In the previous section, we saw parametric functions and their differentiation. We saw a solved example also. In this section, we will see a few more solved examples.
Solved example 21.56
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~x = a t^2,~y = 2at}$.
Solution:
Solved example 21.57
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~x = a(\theta + \sin \theta) ,~y = a(1 - \cos \theta)}$.
Solution:
◼ Remarks:
• 7 (magenta color):
♦ In the numerator, we use the identity 15.
♦ In the denominator, we use the identity 14.
(List of identities can be seen here)
Solved example 21.58
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~~x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}}$.
Solution:
1. Let us make the following set of substitutions:
$\rm{x^{\frac{1}{3}} = u,~~y^{\frac{1}{3}} = v,~~ a^{\frac{1}{3}} = k}$.
2. So the given expression becomes: $\rm{u^2 + v^2 = k^2}$
3. Now we make another set of substitutions:
u = k cos 𝜃, v = k sin 𝜃.
4. So the expression in (2) becomes:
$\rm{(k \cos \theta)^2 + (k \sin \theta)^2 = k^2}$, which is true.
5. So from (1) and (3), we get:
• $\rm{x^{\frac{1}{3}} = u = k \cos \theta}$
⇒ $\rm{x = k^3 \cos^3 \theta = a \cos^3 \theta}$
• $\rm{y^{\frac{1}{3}} = v = k \sin \theta}$
⇒ $\rm{y = k^3 \sin^3 \theta = a \sin^3 \theta}$
6. Thus we have a parametric equation:
$\rm{x = a \cos^3 \theta, ~ y = a \sin^3 \theta}$
• Now we can find $\rm{\frac{dy}{dx}}$ as follows:
Solved example 21.59
$\rm{\text{Find}~\frac{dy}{dx}~\text{if}~~x = 10(t - \sin t),~y = 12(1 - \cos t)}$.
Solution:
The link below gives a few more solved examples:
In the next section, we will see derivatives of
functions in parametric forms.
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