CalcLabs with Maple V

Solutions for Assignment 2

Jacobowitz

Page 31 (1, 3, 7, 8, 15, 17, 23)

1.       For a square, we know that each side is of equal length. To solve the problem, we draw a line segment connecting diagonal corners. This line segment is the square's diagonal length d. Now, the square has been split into two triangles. Working with one triangle, we can obtain an expression for the length of a side by using trigonometry. The length of the side s is equal to the length of the hypotenuse multiplied by the cosine or sine (for right triangle with sides of equal length) of the angle (theta) between the hypotenuse and the side.

> theta := Pi/4;

> s := d * cos(theta);

> area := s^2;

> subs(d=2, area);

> subs(d=3.7, area);

3.       To find the length of the diagonal of a cube, draw a three dimensional cube. Draw the diagonal from one corner diagonally to the opposite corner. Notice that the diagonal and vertical piece almost make a triangle. To complete the triangle, draw a line from the end of the diagonal to the opposite end of the vertical piece. Now you have formed a triangle. We know that the vertical side of the triangle has length s. Now we can find out the length of the horizontal side of the triangle in terms of s be recognizing their is another triangle of sides s and s whose hypotenuse is the length of the horizontal side. Notice that this new triangle is a right triangle whose sides are of equal length. Therefore the theta between its hypotenuse and adjacent side is 45 degress. Since sin(theta)=s / length(hypotenuse) and theta = 45, we know that the length of the hypotenuse for the new triangle = s / sin(theta). Remember that the hypotenuse of the new triangle is the same length as the horizontal side length in our original triangle. So our original triangle has sides of length s, s/sin(theta), and d.

> theta := Pi/4;

> phi := arctan(sin(theta));

> s := d*sin(phi);

> volume := s^3;

> subs(d=2, volume);

> subs(d=3.7, volume);

7.   Enter and plot the following expression and evaluate its limit

> f := (x^5 + 1)/(x^2 - 1);

> plot(f, x=-2..0);

> limit(f,x=-1);

8.   Enter function, plot and evaluate limit

> f:=x -> (x^4 - 16)/(x^2 - 4);

> plot(f(x),x=1..3);

> limit(f(x),x=2);

15. Use plot and fsolve to find all solutions of the following

(a)

> plot((sin(x))^2 - cos(3*x), x=0..Pi);

> fsolve((sin(x))^2 - cos(3*x)=0,x,x=0..1);

> fsolve((sin(x))^2 - cos(3*x)=0,x,x=1..2);

> fsolve((sin(x))^2 - cos(3*x)=0,x,x=2..3);

(b)

> plot((cos(x))^2 - sin(3*x), x=0..Pi);

> fsolve((cos(x))^2 - sin(3*x), x=0..0.5);

> fsolve((cos(x))^2 - sin(3*x), x=0.5..1);

> fsolve((cos(x))^2 - sin(3*x), x=0.5..1);

> fsolve((cos(x))^2 - sin(3*x), x=2..2.5);

> fsolve((cos(x))^2 - sin(3*x), x=2.5..3);

(c)

> plot(8*cos(x)-x,x=-infinity..infinity);

> fsolve(8*cos(x)-x, x=-2..0);fsolve(8*cos(x)-x, x= -infinity..-2);

> fsolve(8*cos(x)-x, x=0..2);fsolve(8*cos(x)-x, x=2..6);fsolve(8*cos(x)-x, x=6..infinity);

17. Find decimal approx for all real roots of the following equation.

> plot ((1/x^4)-3+x^2, x=-infinity..infinity, y=-10..infinity);

> fsolve((1/x^4)-3+x^2, x=-1..0);

> fsolve((1/x^4)-3+x^2, x=-101..-1);

> fsolve((1/x^4)-3+x^2, x=0..1);

> fsolve((1/x^4)-3+x^2, x=1..10);

23. Perform the following steps to illustrate that plots can be deceiving.

> f:=x^3 - x^2 - x + 1.001;

> plot(f,x=-2..2);

From this plot it appears as if f has 2 real solutions, at -1 and at 1.

> solve(f=0,x);

Solve command returns 1 real solution and two imaginary solutions

> plot(f, x=0.9..1.1);

The graph does NOT cross the x-axis at x=1.

The purpose of this exercise is to make students more familiar with the limitations of a graphing program. When

the scale of a graph is used incorrectly, mathematical errors can occurr