21.17 - Derivatives of Parametric Functions

In the previous section, we saw logarithmic differentiation. In this section, we will see parametric functions.

We have learned about projectile motion in our physics classes. An example in projectile motion, can be used to learn about parametric functions. It can be written in 6 steps:
1. An object is projected with a velocity of 30 ms⁻¹ at an angle of 60^o with the horizontal. Trace the path of the object. [g = 9.81 ms⁻²]

2. We know that:
• The point of projection is considered as the origin.
• The x-coordinate is the horizontal distance of the object from the origin.
• The y-coordinate is the vertical distance of the object from the origin.
• At the instant when the object is projected, the reading in the stop-watch will be: t = 0.

3. We know how to find the x and y coordinates:
• The x-coordinate is given by: $x = \rm{v_0 \cos \theta_0 t}$
        ✰ Eq.4.13 in section 4.10 of physics notes. 
• The y-coordinate is given by: $y = \rm{v_0 \sin \theta_0 t \,-\,\frac{1}{2}gt^2}$
        ✰ Eq.4.14 in section 4.10 of physics notes. 
• Where,
    ♦ v₀ is the initial velocity
    ♦ 𝜃₀ is the angle of projection.
    ♦ t is the time at which we want the coordinates.

4. We see that, both x and y coordinates are given in terms of t.
    ♦ So x is a function of a third variable ‘t’.
    ♦ y is also a function of the third variable ‘t’.
• We can write: x = f(t) and y = g(t).
• Such functions are called parametric functions.
    ♦ The third variable ‘t’ is the parameter.

5. Let us plot the various x and y positions.
• The table below gives the x and y coordinates:

(In this table, the last entry of the "time column" is obtained by finding the “time duration of the flight”) 
• The coordinates from the table, are plotted in the graph in fig.21.22 below:

Fig.21.22

• The pink smooth curve is drawn through the plotted points. Two sample plotted points are shown in the fig.
• The pink curve represents the path of the projected object. We obtained this path using a parametric function.

6. When we learned about projectile motion in the physics classes, we were not aware about parametric functions. There we plotted the pink curve by eliminating ‘t’ from the expressions for x and y. Such an elimination may not be possible in some complicated problems. For those complicated problems, parametric functions will help us to obtain the required results easily.

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Now we have a basic idea about parametric functions. So we will learn the method to find the derivatives of such functions. It can be written in steps:
1. Consider the parametric equation:
x = f(t), y = g(t)

2. Assume that, by eliminating t, a direct relation F can be obtained between x and y. We can write it as: y = F(x).

3. Differentiating both sides w.r.t 't', we get:
$\rm{\frac{dy}{dt} ~=~\frac{d}{dt}(F(x))}$    

4. In the L.H.S. of (3), y is equal to g(t). So 't' is the variable. We can easily calculate $\rm{\frac{dy}{dt}}$

5. In the R.H.S. of (3), F(x) does not have t. This is because, F(x) is the direct relation between x and y. It is obtained by eliminating 't'.
• So we cannot find $\rm{\frac{d}{dt}(F(x))}$ directly. We need to use the chain rule. We get:
$\rm{\frac{d}{dt}(F(x))}~=~\frac{dF}{dx}.\frac{dx}{dt}$

6. So the equation in (3) becomes:
$\rm{\frac{dy}{dt} ~=~\frac{dF}{dx}.\frac{dx}{dt}}$
• Consider the R.H.S. of the above equation.
• From (2) we see that, F is same as y. So this equation becomes:
$\rm{\frac{dy}{dt} ~=~\frac{dy}{dx}.\frac{dx}{dt}}$
• From this we get:
$\rm{\frac{dy}{dx}~=~\frac{\frac{dy}{dt}}{\frac{dx}{dt}}}$

7. We can use the result in (6) above, to find $\rm{\frac{dy}{dx}}$, when x and y are given in terms of a third variable 't'.

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Now we will see a solved example.

Solved example 21.55
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~x = a \cos \theta,~y = a \sin \theta}$.
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{a \cos \theta}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dx}{d \theta}}    & {~=~}    &{-a \sin \theta}    \\
{~\color{magenta}    3    }    &{{}}    &{y}    & {~=~}    &{a \sin \theta}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{dy}{d \theta}}    & {~=~}    &{a \cos \theta}    \\
{~\color{magenta}    5    }    &{\text{Therefore}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{a \cos \theta}{-a \sin \theta}}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{-\cot \theta}    \\
\end{array}$

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In the next section, we will see a few more solved examples.

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