CHAPTER 4
Parametric Differentiation
Note that some text books adopt the notation for differentiation with regards to time
and referred to as the rate of change in
as
. When we differentiate with respect to displacement
(with the dash) and referred to as the change in
with regards to
(with the dot)
, it is shown as
. In previous chapters, the dash
notation was used when differentiating functions containing only one independent variable irrespective of its type but
in this chapter we will indicate time derivatives with a dot and space derivatives with a dash as both types will be
present in the same question and using the dash notation for all can lead to confusion.
Parametric equations vs Cartesian Equations
Depending on the application, the circle curve in pic a) with reference point
• cartesian equation
• parametric pair
by the tip of the
• vector
can be observed in terms of its
or in terms of the
. It can even be expressed as the curve (locus) traced
A locus is a curve or figure formed by all the points satisfying the cartesian equation. A parametric equation
expresses the cartesian coordinates
as separate functions in terms of polar coordinates
which
defines the location of each coordinate on the locus in terms of a chosen reference point (in this case the origin).
The parametric equations must satisfy the cartesian equation, that is, if you substitute (2) into the left hand side of
(1), you must find that the left hand side equates to . Complicated cartesian equations can be difficult to
differentiate which makes determining important information such as the location of critical coordinates or equations
of lines tangent to the cartesian equation challenging. In some cases, the associated parametric pair is easier to
differentiate and they can instead be used to determine equivalent expressions for the derivative. This Chapter will
derive alternative formulas for calculating
using the parametric equations instead.
pic a)
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4.1 Derivatives in the Cartesian case
It describes the direct relationship between the dependent variable
a cartesian relationship useful is
• The ability to determine the instantaneous rate of change in
and the independent variable
. What makes
( slope of the line tangent to the curve of )
as
• The coordinate where reaches an optimum (critical value) can be determined by solving
• The type (maximum, minimum or horizontal inflection) of critical coordinate can be determined by the sign
of
when it is evaluated at the critical coordinate.
Example
For the cartesian circle locus
, calculate
Calculating
Recall that
represents the gradient/slope of the line tangent to the curve of . From the above, the critical
coordinate is located at
which is when
. The corresponding coordinates (substitute
back into
) are at
as expected. The two critical coordinates are therefore
.
Calculating
The nature of the critical coordinates are determined by the second derivative test.
We see that
will result in a negative
which indicates a maximum critical coordinate and
minimum critical coordinate as here
is positive.
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a
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4.2 Derivatives in the parametric case
In a parametric pair,
are separately defined in terms of a reference parameter such as . When an object
is moving at a constant angular velocity radians/second, the radian measure can be expressed as the dynamic
parameter
. What makes parametric equations useful is
• Equation (2) and (3) can be expressed in terms of time as the dynamic (indicating movement) parametric
pair
and
with locus vector
• It will be shown in the next section how the time derivatives
of the parametric pair can be used to
calculate equivalent expressions to those of
. These versions are often simpler than attempting to
calculate
using the cartesian equation.
4.2.1 The first derivative equivalent
These are best derived using an example such as the above with parametric equivalent pair
. We will assume that
We aim to derive equivalent formula for
using the parametric pair
Step 1: Calculate the time derivatives
cartesian
and
and
instead.
for the given parametric pair and relate the expression
to the
in (4).
The First derivatives are
Note that from
with
and
,
and
. Substituting this into the above
expression,
as before.
Therefore, the time expression
Using the parametric equivalent
DR ERASMUS
is equivalent to the cartesian (in terms of
) expression
in determining the critical coordinate, we also evaluate where
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when
which is when
.
Note that
and therefore
as this will result in division by zero. To determine if these
critical time values agree with the same cartesian critical coordinates calculated for the cartesian equation, we must
substitute
back into the parametric pair:
You will find that the remaining (infinitely many) critical values will alternate between
that there are only two critical coordinates and that they are located at
.
To understand why
is compatible with
, let's simplify the expression
. This confirms
symbolically:
or rather
Examples
i) Given the parametric pair
, calculate an expression for
cartesian equivalent for the locus plotted by the parametric pair?
Calculate
. Can you formulate the
, the first derivative is then
In determining the cartesian equivalent, at least one of the expressions in the parametric pair must be suited to
expressing
in terms of either
or
. For this pair
is suited where
, we will find that the cartesian equivalent is the curve
. If we substitute this into
. Calculating
using the cartesian equation will lead to equivalent results.
ii) Calculate the slope/gradient of the line tangent to the cartesian curve at
are known where
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if only the parametric equations
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We require
:
We must calculate
:
at
. To calculate an equivalent value for
to substitute into the above, use
. The gradient of the tangent line at
iii) Calculate an expression for
this parametric pair?
Calculate
is therefore
.
from the given parametric pair. Can you determine the cartesian equivalent for
(quotient rule)
(quotient rule)
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In determining the cartesian equivalent, neither of the two parametric expressions can be used to express in terms
of or . Not compulsory: Let's try adding the square of the two parametric equations (this is inspired by the fact
that this will lead to
)
From the above expression,
can be expressed in terms of
Use any of the two parametric expressions and substitute
:
with
:
The cartesian equivalent is then
Not as elegant as the parametric pair!
iv) The x and y coordinates of a particle on a bicycle wheel change according to the parametric set
The curve traced is called a cycloid or 'Helen of Geometry' - the most beautiful curve in the world. Use the
parametric set to determine the location of the critical coordinates. Can you use this parametric pair to determine the
cartesian equivalent?
Using Matlab to draw this dynamic graph over the interval
is shown below:
syms t
y(t)=1-cos(t);
x(t)=t-sin(t);
fplot(x,y,[0,4*pi])
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Extending the plot range will add more curves with time.
This example is suited to expressing the cartesian equivalent as
can be expressed in terms of
:
This is a very complicated cartesian equation which is the reason the parametric set will be more useful when
determining
.
In calculating the critical coordinates for the cycloid using the parametric pair over the time range
Note that the numerator 1
The permitted
values are therefore
which excludes values
(infinitely many of these)
syms t
x(t)=t-sin(t); y(t)=1-cos(t);
dy=diff(y)
dy(t) =
dx=diff(x)
dx(t) =
dash=dy/dx
% y'
dash(t) =
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Paced Exercise (4A)
1. For each of the following graphs, calculate expressions for
equation.
1.1
using the parametric pair and the Cartesian
1.2
1.1 Asteroid: This shape is formed by tracing a point on a circle with radius
radius
while it is rolling inside a circle with
.
parametric pair:
Cartesian equation:
• You should find that
with
• Using the cartesian equation
1.2 Cardioid: This shape is formed by tracing a point on a circle (radius
circle with radius .
) while it is rolling on the outside of another
parametric pair:
Cartesian equation:
• You should find that
with
• Using the cartesian equation
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2. The Folium of Descartes (shown below) simulates a flower petal with different sized petals depending on the
value of .
parametric pair:
Cartesian equation:
• For the parametric equation you should find
• For the cartesian equation you should find that
3. The path of a projectile propelled at an inclination
radians to the horizontal axes with initial speed
and from
an initial height h above the horizontal axes subjected to gravity
is given by the parametric set
If a baseball is projected with initial values
, describe the motion of the ball (critical
coordinates) until it hits the ground again (
solutions below.
). Do your own calculations and compare your results to the Matlab
syms t
vo=90; h=1; theta=pi/4; g=9.8;
x(t)=vo*cos(theta)*t;
y(t)=-(1/2)*g*t^2+vo*sin(theta)*t+h;
t_ground=eval(poles(1/y)) % t values when y=0
t_ground = 2×1
13.0034
-0.0157
% the above shows that the ball will hit the ground after 13 seconds
% this defines the time range as being from t=0 till t=13
dx=diff(x)
dx(t) =
dy=diff(y)
dy(t) =
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dash=dy/dx
% this is y'
dash(t) =
Simpledash=vpa(dash,3)
% simplified version of dash
Simpledash(t) =
t_critic=eval(poles(1/dash)) % when is y'=0
t_critic = 6.4938
% no range is specified so poles argument contains only 1/y'
x_critic=eval(x(t_critic)) % insert t_critic into x
x_critic = 413.2653
y_critic=eval(y(t_critic))% insert t_critic into y
y_critic = 207.6327
% the corresponding critical (x,y) coordinate is at (413.27, 207.63)
fplot(x,y,[0,13]) % plot the cartesian equation over time range
hold on
plot(413.27,207.63,'*') % show critical (x,y) using *
hold off
xlabel('x')
ylabel('y')
Note that when using Matlab, the eval command is used to simplify an expression. The expression must not contain
unknowns (symbols) such at . It is not compulsory to use this command but every now and then Matlab presents a
solution in terms of a fraction instead of its associated decimal value. In these cases, precede the calculation with
the eval command.
If the expression does contain unknowns (symbols) and you would like to simplify it, use the command vpa. The
first entry into vpa is the expression you would like to simplify and the second entry the amount of significant digits it
should show in the output. In the above 3 significant digits are specified in the simplification of the dash expression:
You will notice that Matlab keeps more digits which is why your hand calculated solutions are slightly different.
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4. Determine where the locus described by the parametric pair
optimum in the time range
reaches an
.
• Calculate
• Express
. You should find it to be
• A critical (optimum) is situated at
values for which
, that is, for which
. You should find that a suitable zero in the range
which solves as
coordinate corresponding to
is
but
corresponds to
. Only the positive value is in our range. The critical
.
syms t
y(t)=sin(2*t)*cos(2*t);
x(t)=sin(2*t);
fplot(x,y,[0,pi/2]) % plot continuous (x,y) corresponding to all t in [0,pi/2]
hold on
plot(0.7,0.5,'*') % indicate the the maximum coordinate with a *
hold off
xlabel('x')
ylabel('y')
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5. In a projectile motion with
and parametric set
the values of the constants are
,
. Describe the motion of the ball (critical coordinates)
until it hits the ground again (
).
• See if you agree with the Matlab generated solutions
syms t
vo=30; h=0; theta=pi/3; g=9.8;
x(t)=vo*cos(theta)*t;
y(t)=-(1/2)*g*t^2+vo*sin(theta)*t+h;
t_ground=eval(poles(1/y)) % t values when y=0
t_ground =
5.3022
and
0
% the above shows that the ball will hit the ground after 5.3022 seconds
% this defines the time range as being from t=0 till t=5.3022
dx=diff(x)
dx(t) = 15
dy=diff(y)
dy(t) =
dash=dy/dx
% this is y'
dash(t) =
Simpledash=vpa(dash,3)
% simplified version of dash
Simpledash(t) =
t_critic=eval(poles(1/dash)) % when is y'=0
t_critic = 2.6511
% no range is specified so poles argument contains only 1/y'
x_critic=eval(x(t_critic)) % insert t_critic into x
x_critic = 39.7665
y_critic=eval(y(t_critic))% insert t_critic into y
y_critic = 34.4388
% the corresponding critical (x,y) coordinate is at (39.7665, 34.4388)
fplot(x,y,[0,5.3022]) % plot the cartesian equation over time range
hold on
plot(39.7665,34.4388,'*') % show critical (x,y) using *
hold off
xlabel('x')
ylabel('y')
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4.2.2 The second derivative equivalent
Continuing with the above example, use the first derivative
to the cartesian
in (5)
Start by first calculating the numerator
so before calculating
. If further simplification of
. In our case
We must then divide this expression by
Substituting
and relate the expression
is possible, it is advised you do
is the simplest form.
:
into the above:
as before.
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This proves that the time expression
is equivalent to the cartesian (in terms of
To confirm that the critical time values
respectively represent a maximum and minimum critical
coordinate, substitute them into the expression
When
) expression
:
(negative) indicating that
corresponding to the coordinate
, represents a
(positive) indicating that
corresponding to the coordinate
, represents
maximum.
When
a minimum.
To understand why
agrees with
, let's simplify the expression
symbolically:
or rather
Examples
i) Given the parametric pair
, calculate an expression for
.
From the previous section,
The numerator of the second derivative is
and the denominator is
which results in the
expression
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ii) The x and y coordinates of a cardioid changes according to the parametric set
Use the parametric set to determine if the critical coordinates determined in the previous subsection's example set
represent maximums or minimums.
It was determined that
with critical coordinates occurring at
The numerator of
requires the quotient rule
The second derivative is then:
Insert the permitted values
:
Using Matlab to do the calculations for you is shown below:
syms t
x(t)=t-sin(t);
y(t)=1-cos(t);
dy=diff(y)
dx=diff(x)
dx(t) =
dash=dy/dx
% y'
dash(t) =
tval=poles(1/dash,0,4*pi) % y'=0
tval =
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dash2(t)=diff(dash)/dx
% y''
dash2(t) =
maxmin=dash2(tval) % substitute tval into y''
maxmin =
Note that the poles command requires a fraction as its argument. This command calculates the zeros of the
denominator which is why we insert
as the denominator of the fraction 1/y'. Note also that you can specify the
range in which this command must calculate the zeros, in this case, between 0 and pi/4.
Paced Exercise (4B)
Determine the nature of the optimums calculated in Paced Exercise (4A) for the locus described by the parametric
pair
• From Paced Exercise (4A),
• We require
formula for
substituting
with critical coordinate corresponding to
at
.
to determine if this critical coordinate is a maximum or minimum. The numerator in the
is the time derivative of . First simplify the
expression (before you differentiate) by
. This should bring you to the alternative representation
. The time derivative is then
• You should find that
and that the coordinate
represents a maximum. See the asterix * in the graph below
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corresponding to
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4.3 Derivatives of the locus expression (vector)
Staying with the circle curve example, when movement along the curve/locus is implied, the position of the particle
moving along the curve can be represented by the displacement vector
).
The first derivative of
can be related to
but the second derivative of
of the vector equation are interpreted differently as discussed below.
Step 1: The time derivative of the vector
The velocity vector
(assuming
can't be related to
. The derivatives
is defined as the velocity vector
where
is shown in pic b). It is representable by a line tangent to the displacement curve with
gradient/slope
which is
. Recall that velocity is defined as the instantaneous gradient
of the line tangent to the displacement curve.
Step 2: The second derivative of
is defined as the acceleration vector
where
These are the components of the resultant acceleration experienced by the particle while moving around the rim of
the circle. It does not indicate if the critical coordinates are maximums and minimums as was the case for
in the
cartesian expression. The acceleration vector in this case points towards the origin as can be seen in pic c) below.
pic b)
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pic c)
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Example
The x and y coordinates of a particle on a bicycle wheel change according to the parametric set (cycloid)
Express the cycloid as a vector and calculate the displacement, velocity and acceleration vectors at
Displacement
Velocity
Acceleration
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Paced Exercise (4C)
Calculate expressions for the displacement, velocity and acceleration vectors at time
parametric equations are given by:
seconds if the
• You should find that
Displacement
Velocity
Acceleration
syms t
x(t)=t^2 - t^3; y(t)=5*t^2-t^3;
fplot(x,y,[0,4]) % plot continuous (x,y) corresponding to all t in [0,4]
hold on
plot(-4,12,'*') % (x,y) corresponding to t=2
hold off
xlabel('x')
ylabel('y')
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