Maple project
Math 180
2
2
3
I. Defining functions: f x   3x  5x  3 sin x  1
> f:=x->(3*x^2-5*x+3)*(sin(x^3+1))^2;



Name:

2
2
3
f := x ® (3 x - 5 x + 3) sin(x + 1)
Evaluate the function at x 
> f(Pi/2.0);

2
2
3
2
(0.7500000000 p - 2.500000000 p + 3) sin(0.1250000000 p + 1)
Approximate the above value with 10 decimal places.
> evalf[10]((.7500000000*Pi^2-2.500000000*Pi+3)*sin(.1250000000*Pi^3+1)^2);
2.480792228
Sketch the graph over the interval  2 ,2 
> plot(f(x),x=-Pi..Pi);
Take limit function for x approaches to
> a:=limit(f(x),x=Pi/2);
a := -
2
5
5
æ 1 3ö
p + p cosç p ÷
2
2
è 8
ø
æ 1 3ö
+ 3 - 3 cosç p ÷
è 8
ø
2

2
2
æ 1 3ö
æ 1 3ö
5
æ 1 3ö
cos(1) - 5 p cosç p ÷ cos(1) sinç p ÷ sin(1) + p sinç p ÷
2
è 8
ø
è 8
ø
è 8
ø
2
æ 1 3ö
æ 1 3ö
æ 1 3ö
cos(1) + 6 cosç p ÷ cos(1) sinç p ÷ sin(1) - 3 sinç p ÷
è 8
ø
è 8
ø
è 8
ø
2
3 2 3 2
æ 1 3ö
+ p - p cosç p ÷
4
4
è 8
ø
-
3 2 æ 1 p 3ö
p sinç
÷
4
è 8
ø
2
cos(1) +
3 2
æ 1 3ö
æ 1 3ö
p cosç p ÷ cos(1) sinç p ÷ sin(1)
2
è 8
ø
è 8
ø
2
2
sin(1)
Approximate the above with five digits:
> evalf[5](a);
2.4806
1.
Use the above technique to evaluate the following:
a)

 
 sin 3 x 2  5 x  1
lim 
x 
x2 1



b)
 tan( x 2  5 x  3 

lim 
x  3

4
x 3 1
e sin x 

2
sin(1)
2
sin(1)
2
2


 5x 6  2 x 3  4 x  1 
lim e x 5 x 3 cos x 2  4


x 0
c)
d)
lim
x  3 x 3  2 x 2  4 x  5 


II.
Differentiation
> f:=theta->sin(theta^2-3*theta+1)*(cos(theta-2))^3;
f := sin( 23 1 ) cos( 2 )3
2
Define the first derivative:
> fprime:=diff(f(theta),theta);
fprime :=
cos( 23 1 ) ( 2 3 ) cos( 2 ) 33 sin( 23 1 ) cos( 2 ) 2 sin( 2 )
Find critical points:
> solve(fprime,theta);
1
2 
2
To determine whether the critical point is min or max, evaluate the second derivative at the critical point.
Define the second derivative function:
> fdouble:=diff(fprime,theta);
fdouble := sin( 23 1 ) ( 2 3 ) 2 cos( 2 ) 32 cos( 23 1 ) cos( 2 ) 3
6 cos( 23 1 ) ( 2 3 ) cos( 2 ) 2 sin( 2 )
6 sin( 23 1 ) cos( 2 ) sin( 2 ) 23 sin( 23 1 ) cos( 2 ) 3
> eval(fdouble,theta=2+0.5*Pi);
0
So the critical point here is an inflection point.
2.
Do problems:
Section 4.3: 59 – 62.
3.
A cone of height h and radius r is constructed from a flat circular disk of radius a in. by removing a section
AOC of arch length x in. and then connecting the edges OA and OC.
a)
b)
c)
Find a formula for the volume V of the cone in terms of x and a.
Find radius r and height h in the cone of maximum volume for a = 4, 5, 6 and 8.
How should we cut the disk of radius 4in (a=4in.) so that the volume of the cone maximum?
III. Analyzing functions:
Define a function
> f:=x->5*x^3-7*x^4+1;
f := x5 x37 x41
> plot(f(x),x=-infinity..infinity);
Define the first derivative function
> fprime:=diff(f(x),x);
fprime := 15 x228 x3
Find critical points
> solve(fprime,{x});
15
{ x0 }, { x0 }, { x }
28
Define the second derivative
> fdouble:=diff(fprime,x);
fdouble := 30 x84 x2
Check whether the critical point is min/max or inflection point.
> eval(fdouble,x=0);
0
> eval(fdouble,x=15/28);
-225
28
Second derivative is negative then it is a maximum. Now then we need to find the maximum point.
> eval(f(x),x=15/28);
104683
87808
 15 104683 
So the maximum is  ,

 28 87808 
4.
Do problem 2, Section 4.4:
36, 44 and 97
IV.
Evaluate Riemann Sum using MAPLE
Now, using Riemann Sum to evaluate integral
1
 x dx .
3
0
First define the interval of the integral:
> a:=0;b:=1;
a := 0
b := 1
Define delta x:
> deltax:=(b-a)/n;
deltax :=
1
n
deltax :=
1
n
Now, define function:
> deltax:=(b-a)/n;
> f:=x->x^3;
3
f := x ® x
Find xi  a  x
> x[i]:=a+deltax*i;
xi :=
i
n
Define the Riemann Sum:
> RiemanSum:=Sum(f(x[i])*deltax, i=1..n);
n
RiemanSum :=
S
i3
i= 1 n4
Then the actual area is:
> Area:=limit(RiemanSum,n=infinity);
Area :=
5.
1
4
Follow the above outline to do the following problems:
a)
c)
e)

 cos
xe 2 x 1 dx
b)
cos x sin xdx
d)
cos 3 x sin 3 xdx
f)


2

2
0
0
3
3
3x  5
dx
5 x 2  3

5

1
0


0
arcsin x
dx
1 x2
cos2 x e sin2 x  dx
V.
Find the integrals using MAPLE With Fundamental Theorem of Calculus
Example 3
Define a function
> f:=x->5*x^3-2*sin(2*x)+1;
f := x5 x32 sin( 2 x )1
Define an integral
> Int(f(x),x=0..1);
1
3

 5 x 2 sin( 2 x )1 d x
0
> int(f(x),x);
5 4
x cos( 2 x )x
4
> int(f(x),x=0..1);
5
cos( 2 )
4
> F:=int(f(x),x=0..1);
5
F := cos( 2 )
4
> evalf(F);
.8338531635
7.
Do the following functions:
Section 5.4: #24, 27, 28, and 32
Section 5.5: #36, 63, 64, and 65