Question

Find $\dfrac{{dy}}{{dx}}$ ,
If $\dfrac{{dx}}{{dt}} = ap\cos pt,\,\,\dfrac{{dy}}{{dt}} = - bp\sin pt$

Answer 1

Hint : Here, derivative of two parametric equations are given. To find a solution to the required problem or required value of $\dfrac{{dy}}{{dx}}$ from them we divide derivatives of parametric equations having y variables with derivatives of parametric equations having x variables.

Complete step-by-step answer :
Given, derivative of two parametric equations. Which are as follows:
$\dfrac{{dx}}{{dt}} = ap\cos pt\,\,and\,\,\,\dfrac{{dy}}{{dt}} = - bp\sin pt$
To find $\dfrac{{dy}}{{dx}}$ . We divide two parametric equations. As, we required $\dfrac{{dy}}{{dx}}$ as an answer. We divide derivatives of parametric equations having y variables with derivatives of parametric equations having x variables.
Which implies we have

$$\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{ - bp\sin pt}}{{ap\cos pt}} \\\ \Rightarrow \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} = \dfrac{{ - b\sin pt}}{{a\cos pt}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - b\sin pt}}{{a\cos pt}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{b}{a}\tan pt \;$$

Hence, from above we see that the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{{ - b}}{a}\tan pt$ .

Note : To find the value of $\dfrac{{dy}}{{dx}}$ . We first see that what function is given in the problem if there is a function y given in terms of ‘x’ then we can find the value of $\dfrac{{dy}}{{dx}}$ directly. But, in case when derivative of two parametric equations are given then on dividing derivative of parametric equation having y variable with another equation we can find the value of $\dfrac{{dy}}{{dx}}$ .