PH36010 MathCAD worksheet 3
Roots, solve blocks and symbolic mathematics
The aim of this worksheet is to introduce you to different solution techniques in
MathCAD.
By the end of the sheet you will know how to:
 Use the root() function to solve equations of one variable
 Use the polyroots() function to find multiple roots of a polynomial
 Use a solve block to solve more complex systems of equations
 Use the symbolic processor to manipulate and solve equations
This worksheet takes the form of a number of examples which illustrate and introduce
these principles. Work through as many of them as you are able. At least you should read
through the entire sheet to get an overview of the topic.
Exercise 1 – Using the root() function
In this exercise we will use the root() function to solve some simple equations.






Start MathCAD and create a new, blank worksheet
The first equation we will try and find a solution for is
ex = x2
In order to use the root function, this must be in the form of f(x)=0
so re-write the equation in this form, and plot it.
Add grid lines to the graph to help estimate the point where the curve crosses the
y=0 line.
You should adjust the axes limits of the plot to zoom in on the region where the
line crosses the axis.
When you have finished, your plot should look like the one below:
f( x)
x
e
2
x
2
1
f( x)
0
1
1
0.5
0
0.5
x
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
x
MathCAD Example Sheet 3
By examining the graph, you can see that the solution will lie between –1 and –
0.5, so pick –0.75 as the guess and use the root function to get the exact value.
0.75
root( f( x)  x)  0.703
Having successfully solved this equation, use the same technique to solve the following
equations. In each case, use a plot of the function to find approximate solutions before
solving with root().
1.423  x
3
2
8.114238  x 8.537238 0


sin( x) cos( x)
A cubic equation will have 3 roots.
This will have many solutions

7 x
Complex roots need a complex guess
x
2
2 x 6
Exercise 2 – The effect of tolerance on solutions
The root() function uses an iterative technique which takes successive guesses to,
hopefully, converge on a solution. There is a system variable, TOL, when determines
how close the function has to evaluate to 0 before mathCAD assumes that it has reached a
solution. A small value of TOL will, in general, produce more accurate solutions, but
may take longer to arrive at a result. In this exercise we will see how the value of TOL
changes the result of solutions.
The following quadratic equation has two co-incident roots:
9 x
2
12 x 4
Plot the quadratic and zoom in on the area of the roots.
Use the following segment of worksheet to enable you to see how varying the value of
TOL changes the solutions offered by MathCAD.
TOL
0.0001
x
1
s oln1 root( f( x)  x)
s oln1 0.66898
f( s oln1)  4.82364  10
x
0.5
s oln2 root( f( x)  x)
s oln2 0.66364
f( s oln2)  8.2614  10
5
5
It is possible to change the value of TOL by means of a menu option within mathCAD
(Math|Options|Built-In variables|TOL) but the method of explicitly setting the system
variable on the worksheet is preferable as it shows what is being changed.
Since TOL is already defined by the system, you will have to force the system to put up
an assignment operator by using the ‘:’ key or selecting := from the evaluation palette.
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MathCAD Example Sheet 3
Exercise 3 – Using Polyroots() function to get roots of
polynomials
MathCAD provides a specialist function, polyroots(), for funding all the roots of a
polynomial in one operation. It has the advantage that all roots, both real and complex,
are returned at once and that it is not necessary to provide guess values.
As an example, to find all the roots of the cubic equation from exercise 1, do the
following:
3
x
1.423  x
2
8.114238  x
8.537238
polyroots
8.114238
1.423
1
8.537238 0
Equation to solve
2.718

1
3.141
You will need to generate the argument for the polyroots() function by inserting a matrix
with 4 rows and a single column.
Now use polyroots() to find solutions to the following polynomials.




2
7 x 2 x 6
2
9 x 12 x 4
4
2
3
9 x 13 x 12 x 12 x 4
4
3
2
x 20  x 87  x 180  x 864
Can you use polyroots to find the 3 cube roots of –1 and the 4 fourth roots of –1 ?
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Exercise 4 – Using a solve block to solve more difficult
problems.
The solve block is another iterative technique available within mathCAD which will
solve up to 50 equations in 50 unknowns. The equations do not have to be polynomials
and can include inequalities (<,>,,). Each of the unknowns must be given an initial
guess value.
We will start with some simple examples and progress to more difficult situations. The
first example will use a solve block to find the square root of 3, starting with a guess
value of 2.
x
2
Given
2
x
3 0
find( x)  1.73205
Initial guess value is defined first
Solve block starts with keyword 'given'
Use <ctrl>= to get bold equals sign
The find() function finishes the solve block and
returns solution(s)
Note the use of the bold equals sign (=) or ‘equals to’ operator in the solve block. This is
a different symbol to either the assignment operator (:=) or the ‘print the value of a
variable’ operator (=).
The ‘equals to’ operator is used extensively within solve blocks and can also be used in
Boolean logic.
In the next example, we will use a solve block to find the places where the unit circle
intersects with the line x+y=0.
Unit circle (x2+y2=12)
Solutions
Line of x+y=0
As before, we start our solve block with guess values for x- and y-
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x
1
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y
1
Guess values
given
2
x
Equations in solve block
2
y 1
x y 0
find( x  y) 
0.70711
Find() returns vector of solutions
0.70711
this returns the x,y values for one of the intersection points. In order to return the coordinates of the other point, we can either start with guess values closer to it, or put a
condition in the solve block to force the solution.
x
1
y
1
given
2
x
2
y 1
Extra inequality added to solve block
x y 0
x 0
find( x  y) 
0.70711
0.70711
MathCAD has restrictions on what may be written in the solve block (between ‘given’
and ‘find()’ ), in particular it is not allowed to put a definition of either a variable or a
function.
In the above examples, we have used the ‘print value of’ equals sign to print the values of
x and y at the solution. By creating a vector to hold the results, we can set variables to
hold the solution in one step, as shown below:
xval
yval
find( x  y)
Multiple assignments in single step
xval  0.70711
yval 0.70711
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Use a solve block to solve the following systems of equations:
3  x 7  y 90
2
x
7  x 2  y 25
x 2 y z 4
2
y 3
3 x 4 y 2 z
x y 1
You should have 4 solution
points
3
6 x 2 y z 4
For the first and last of these, you should check your result using the matrix inversion
technique from a previous worksheet.
At the back of this handout is an exercise, taken from a previous PH15010 example sheet,
to plot the trajectory of a cannon ball. Create this exercise in a fresh worksheet and use
one of the above solution techniques to determine the time at which the cannon ball hits
the ground after it has been fired.
Exercise 5 – Symbolic Maths and manipulation of equations
As well as the techniques for providing numeric solutions to equations and systems of
equations we have used so far, MathCAD includes a power symbolic processor, which
can manipulate and solve equations using algebraic techniques.
When working with the symbolic processor, it is helpful to bring up the symbolic toolbar.
To see how the symbolic processor can solve equations and evaluate expressions
symbolically enter the following exercises.
Simplifying Expressions
To see how the symbolic processor can help simplify expressions, enter the following
expression in MathCAD.
6x
2
2x a 7x 3a x
2
with the selection box In the expression, press ‘simplify’ from the symbolic palette and
see how mathCAD simplifies the expression, collecting common terms and cancelling
out where possible.
6 x
2
2 x a
7 x 3 a
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x simplify
7 x
2
5 x 4 a
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Use the symbolic processor to simplify the following expressions:
3 x
7 a
2
6.5  p
2 x
6 a
2
5.7  q
2
6.9  q
2
2
3 x
2
2
x
1.2  p
7 a
( 2 q)
2
2
x
2
3 3 6 3
r s m n
4 3
r m n
7
3
3
7
Expanding Expressions
To see how the symbolic processor can expand expressions, put the selection rectangle in
the expression and press ‘expand’ from the symbolic palette.
The expand keyword has a placeholder after it, you should either fill in the variable you
wish to expand on or, delete the placeholder. There appears to be a bug in mathCAD, as it
seems to make no difference what is in the placeholder.
3
( g h) expand
3
g
3
h
3 g h
2
2
3 g  h
Expand the following examples:
2
y  ( x 7  y)
3
x
( p  x q )  ( r x s)
1.23  a
2
2 2
2
2
4.56  b  7.89  b
6.66  a
2.7  a
2
4.2  b
2
‘Undefining’ Variables
When processing equations symbolically, mathCAD will use the numeric value of any
variables that have been already defined. As an example, consider the following binomial
expansion:
x
1
2
( x y) expand
1
2 y y
2
Here mathCAD has used the value of 1 in the place of x before performing the expansion.
Depending on your problem, this may not always be what you want to achieve.
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In order to undefined a variable, use an assignment operator to set the variable equal to
itself. MathCAD understands this recursive definition to undefine the variable. This is
illustrated below:
x
1
2
( x y) expand
x
1
2 y y
2
x
2
( x y) expand
2
x
2
2  x y y
Solving Equations
To solve an equation using the symbolic processor, enter the equation you wish to solve
and then press ‘solve’ on the symbolic toolbar. Fill in the variable you wish to solve for
in the placeholder and press <enter> to see the result.
2
x
x 2 0 s olve x
1
2
Note that MathCAD cannot find a solution to the above case where x is already defined,
you must undefine x using the recursive definition introduced above.
Now solve the following equations using the symbolic processor:
( x 1)  ( x 1) 0
2
x
1 0
2
x
1 0
( y 2)  ( y 3) 0
3 z
2
2 z
8 0
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Exercise 6 – The tin can problem
This is an exercise taken from a first year example sheet, which shows how we can use
the symbolic processor in MathCAD to solve a real world problem. It is presented as a
worked example, which you should copy and satisfy yourself that you understand the
steps to reach a solution.
A manufacturer of tin cans wishes to maximise the volume contained in a can, whilst
minimising the amount of metal used to construct the can. Show that, for given amount of
metal, the volume of a can is maximised when the radius is half the height.
Start by writing down expressions for the Volume and Area of the can, in terms of the
radius, r, and height, h.
2
Vol   r  h
Area ( 2    r  h)
2   r
2
Use the bold symbolic equals sign in constructing these two equations, as we will be
manipulating them with the symbolic processor.
Copy the expression for area and use the symbolic solver to derive an expression for h in
terms of the area and r.
Area ( 2    r  h)
2    r solve h
2
1
2
Area 2    r
(   r)
2
Copy the expression for the volume and use the symbolic engine to substitute h in the
expression with the expression you have just derived:
2
Vol   r  h s ubs tituteh

Area 2    r
(   r)
2
1
2
Vol
1 
2
r Area 2    r
2
We now have an expression for the volume of the can in terms of the Area and the radius.
Given that the area is constant, dVol/dr will be zero at the maximum volume. Copy and
paste the expression for volume and use symbolic differentiation to differentiate it with
respect to r.
d 1 
2
r Area 2    r
dr 2
1
2
Area 3    r
2
Use the  symbol from the symbolic palette, or type <ctrl>.
Now we have an expression for dVol/dr, we can find the value of r where this is zero.
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1
1
2
Area 3    r
2
0 solve r
1   
2
6 (  Area)
( 6  )
1
1   
2
6 (  Area)

(6 )
MathCAD has provided two solutions corresponding to positive and negative values of r.
We are interested in positive values of r, so we will copy the first solution and substitute
back the expression for the area, to arrive at an expression for r at the point of maximum
volume.
1
1
1   
2
6 (  Area)
substituteArea

( 2    r  h)
( 6  )
1      
6  2rh
( 6  )
2   r
2
2   r
2
2
We now have an expression for r in terms of r and h which maximises the volume. So we
can now solve for r in terms of h.
1
1      
2
6  2  r h 2   r

(6 )
2
0
r s olve r
1
2
h
Neglecting the r=0 case, we now have the solution that the maximum volume is attained
when r=h/2.
Note that it is possible to combine together several symbolic operations in one step, as
shown below:
1
1 
2
r Area 2    r
dr 2
d
1      
6  2rh
( 6  )
solve r
0
substituteArea

( 2    r  h)
2   r
2
2   r
2
2
1
1      
6  2rh
( 6  )
2   r
2
2
As an exercise, return to the cannon ball problem mentioned earlier, and, using the
symbolic processor, derive an expression for the distance the cannon ball will travel
before it hits the ground in terms of initial velocity, V0 and angle of the cannon, .
Use symbolic differentiation to find the angle of that maximises the range of the
cannon.
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Appendix 1- Plotting the path of a cannon ball.
This exercise is taken from a previous PH15010 course and shows how MathCAD can
predict the simple trajectory of a cannon ball.
So far we have used mathCAd’s plotting abilities to plot a simple function, such as sin(x)
vs x and also seen how we can plot one vector against another. Frequently we wish to
plot one function against another, for example we may have functions which describe the
height and distance downrange of a projectile as a function of time. By plotting one
function against the other we can plot the trajectory of the projectile.
The following exercise plots the trajectory of a cannon ball fired at a particular angle with
a particular initial velocity.

30 deg
VInitial 50
m
s
 ()
VXInitial VInitialcos
Seperate initial velocity into X- & Y- components
 ()
VYInitial VInitialsin
The sheet starts by describing the angle of the cannon and the initial velocity and then
goes on to split the velocity into components in the x- and y- directions.
Once you have created these statements, it is possible to create two functions to describe
the position of the cannonball in the X- and Y- direction, using the simple equations of
motion under constant acceleration. We ignore air resistance, and the position in the xaxis is readily calculated from the initial velocity in the x-axis. Similarly, the position in
the y direction may be calculated, using the built-in value ‘g’ for acceleration due to
gravity. This acts downwards, whereas our reference direction for the y-axis is upwards,
hence the minus sign in the equation.
Pos X( t )

VXInitialt
Pos Y( t )

VYInitialt
1 2
gt
2
All that remains is to select a suitable range of time steps and to plot the functions.
A good range is from 0 to 6 seconds in steps of 0.1 second.
Note that time must have the units of seconds and see how this must be incorporated into
all 3 terms defining the range variable.
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MathCAD Example Sheet 3
0 s  0.1 s  6 s
40
30
20
10
PosY( t )
0
10
20
30
0
50
100
150
200
250
300
PosX ( t )
I’ve added grid lines on the y-axis which allow an estimate to be made of when the
cannonball will hit the ground, which in this case will be about 220m down range.
As an exercise, determine the value of  which gives the maximum range – is this value
what you expected ?
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