Question

Find the value of \[\dfrac{dy}{dx}\], if it’s given that $x=a{{t}^{2}}$ and $y=2at$.

Answer 1

Hint: We need to find the respective differentiations. We have been given the value of $x$ and $y$ as a function of $t$. So, we find the differentiation of both $x$ and $y$ with respect to $t$ to get the value of \[\dfrac{dx}{dt}\] and \[\dfrac{dy}{dt}\] respectively. Then we use both values of \[\dfrac{dx}{dt}\] and \[\dfrac{dy}{dt}\] to get the value of \[\dfrac{dy}{dx}\]. In that case we are actually going to use the values of \[\dfrac{dy}{dt}\] and \[\dfrac{dt}{dx}\].

Complete step-by-step answer:
We have been given the value of $x$ and $y$ as a function of $t$.
Let, $x=f(t)=a{{t}^{2}}$ and $y=g(t)=2at$.
First, we find the value of \[\dfrac{dx}{dt}\] by differentiating $x$ with respect to $t$.
So, we take both side differentiation.
$\begin{align}
  & x=a{{t}^{2}} \\
 & \Rightarrow \dfrac{d}{dt}(x)=\dfrac{d}{dt}(a{{t}^{2}}) \\
\end{align}$
We know that $\dfrac{d}{dt}({{t}^{n}})=n{{t}^{n-1}}[n\in \mathbb{N}]$
Here, we place $n=2$.
So, $\dfrac{dx}{dt}=\dfrac{d}{dt}(a{{t}^{2}})=a\dfrac{d}{dt}({{t}^{2}})=a(2t)=2at$
‘a’ being a constant it comes out of the differentiation as it is.
Now, we find out the value of \[\dfrac{dy}{dt}\] by differentiating $y$ with respect to $t$.
So, we take both side differentiation.
$\begin{align}
  & y=2at \\
 & \Rightarrow \dfrac{d}{dt}(y)=\dfrac{d}{dt}(2at) \\
\end{align}$
We know that $\dfrac{d}{dt}({{t}^{n}})=n{{t}^{n-1}}[n\in \mathbb{N}]$
Here, we place $n=1$.
So, $\dfrac{dy}{dt}=\dfrac{d}{dt}(2at)=2a\dfrac{d}{dt}(t)=2a(1)=2a$
‘2a’ being a constant it comes out of the differentiation as it is.
Now, we use the value of \[\dfrac{dx}{dt}\] and \[\dfrac{dy}{dt}\] to get the value of \[\dfrac{dy}{dx}\].
We can express \[\dfrac{dy}{dx}\] as a multiplication of \[\dfrac{dy}{dt}\] and \[\dfrac{dt}{dx}\].
So, $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$ ...(i)
We need the value of \[\dfrac{dt}{dx}\] which is the inverse value of \[\dfrac{dx}{dt}\].
So, \[\dfrac{dt}{dx}=\dfrac{1}{\dfrac{dx}{dt}}=\dfrac{1}{2at}\]
Now, we place the values in equation (i).
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}=2a\times \dfrac{1}{2at}=\dfrac{1}{t}$.
Thus, the value of \[\dfrac{dy}{dx}\] is $\dfrac{1}{t}$.

Note: We need to remember that when we are finding the value of \[\dfrac{dy}{dx}\] as a multiplication of \[\dfrac{dy}{dt}\] and \[\dfrac{dt}{dx}\], we are not cancelling out $dt$ to get that value. It’s not possible to cancel out a single differential unit. It’s just a notion of expression. We are finding the respective differentiation using chain rule.
Also, we can find the value of $y$ as a function $x$ at the start of the solution by eliminating t.
So, $y=2at\,\,\,\Rightarrow t=\dfrac{y}{2a}$
Now, $x=a{{t}^{2}}=a{{\left( \dfrac{y}{2a} \right)}^{2}}=\dfrac{{{y}^{2}}}{4a}$
So, we get $y$as a function $x$ where ${{y}^{2}}=4ax$.
So, we differentiate to get
$\begin{align}
  & 2y\dfrac{dy}{dx}=4a \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}=\dfrac{2a}{y}=\dfrac{2a}{2at}=\dfrac{1}{t} \\
\end{align}$
This is another way to solve the problem.