=3^{2}+4^{2}-2\cdot 3\cdot 4\cdot \cos \theta =25-24\cos \theta .}
2
a
⋅
d
a
d
θ
=
24
sin
θ
.
{\displaystyle 2a\cdot {\frac {da}{d\theta }}=24\sin \theta .}
d
a
d
θ
=
24
sin
θ
2
a
=
24
sin
90
∘
2
⋅
5
=
24
10
=
2.4.
{\displaystyle {\frac {da}{d\theta }}={\frac {24\sin \theta }{2a}}={\frac {24\sin 90^{\circ }}{2\cdot 5}}={\frac {24}{10}}=2.4.}
d
a
d
t
=
2.4
⋅
11
π
6
=
4.4
π
{\displaystyle {\frac {da}{dt}}=2.4\cdot {\frac {11\pi }{6}}=4.4\pi }
inches/hour.
Figure 2: Diagram showing position of piston as a function of rotation of crankshaft in reciprocating engine. Piston moves up and down between
8
{\displaystyle 8}
and
18
{\displaystyle 18}
inches.
18
≥
y
≥
8
{\displaystyle 18\geq y\geq 8}
a
2
=
y
2
+
c
2
−
2
y
c
cos
(
x
)
{\displaystyle a^{2}=y^{2}+c^{2}-2yc\cos(x)}
y
2
−
2
c
cos
(
x
)
y
+
c
2
−
a
2
=
0
{\displaystyle y^{2}-2c\cos(x)y+c^{2}-a^{2}=0}
y
2
−
2
(
5
)
cos
(
x
)
y
+
(
5
)
2
−
(
13
)
2
=
0
{\displaystyle y^{2}-2(5)\cos(x)y+(5)^{2}-(13)^{2}=0}
y
2
−
10
cos
(
x
)
y
−
144
=
0
{\displaystyle y^{2}-10\cos(x)y-144=0}
y
=
10
cos
(
x
)
+
(
10
cos
(
x
)
)
2
−
4
(
−
144
)
2
{\displaystyle y={\frac {10\cos(x)+{\sqrt {(10\cos(x))^{2}-4(-144)}}}{2}}}
y
=
10
cos
(
x
)
+
100
(
cos
(
x
)
)
2
+
576
2
{\displaystyle y={\frac {10\cos(x)+{\sqrt {100(\cos(x))^{2}+576}}}{2}}}
y
=
5
cos
(
x
)
+
25
(
cos
(
x
)
)
2
+
144
{\displaystyle y=5\cos(x)+{\sqrt {25(\cos(x))^{2}+144}}}
Code supplied to grapher (without white space) is:
( 5 )( cos ( x )) + ( (( 25 )(( cos ( x )) ^ 2 ) + 144 ) ^ ( 0.5 ) )
When expressed in this way, it's easy to convert the code to python code:
( 5 ) * ( cos ( x )) + ( (( 25 ) * (( cos ( x )) ** 2 ) + 144 ) ** ( 0.5 ) )
Positions of interest:
Graph and diagram of piston at top dead center.
Graph and diagram of piston at bottom dead center.
Graph and diagram of piston half-way between TDC and BDC.
Graph and diagram of crank half-way between TDC and BDC.
Graph and diagram of connecting rod tangential to crank.
Figure 4a: Diagram showing acceleration of piston in reciprocating engine. Negative acceleration has a greater absolute value than the positive, but it does not last as long.
Acceleration introduces the second derivative. While velocity was the first derivative of position with respect to time,
acceleration is the first derivative of velocity or the second derivative of position.
From velocity above
y
y
′
−
5
cos
(
x
)
y
′
+
5
sin
(
x
)
y
=
0
{\displaystyle yy'-5\cos(x)y'+5\sin(x)y=0}
By implicit differentation:
y
y
″
+
y
′
y
′
−
5
(
cos
(
x
)
y
″
+
y
′
(
−
sin
(
x
)
)
)
+
5
(
sin
(
x
)
y
′
+
y
cos
(
x
)
)
=
0
{\displaystyle yy''+y'y'-5(\cos(x)y''+y'(-\sin(x)))+5(\sin(x)y'+y\cos(x))=0}
y
y
″
+
y
′
y
′
−
5
cos
(
x
)
y
″
+
5
y
′
sin
(
x
)
+
5
sin
(
x
)
y
′
+
5
y
cos
(
x
)
=
0
{\displaystyle yy''+y'y'-5\cos(x)y''+5y'\sin(x)+5\sin(x)y'+5y\cos(x)=0}
y
y
″
−
5
cos
(
x
)
y
″
=
−
(
y
′
y
′
+
5
y
′
sin
(
x
)
+
5
sin
(
x
)
y
′
+
5
y
cos
(
x
)
)
{\displaystyle yy''-5\cos(x)y''=-(y'y'+5y'\sin(x)+5\sin(x)y'+5y\cos(x))}
y
″
(
y
−
5
cos
(
x
)
)
=
−
(
y
′
y
′
+
10
sin
(
x
)
y
′
+
5
y
cos
(
x
)
)
{\displaystyle y''(y-5\cos(x))=-(y'y'+10\sin(x)y'+5y\cos(x))}
y
″
=
y
′
y
′
+
10
sin
(
x
)
y
′
+
5
y
cos
(
x
)
5
cos
(
x
)
−
y
{\displaystyle y''={\frac {y'y'+10\sin(x)y'+5y\cos(x)}{5\cos(x)-y}}}
Substitute for
y
{\displaystyle y}
and
y
′
{\displaystyle y'}
as defined above, and you see the code input to grapher
at top of diagram to right.
Figure 4b: Diagram showing "irregularities" in the curve of velocity while velocity is increasing.
y
{\displaystyle y}
axis compressed to illustrate shape of curves.
"Kinks" in the curve:
It is not obvious by looking at the curve of velocity that there are slight irregularities in the curve when velocity is increasing.
However, the irregularities are obvious in the curve of acceleration.
During one revolution of the crankshaft there is less time allocated for negative acceleration than for positive acceleration.
Therefore, the maximum absolute value of negative acceleration is greater than the maximum value of positive acceleration.
Figure 5: Diagram showing positions of maximum velocity of piston in reciprocating engine. In the first quadrant, from
x
=
0
∘
{\displaystyle x=0^{\circ }}
to
x
=
90
∘
,
{\displaystyle x=90^{\circ },}
the piston moves through
6
{\displaystyle 6}
inches. In the second quadrant, from
x
=
90
∘
{\displaystyle x=90^{\circ }}
to
x
=
180
∘
,
{\displaystyle x=180^{\circ },}
the piston moves through
4
{\displaystyle 4}
inches. Therefore, in the first quadrant, acceleration must be greater than in the second quadrant.
Velocity is rate of change of position. See also Figure 3 above.
Minimum velocity:
Velocity is zero when slope of curve of position is zero.
This occurs at top dead center and at bottom dead center, ie, when
x
=
0
{\displaystyle x=0}
and
x
=
π
.
{\displaystyle x=\pi .}
Maximum velocity:
Intuition suggests that the position of maximum velocity might be the point at which the connecting rod is tangent
to the circle of the crankshaft. In other words:
x
=
arctan
(
13
5
)
=
68.96
…
∘
{\displaystyle x=\arctan({\frac {13}{5}})=68.96\dots ^{\circ }}
or, that the position of maximum velocity might be the point at which the piston is half-way between top dead center and bottom dead center.
In other words:
x
=
arccos
(
2.5
13
)
=
78.9
…
∘
{\displaystyle x=\arccos({\frac {2.5}{13}})=78.9\dots ^{\circ }}
However, velocity is maximum when acceleration is
0
,
{\displaystyle 0,}
which occurs when
x
=
±
71.26
…
∘
.
{\displaystyle x=\pm 71.26\dots ^{\circ }.}
Suppose that the engine is rotating at
100
{\displaystyle 100}
radians/second or approx.
955
{\displaystyle 955}
RPM.
v
m
a
x
=
y
′
i
n
c
h
e
s
r
a
d
i
a
n
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
{\displaystyle v_{max}=y'{\frac {inches}{radian}}\cdot 100{\frac {radians}{second}}}
abs
(
v
m
a
x
)
=
5.36
(
100
)
{\displaystyle (v_{max})=5.36(100)}
inches/second or approx.
30.5
{\displaystyle 30.5}
mph.
Figure 6: Diagram showing positions of minimum and maximum acceleration of piston in reciprocating engine.
Acceleration is rate of change of velocity. See also Figures 4a and 4b above.
Minimum acceleration:
Acceleration is zero when slope of curve of velocity is zero. This occurs at maximum velocity or when
∠
x
{\displaystyle \angle x}
is approx.
±
71.3
∘
.
{\displaystyle \pm 71.3^{\circ }.}
Maximum acceleration:
Acceleration is maximum when slope of curve of velocity is maximum.
Maximum negative acceleration occurs when slope of curve of velocity is maximum negative. This happens at top dead center
when
∠
x
=
0.
{\displaystyle \angle x=0.}
Maximum positive acceleration occurs when slope of curve of velocity is maximum positive. This happens before and after bottom
dead center when
∠
x
{\displaystyle \angle x}
is approx.
±
127
∘
.
{\displaystyle \pm 127^{\circ }.}
Let the engine continue to rotate at
100
{\displaystyle 100}
radians/second.
a
c
c
m
a
x
=
y
″
i
n
c
h
e
s
r
a
d
i
a
n
2
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
⋅
100
r
a
d
i
a
n
s
s
e
c
o
n
d
{\displaystyle acc_{max}=y''{\frac {inches}{radian^{2}}}\cdot 100{\frac {radians}{second}}\cdot 100{\frac {radians}{second}}}
abs
(
a
c
c
m
a
x
)
=
6.92
∗
100
∗
100
{\displaystyle (acc_{max})=6.92*100*100}
inches/second
2
{\displaystyle ^{2}}
or approx.
180
{\displaystyle 180}
times
the acceleration due to terrestrial gravity.
This maximum value of acceleration is maximum negative when
∠
x
=
0.
{\displaystyle \angle x=0.}
According to Newtonian physics
f
=
m
a
{\displaystyle f=ma}
, force = mass*acceleration, and
w
=
f
s
{\displaystyle w=fs}
, work = force*distance. In this engine energy expended in just accelerating piston to maximum velocity
is proportional to rpm
2
{\displaystyle ^{2}}
.
Perhaps this helps to explain why a big marine diesel engine rotating at low RPM can achieve efficiency of
55
%
.
{\displaystyle 55\%.}
Let a body move in accordance with the following function of
t
,
f
(
t
)
{\displaystyle t,\ f(t)}
where
t
{\displaystyle t}
means time:
p
t
=
f
(
t
)
=
a
t
2
+
b
t
+
c
{\displaystyle p_{t}=f(t)=at^{2}+bt+c}
where
p
t
{\displaystyle p_{t}}
is position at time
t
.
{\displaystyle t.}
p
t
{\displaystyle p_{t}}
has the dimension of length. Therefore, each component of
f
(
t
)
{\displaystyle f(t)}
must have the dimension of length.
For
b
t
{\displaystyle bt}
to have the dimension of length,
b
{\displaystyle b}
must have the dimensions of
l
e
n
g
t
h
t
i
m
e
{\displaystyle {\frac {length}{time}}}
or velocity.
For
a
t
2
{\displaystyle at^{2}}
to have the dimension of length,
a
{\displaystyle a}
must have the dimensions of
l
e
n
g
t
h
t
i
m
e
2
{\displaystyle {\frac {length}{time^{2}}}}
or acceleration.
If
t
==
0
,
p
0
=
c
{\displaystyle t==0,\ p_{0}=c}
and
p
t
=
f
(
t
)
=
a
t
2
+
b
t
+
p
0
.
{\displaystyle p_{t}=f(t)=at^{2}+bt+p_{0}.}
The derivatives enable us to assign specific values to
a
,
b
.
{\displaystyle a,\ b.}
d
(
p
t
)
d
t
=
f
′
(
t
)
=
v
t
=
2
a
t
+
b
{\displaystyle {\frac {d(p_{t})}{dt}}=f'(t)=v_{t}=2at+b}
where
v
t
{\displaystyle v_{t}}
is velocity at time
t
.
{\displaystyle t.}
If
t
==
0
,
v
0
=
b
{\displaystyle t==0,\ v_{0}=b}
and
p
t
=
f
(
t
)
=
a
t
2
+
v
0
t
+
p
0
.
{\displaystyle p_{t}=f(t)=at^{2}+v_{0}t+p_{0}.}
d
(
v
t
)
d
t
=
f
″
(
t
)
=
2
a
,
{\displaystyle {\frac {d(v_{t})}{dt}}=f''(t)=2a,}
a constant equal to the acceleration to which the body is subjected.
For convenience let us say that
p
t
=
1
2
a
t
2
+
v
0
t
+
p
0
{\displaystyle p_{t}={\frac {1}{2}}at^{2}+v_{0}t+p_{0}}
where
a
{\displaystyle a}
is the
(constant) acceleration to which the body is subjected.
Then
v
t
=
1
2
⋅
2
a
t
+
v
0
=
a
t
+
v
0
{\displaystyle v_{t}={\frac {1}{2}}\cdot 2at+v_{0}=at+v_{0}}
and
d
(
v
t
)
d
t
=
a
.
{\displaystyle {\frac {d(v_{t})}{dt}}=a.}