Question

How do you find $\dfrac{dy}{dx}$ given $y={{t}^{3}}-3t$ and $x={{t}^{2}}$?

Answer 1

To solve the above question we will use the concept of parametric differentiation from calculus. We can see from the question that the parametric coefficient is ‘t’. So, we will first find the first derivative of x and y with respect to ‘t’ and then we will substitute the value of $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ in $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$ to get $\dfrac{dy}{dx}$.

Complete step-by-step solution:
We will use the concept of parametric differentiation from calculus to solve the above question. A parametric derivative is a derivative of a dependent variable with respect to another dependent variable that is taken when both variables depend on an independent third variable, usually thought of as "time" (that is, when the dependent variables are x and y and are given by parametric equations in t ).
Since, we have x and y in terms of ‘t’ .
From question we have:
$x={{t}^{2}}$ and $y={{t}^{3}}-3t$.
And, we know that first derivative for the above parametric equation is given by:
$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$
So, first we will find the first derivative of y with respect to t:
So, first derivative of equation $y={{t}^{3}}-3t$ with respect to ‘t’ is given as:
$\dfrac{dy}{dt}=\dfrac{d\left( {{t}^{3}}-3t \right)}{dx}$
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d\left( {{t}^{3}} \right)}{dt}-\dfrac{d\left( 3t \right)}{dt}$, using the sum rule of differentiation. $\dfrac{d\left( A+B \right)}{dx}=\dfrac{dA}{dx}+\dfrac{dB}{dx}$
$\Rightarrow \dfrac{dy}{dt}=3{{t}^{3-1}}-3\times 1$, using the power rule of differentiation $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
$\Rightarrow \dfrac{dy}{dt}=3{{t}^{2}}-3--(1)$
Now, we will find the first derivative of equation $x={{t}^{2}}$ with respect to t. So, we will get:
$\dfrac{dx}{dt}=\dfrac{d\left( {{t}^{2}} \right)}{dt}$
$\Rightarrow \dfrac{dx}{dt}=2t---(2)$
Now, we know that $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$, so after putting the value of $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ from equation (1) and (2), we will get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{3{{t}^{2}}-3}{2t}$
Hence, the first derivative $\dfrac{dy}{dx}$ is equal to $\dfrac{dy}{dx}=\dfrac{3{{t}^{2}}-3}{2t}$.
This is our required solution.

Note: When we are given the value of x and y with respect to “t” i.e. the equation is parametric then to find the first derivative of y with respect to x we always first find the derivative of x and y with respect to t and then we put the value of both the derivatives in the equation $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$.