QUESTION

# If $x = ct$ and $y = \dfrac{c}{t}$ , find $\dfrac{{dy}}{{dx}}$ at $t = 2$ . A. $\dfrac{1}{4}$  B. $4$  C. $- \dfrac{1}{4}$  D. $0$

Hint: In order to solve such a problem, find some common variable in the terms and differentiate both the given terms with respect to the common variable and finally divide the differentiated terms. And then put the value given.

Given that $x = ct$ and $y = \dfrac{c}{t}$
To find $\dfrac{{dy}}{{dx}}$
As t is the common variable in x and y term.
So we will first find $\dfrac{{dx}}{{dt}}{\text{ and }}\dfrac{{dy}}{{dt}}$
So for $x = ct$
Differentiating the equation with respect to $t$
$\dfrac{{dx}}{{dt}} = c$ ---- (1)
And for $y = \dfrac{c}{t}$
Again differentiating the equation with respect to $t$
$\dfrac{{dy}}{{dt}} = \dfrac{{ - c}}{{{t^2}}}$ --- (2)
Now diving equation (2) by equation (1) we get
$\Rightarrow \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{\dfrac{{ - c}}{{{t^2}}}}}{c} \\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{t^2}}} \\ \left[ {\because t = 2\left( {{\text{given}}} \right)} \right] \\ \therefore \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{2^2}}} = \dfrac{{ - 1}}{4} \\$
Hence the value of the above problem is $\dfrac{-1}{4}$ and option C is the correct option.

Note: In order to find the derivative in such a question when there are 2 equations, try to derive both the equations by the dependent variable common in both the equations and then try to find the link with the problem by dividing or multiplying the derivative. Never try to put the value of before differentiating as in that case the answer will be wrong.