If $dy/dx$ is inversely proportional to $x^2$, can we find $y$?

If $dy/dx$ is inversely proportional to $x^2$, can we find $y$?

Question

1. Show that the curve $y=x(5-x)^4$ has a turning point where $x=5$. Determine whether this turning point is a maximum or a minimum.

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1. It is given that that $\dfrac{dy}{dx}$ is inversely proportional to $x^2$ and that $y$ and $\dfrac{dy}{dx}$ are each equal to $1$ when $x=2$. Express $y$ in terms of $x$.

Suggestion

How can we find the turning points of a function?

What methods do we know for determining the nature of a turning point?

Which method works best in this case?

What does it mean for $a$ to be inversely proportional to $b$? Can you write this mathematically?

Can we now replace $a$ with $\dfrac{dy}{dx}$, and $b$ with $x^2$, and use the information in the question?

Solution

Using the Product Rule, we have that

$$\begin{align*} \frac{dy}{dx}&=(5-x)^4-4x(5-x)^3\\ &=(5-x)^3[5-x-4x] \\ &=5(5-x)^3(1-x). \end{align*}$$

We see that $\dfrac{dy}{dx}$ vanishes when $x=5$, therefore $(5,0)$ is a turning point.

We can see from the expression for $\dfrac{dy}{dx}$ that $\dfrac{d^2y}{dx^2}$ will also vanish at $(5,0)$, so calculating this won’t help decide if we have a maximum or a minimum.

Let’s instead look at the expression for $y$.

We know that when $x=5$, $y=0$.

When $x$ is slightly smaller than $5$, $y$ is positive, and the same is true when $x$ is slightly larger than $5$, as $x$ and $(5-x)^4$ are both positive.

Therefore $(5,0)$ is a local minimum for $y$.

If $\dfrac{dy}{dx}$ is inversely proportional to $x^2$, then we can write $\frac{dy}{dx}=\frac{c}{x^2}.$ Since $\dfrac{dy}{dx}=1$ when $x=2$, we see that $c=4$, and so $\frac{dy}{dx}=\frac{4}{x^2}.$ We now integrate this, and see that $y=-\frac{4}{x}+d.$ Using that $y=1$ when $x=2$, we see that $d=3$.

Therefore $y=3-\dfrac{4}{x}$.