MATHCAD TUTORIAL
LESSON 1: Basics
Page 1
!
" #
% #
$
$
&
____________________________________________________________________________________
!"
mass := 3 ⋅ kg
accel := 4.0⋅
m
Force := mass⋅ accel
2
Force = 12 N
s
#
$
% &
'
()
*
+
*
,
+
.
-
*
/
+
0 + 0
_____________________________________________________________________________________
,
"
$
! #
+
'
1112
/
2111
(
3
*
$
4
.
3
5
% 8
;
< 8
6
9"9
9:9
9=9 9 9 9>9
/
9?9
7
9":9
MATHCAD TUTORIAL
LESSON 1: Basics
*
Page 2
"
____________________________________________________________________________________
("
&
:(
:5
"
a := 2
b := 4
a⋅ b = 8
&
:D
c := 9
x :=
a
b+c
: B ?- = .C>
"
⋅b
9 " ? = @ A@
=
x = 0.615
A> 9 #
@ A
?- = .
*
&
y :=
9 :9
*
a+b
c
:B *
- = .C?
y = 0.272
9 "9
*
@
A@
A? 9 #
A
=
@
@
*
9 = @
A0 +
A
*
+
0
?
_____________________________________________________________________________________
#
*
*
7
*
3
*
MATHCAD TUTORIAL
LESSON 1: Basics
3
Page 3
*
*
E
9
&
*
&
4
'
?
&
+
_____________________________________________________________________________________
$"
&
4
type of font
&
F
χηαραχτερσ -
F
. G
'
-
EH
# IJ=
F
?
.
F
αβ
_______________________________________________________________________________________
MATHCAD TUTORIAL
5"
LESSON 1: Basics
&
;9
9 @ # IJ= A
)
.
α! $)9/ 9
( 5)/
" 9 @ # IJ= AB!"$)9/ 9 @ # IJ= AB("5)/
"9 # "
9B9
*
9 @ # IJ= A
!
$)/
(
5) E
.
99*
)/
Page 4
9
9B9
*
α 1 := 30
α 2 := 40
!
α general := 6
α
(
α
0
α:
α = 30
α
40
_____________________________________________________________________________________
,
3
,
'
"
item1 := 1
α item1 = 30
α item1+item1 = 40
&
&
K
(
α
.
MATHCAD TUTORIAL
! 8
LESSON 1: Basics
*
*
(
Page 5
% $> (=5 $> D !:)
%
α
$ 6
α⋅γ
ψ
γ
$D γ
!) (
=
α−1
α+
=
ψ
γ
8
⋅ 3 + 6⋅
M
α+ψ
3
2
=
ψ
L
"
MATHCAD TUTORIAL
LESSON 1: Basics
Page 6
5
a := 23
b := 46
ans := a +
b
− b⋅ c
c
c
b+
ans :=
c := 92
a
d := 5
ans = −4.208 × 10
a − b⋅ c + b⋅ (c − 1)
*
⋅d
ans = 10.87
a
b
*
a
d
6
ans = 3.052 × 10
c+d
#
#
" #
c⋅ d
+
a
a
a ⋅b
ans :=
3
d
b
*
2
+d
c
'
O
'
"
-
. &
5
K
3
-
'
.
N
-
.
5! = 120
-
.
ln( 9 ) = 2.197
-
−6
e
-
6
5
−1
.
.
-
= 0.2
-
.
.
MATHCAD TUTORIAL
3
9 = 2.08
10
LESSON 1: Basics
-
Page 7
.
*
-
.
sin( 8.5) = 0.798
-
'
-
*
.
*
5>3=1
-! :
5<3=0
-
'
.
/):
./ -
.
-
.
M = 14
.
-
M :=
5 2
3 4
.
v⋅ v =
-
.
'
(
v1 × v2 =
-
.
(
'
!
!
v=
M
0
=
−1
5
0.286 −0.143
−0.214 0.357
#
.
-
2 4
=
.
-
3
5 3
T
M =
M
-
,
-
.
.
slope( x , y) =
intercept( x , y) =
linterp( x , y , 4 ) =
max( x) = 0.615
rank( M ) = 2
.
*
*
-
.
(
$
MATHCAD TUTORIAL
rank( M ) = 2
LESSON 1: Basics
Page 8
min( x) = 0.615
erf ( 0.4) = 0.428
mean( y) = 0.272
stdev( y) = 0
var( x) = 0
corr( x , y) =
median( y) = 0.272
#
#
$
(
!
$
E
*
+
*
/
9":9
*
*
O
*
*
*
IEE '8#
*
F
"
,E#
- .:)
":
*
-- .
.
- .
- . : )/
*
6
0
,
!
&
O
,
____________________________________________________________________________________
MATHCAD TUTORIAL
LESSON 1: Basics
Page 9
S
S
To solve for a variable in an equation using the solve
keyword:
,
*
[Ctrl]=
*
7
*
Boolean
keyword
*
Symbolics toolbar.
P
Q
[Enter]
R
*
MATHCAD TUTORIAL
LESSON 1: Basics
Page 10
"
x := 2
2
!
x − 4 − 21 = 0
(
( a2 − b = 3 )
F
7
F
F
F
'
-
.
*
'
'
E
'
J
F
'
F
*
,
F
*
*
3
*
O
*
,
*
&
*
*
-
.
*
______________________________________________________________________________________
G
*
"
3⋅ x + 4 ⋅ y − 4z = 2
x := 1
y := −1
−7 ⋅ x + y + z = 5
z := 1
F
Given
3⋅ x + 4 ⋅ y − 4z = 2
*
−7 ⋅ x + y + z = 5
6x + 3y − z = 0
−0.522
Find( x , y , z) =
1.12
0.228
'
6x + 3y − z = 0
MATHCAD TUTORIAL
LESSON 1: Basics
Page 11
Finding Derivatives and Integrals
A second derivative can be found using Evaluate
d
2
dz
2
yields
( z⋅ atan( z) )
2
(1 + z2)
− 2⋅
Symbolically:
z
2
( 1 + z 2)
2
Alternatively, click on the variable of differentiation x and
choose Variable
Differentiate from the Symbolics menu.
2
by differentiation, yields 4⋅ x
2⋅ x + y
E
[Ctrl].
*
(
)
2
d
2x + y → 4
dx
Choose Evaluate
Complex from the Symbolics menu:
d
sin( x⋅ i + y)
dx
sin( y) ⋅ sinh( x) + i⋅ cos( y) ⋅ cosh( x)
E
complex
=
d
sin( x⋅ i + y) complex → −sin( 1 ) ⋅ sinh( 1 ) + i⋅ cos( 1 ) ⋅ cosh( 1)
dx
Click on x and choose Variable
Integrate from the Symbolics
menu.
2 x
by integration, yields x2⋅ exp( x) − 2 ⋅ x⋅ exp( x) + 2 ⋅ exp( x)
x ⋅e
I
S