Question

If $ y $ varies directly with $ x $ and $ y=-4.5 $ when $ x=13.5 $ , how do you find $ x $ when $ y=24 $ ?

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 directly 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 proportional to the rate of change of $ x $ which can be represented as $ \dfrac{dx}{dx} $ which is nothing but $ 1 $ .
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 \dfrac{dy}{dx} $ $ \propto $ $ 1 $
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} $ $ \propto $ $ 1 $
\[\Rightarrow \dfrac{dy}{dx}=k\]
Now let us integrate on both sides.
Upon integrating on both sides, we get the following :
 $ \Rightarrow \dfrac{dy}{dx} $ $ \alpha $ $ 1 $
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=k \\
 & \Rightarrow \int{\dfrac{dy}{dx}=\int{k}} \\
 & \Rightarrow y=kx \\
\end{align}\]
From the condition, we have a base condition which is $ y=13.5 $ when $ x=-4.5 $ .
So let us use this piece of information to find the value of $ k $ .
 $ \begin{align}
  & \Rightarrow y=kx \\
 & \Rightarrow -4.5=13.5k \\
 & \Rightarrow -45=135k \\
 & \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 $ x $ when the value of $ y=24 $ .
Let us plug in $ y=24 $ in the equation $ y=\dfrac{-x}{3} $ to find $ x $ .
Upon doing so, we get the following :
 $ \begin{align}
  & \Rightarrow y=\dfrac{-x}{3} \\
 & \Rightarrow 24=\dfrac{-x}{3} \\
 & \Rightarrow x=-72 \\
\end{align} $
 $ \therefore $ Hence, the value of $ x $ when $ y=24 $ is $ -72 $ .

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 $ x $ when $ y=24 $ .