Question

If $y$varies inversely with $x$ and $y=6$ when $x=18$, how do you find $y$ when $x=8$?

Answer 1

Hint: For these kinds of questions, we need to make use of the small concept of differential equations. We are told that $y$ varies inversely with $x$ . But not $y$ varies the same as the variation in $x$ . We should be able to infer that the rate of change of $y$is proportional to rate of change of $x$ but not equal to the rate of change of $x$. So let us write down an equation to mathematically show that and use integration to find out the exact equation relating both $x,y$.

Complete step by step solution:
The rate of change of $y$which can be represented as $\dfrac{dy}{dx}$ is inversely proportional to the rate of change of $x$ .
Here, we should keep it mind that we are differentiating with respect to $x$.
So now let's put down the equation which represents that the rate of change of $y$is proportional to rate of change of $x$.
It is as follows :
$\Rightarrow y\propto \dfrac{1}{x}$
Upon differentiating, we get the following :
$\Rightarrow \dfrac{dy}{dx}$$\propto $ $-\dfrac{1}{{{x}^{2}}}$
Now, we all know that when we try to remove the proportionality symbol, a constant is to be multiplied since the variables always maintain a constant ratio to each other.
So, let us remove the proportionality symbol and multiply any constant, let us say $k$.
Upon doing so, we get the following :
$\Rightarrow \dfrac{dy}{dx}$$=$ $-\dfrac{k}{{{x}^{2}}}$
Upon integrating, we get the following :
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{-k}{{{x}^{2}}} \\
 & \Rightarrow \int{\dfrac{dy}{dx}=\int{\dfrac{-k}{{{x}^{2}}}}} \\
 & \Rightarrow y=kx \\
\end{align}\]
From the question, we have a base condition which is $y=6$ when$x=18$ .
So let us use this piece of information to find the value of $k$.
$\begin{align}
  & \Rightarrow y=kx \\
 & \Rightarrow 6=18k \\
 & \Rightarrow 1=3k \\
 & \Rightarrow k=\dfrac{1}{3} \\
\end{align}$
So the exact equation relating both $x,y$ is \[y=\dfrac{x}{3}\].
Now, let us find out the value of $y$ when the value of $x=8$.
Let us plug in $x=8$ in the equation $y=\dfrac{x}{3}$ to find $y$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow y=\dfrac{x}{3} \\
 & \Rightarrow y=\dfrac{8}{3} \\
 & \Rightarrow y=\dfrac{8}{3} \\
\end{align}$
$\therefore $ Hence, the value of $y$ when $x=8$ is $\dfrac{8}{3}$.

Note: After we integrated our differential expression, we didn’t add an extra constant, $c$, which we usually do. If we did add that constant as well, then we would have two unknown variables in our expression namely $k,c$ but we have only one condition. So we will not get the absolute values of either of the constants. We would only get a relation between them. If that happens, we can’t form an exact equation between $x,y$ and fail to find out the value $y$ when $x=8$.