Calculus Calculations
1. Taking Derivatives of a function:
To calculate
the derivatives of a function for e.g.' f(x)' with respect to 'x' using MATLAB,
there is a function associated with this objective which can be extracted as:
>>
diff (f(x),x)
Where f(x) is
the function expression or predefined function using "inline", and x
is predefined symbol that represent the independent variable of the function
and the derivate respective variable.
Example:
Find the derivate of [ X3 + 3X2 – 5X ], using
MATLAB
Solution:
>>
syms X
>>
function1=inline('X^3+3*X^2-5*X','X')
function1
=
Inline function:
function1(X) = X^3+3*X^2-5*X
>>
diff (function1(X),X)
ans =
3*X^2 + 6*X – 5
Example:
Function: [ log(u) + exp(u) – 3i ]
Solution:
>>
syms u
>>
function2 = inline('log(u)+exp(u)-3i','u')
Function2
=
Inline function:
Function2(u) = log(u)+exp(u)-3i
>>
diff (function2(u),u)
ans =
exp(u) + 1/u
2. Evaluating derivative at a point:
To use the
derivative of a function and evaluate any value or point at its variable, there
is two steps to do that:
·
Define
the original function using inline.
·
Define
other inline function for derivative function
Example:
Function is [X2 + 3X - X ]
Solution:
>>
syms X
>>
function3 = inline ('X^2+3*X-X','X')
Function3
=
Inline function:
Function3(X) = X^2+3*X-X
>>
Diff_F3 = inline (diff (function3(X), X),'X')
Diff_F3
=
Inline function:
Diff_F3(X) = X.*2.0+2.0
>>
Diff_F3 (6)
ans
=
14
>> Diff_F3 (23.6)
ans
=
49.2000
3. Calculate partial derivatives:
When the
equation consists of more than one variable, the partial derivative can be
calculated with the same way using the function and the variable with respect
of.
Example:
Find derivate of [ sine (a)/cosine (b) ] with respect to
"a" and to "b"
Solution:
>>
syms a b
>>
Function4 = inline('sin(a)/cos(b)','a','b')
Function4
=
Inline function:
Function4(a,b) = sin(a)/cos(b)
>>
diff (Function4(a,b),a)
ans =
cos(a)/cos(b)
>> diff (Function4(a,b),b)
ans =
(sin(a)*sin(b))/cos(b)^2
The partial derivative result also can be used as a function to
evaluate any point in it.
Example:
X3 + 2Y2 – 3Z
Solution:
>>
syms X Y Z
>>
F5 = inline('X^3+2*Y^2-3*Z','X','Y','Z')
F5
=
Inline function:
F5(X,Y,Z) = X^3+2*Y^2-3*Z
>>
Diff_F5_X = inline(diff(F5(X,Y,Z),X),'X')
Diff_F5_X
=
Inline function:
Diff_F5_X(X) = X.^2.*3.0
>>
Diff_F5_Y = inline(diff(F5(X,Y,Z),Y),'Y')
Diff_F5_Y
=
Inline function:
Diff_F5_Y(Y) = Y.*4.0
>>
Diff_F5_Z = inline(diff(F5(X,Y,Z),Z),'Z')
Diff_F5_Z
=
Inline function:
Diff_F5_Z(Z) = -3.0
>>
Diff_F5_X_Y =inline(diff(F5(X,Y,Z),Y),'X','Y')
Diff_F5_X_Y
=
Inline function:
Diff_F5_X_Y(X,Y) = Y.*4.0
>>
Diff_F5_X_Y_Z = inline(diff(F5(X,Y,Z),X),'X','Y','Z')
Diff_F5_X_Y_Z
=
Inline function:
Diff_F5_X_Y_Z(X,Y,Z) = X.^2.*3.0
Then
any point can be evaluated in any function of the new derivative functions
created just rely on the independent variable that with respect to.
>>
Diff_F5_X_Y(23,15)
ans
=
1587
>>
Diff_F5_X_Y_Z(12,15.5,22)
ans
=
432
Ordered degree derivatives:
The nth order derivative of the function f(x) along x
can be expressed as follow:
>>
diff(f(x),n), where n is the
nth order of derivative.
Example:
>>
f1 = inline('x^3+2*x^2-3*x-5','x')
f1
=
Inline function:
f1(x) = x^3+2*x^2-3*x-5
%
the function is defined for x independent.
>> syms x
>>
diff(f1(x)) % get the 1st order derivative.
ans
= 3*x^2 + 4*x - 3
>>
diff(f1(x),x)% same result as before.
ans
= 3*x^2 + 4*x - 3
>>
diff(f1(x),2)% the 2nd order derivative.
ans
= 6*x + 4
>>
diff(f1(x),3)% the 3rd order derivative.
ans
= 6
-
Another
example if we have more than one independent variable in a function:
>>
syms a b
>>
Function4 = inline('sin(a)/cos(b)','a','b')
>>
diff(Function4(a,b),2,b)% 2nd order along b.
ans
=sin(a)/cos(b) + (2*sin(a)*sin(b)^2)/cos(b)^3
>>
diff(Function4(a,b),2,a)% 2nd order along a.
ans
= -sin(a)/cos(b)
>>
diff(Function4(a,b),3,a)% 3rd order along a.
ans
= -cos(a)/cos(b)
4. Taking the integrals of a function:
Integral
function in MATLAB is "int(f(x),x)" but in integral there is
no need to define the variable that integral taken with respect to, this in
case there is only one variable in the equation.
Example:
Sin(t) + e(-2t)^2 +2t3
Solution:
>>
syms t
>>
F7 = inline ('sin(t)+exp(-2*t^2)+2*t^3','t')
F7
=
Inline function:
F7(t) = sin(t)+exp(-2*t^2)+2*t^3
>>
INT_F7 = inline(int(F7(t)))
INT_F7
=
Inline function:
INT_F7(t) =
-cos(t)+t.^4.*(1.0./2.0)+sqrt(2.0).*sqrt(pi).*erf(sqrt(2.0).*t).*(1.0./4.0)
>>
INT_F7(10)
ans
=
5.0015e+03
>>
pretty(INT_F7(t))
1/2 4
2822212540896131 erf(2 t) t
---------------------------- - cos(t) + --
4503599627370496 2
Example:
2V3 + 3V2 – T3, find the integral
with respect to V, and evaluate the value (12.5,22.3)
Solution:
>>
syms V T
>>
F6 = inline ('2*V^3+3*V^2-T^3','V','T')
F6
=
Inline function:
F6(V,T) = 2*V^3+3*V^2-T^3
>>
int_F6_V = inline (int(F6(V,T),V),'V','T')
int_F6_V
=
Inline function:
int_F6_V(V,T) =
-T.^3.*V+V.^3+V.^4.*(1.0./2.0)
>>
int_F6_V(12.5,22.3)
ans
=
-1.2446e+05
4.1. Definite integration:
The examples
before demonstrate indefinite integration. To calculate definite integration
for "f(x)" with lower bound "a" and higher bound
"b" using the "int" function we just modify the
syntax by:
int (f(x),a,b)
Which mean
calculate the integral for "f(x)" and evaluate it from "a"
to "b".
Example:
Find
the integral of [X2 + 3X] and evaluate the integration from 2 to 5
Solution:
>>
syms X
>>
F8 = inline('x^2+3*x','x')
F8
=
Inline function:
F8(x) = x^2+3*x
>>
INT_F8 = int(F8(X))
INT_F8
=(X^2*(2*X
+ 9))/6
>>
INT_F8_def = int(F8(X),2,5)
INT_F8_def
= 141/2
Example2:
Find
the integral of [sine(S) + cos(S)] and evaluate the integration from –pi/2 to
pi/2
Solution:
>>
syms S
>>
F9 = inline('sin(S)+cos(S)','S')
F9
=
Inline function:
F9(S) = sin(S)+cos(S)
>>
INT_F9 = int(F9(S))
INT_F9 = sin(S) - cos(S)
>>
INT_F9_def = int(F9(S),-pi/2,pi/2)
INT_F9_def = 2
5. Calculate the limit of a function:
To
calculate limits using MATLAB the command line syntax should be:
limit(f(x),x,a)
where
"f(x)" is the function, and "x" is the independent variable
of the function, and a the limit value or approach.
Example:
>>
syms K
>>
limit(sin(K)/K,K,0)
ans
= 1
>> F10 =
inline('sin(K)/K','K')
F10
=
Inline function:
F10(K) = sin(K)/K
>>
limit(F10(K),K,0)
ans
= 1
>>
syms x
>>
g = inline('(2*x^3+4*x^2+x-3)/(3*x^3-8)','x')
g
=
Inline function:
g(x) = (2*x^3+4*x^2+x-3)/(3*x^3-8)
>>
limit(g(x),x,Inf)
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