Implicit circles

Implicit circles

Excerpt: Students are asked to find the gradient of three points on the graph of a circle, thinking about different ways to solve this problem. The specific problem is then generalised as a way of introducing implicit differentiation.

Problem

Find the gradient of the curve $x^2+y^2=4$ at the following three points.

What is the derivative, $\dfrac{dy}{dx}$, at a general point on the curve?

Next step

What is the derivative at a general point on this curve?

Can you use the same methods as above?

Things you might have done

There are different ways we could think about finding the gradient of a circle at a point.

Geometric approach

Differentiation

We could differentiate the equation of our curve to find the gradient at $x = \frac{1}{2}$.

The difficulty is that we do not have an equation in the form $y = f(x)$. Can we rearrange it?

$$\begin{align*} x^2+y^2 &= 4 \\ y^2 &= 4 - x^2 \\ y &= \pm\sqrt{4-x^2} \end{align*}$$

This final line causes problems as there are two options for the square root. How might we proceed from here?

Given that we have a specific point, $\left(\frac{1}{2}, -\frac{\sqrt{15}}{2}\right)$, we know we want the negative square root as our value of $y$ is negative. Therefore we could differentiate $y = -\sqrt{4-x^2}$ using the chain rule to get $\dfrac{dy}{dx} = -\frac{1}{2} (4-x^2)^{-1/2} \cdot(-2x) = \frac{x}{\sqrt{4-x^2}}.$ At $x = \frac{1}{2}$ this gives us a gradient of $\frac{1}{\sqrt{15}}$.

Notice that our expression for $\frac{dy}{dx}$ is undefined for $x=\pm 2$ and for $\big|x\big|>2$. But here we are only interested in $x = \frac{1}{2}$.

What is the derivative at a general point on the curve?

Again, we can tackle this in a number of different ways.

Geometric approach

We see that the gradient of the radius at $(x,y)$ is $\dfrac{y}{x}$.

As the tangent is perpendicular to the radius, the gradient of the curve at that point will be $\dfrac{-1}{y/x}$ or $-\dfrac{x}{y}$.

Would it matter if our point was in a different quadrant?

Rearranging the equation

If we rearrange the equation, we get $y = \pm\sqrt{4-x^2}$ as above. The two cases make this slightly awkward, but we can treat them separately.

If $y = +\sqrt{4-x^2}$ then using the chain rule we have $\frac{dy}{dx}=\frac{\frac{1}{2}}{\sqrt{4-x^2}}(-2x) = \frac{\frac{1}{2}(-2x)}{y} = -\frac{x}{y}.$

Whereas, if $y = -\sqrt{4-x^2}$ we have $\frac{dy}{dx}=-\frac{\frac{1}{2}}{\sqrt{4-x^2}}(-2x) = \frac{\frac{1}{2}(-2x)}{y} = -\frac{x}{y}.$

So for each case we get the same result as above.

Notice that something strange happens in between the two cases, at $y=0$.

Differentiating directly

What happens if instead we differentiate the original equation $x^2 + y^2 = 4$ directly?

Differentiating both sides we get $\dfrac{d}{dx}\left(x^2+y^2 \right) = \dfrac{d}{dx}\left(4\right)$ which is the same as $\dfrac{d}{dx}\left(x^2\right) + \dfrac{d}{dx}\left(y^2\right) = \dfrac{d}{dx}\left(4\right)$ and therefore $2x + \dfrac{d}{dx}\left(y^2\right) = 0.$

This in itself is not very helpful, but if we define $u=y^2$ we can then use the chain rule. $\dfrac{d}{dx}\left(y^2\right) = \frac{du}{dx} = \frac{du}{dy}\cdot\frac{dy}{dx} = 2y\frac{dy}{dx}$ This is useful because now we have an expression involving the gradient of the curve.

This works because, even though we could not write $y$ as a function of $x$, if you look at just the part of the curve near a given point, $y$ and $u$ are functions of $x$ and the gradients do have sensible meanings.

We can differentiate the $x^2$ and the $4$, but how do we differentiate the $y^2$ term with respect to $x$? Although we can't write $x^2 + y^2 = 4$ explicitly in the form $y = f(x)$, $y$ is still related to $x$. This is what is called an . We can think of $y$ as a function of $x$ that is unknown. That means the $y^2$ term can be thought of as some function of $x$ squared, i.e. $[f(x)]^2$. This is a composite function in $x$ that can be differentiated by the chain rule. Differentiating $[f(x)]^2$ using the chain rule gives us $2f(x)\cdot f'(x)$. If we use the notation of $y^2$ we would write the derivative as $2y \cdot \dfrac{dy}{dx}$. Can you see that these two different ways of writing the derivative are the same?

What we have done is called implicit differentiation and means that when we differentiate $x^2 + y^2 = 4$ we get

$$2x + 2y\dfrac{dy}{dx} = 0.$$

We can now rearrange to find $\dfrac{dy}{dx} = -\dfrac{x}{y}$, the same as we had from the other approaches.

You might notice that the gradient function is in terms of both $x$ and $y$. This makes sense because for each value of $x$, there may be two points on the curve, each of which has a different $y$ and different gradient.

You might check this result by substituting the coordinates given earlier to make sure we get the same gradients.

What happens when $y = 0$? What does this mean geometrically on the circle?

In this case, we had a choice of three methods, but for many other curves the first two approaches do not work and implicit differentiation becomes a valuable tool.

The first approach above relied on the geometry of the circle. It used the fact that a tangent to a circle is at right-angles to the radius. Is this true for an ellipse? Imagine your ellipse as a circle that has been stretched in one direction. What happens to the angle between the tangent and the radius?

Here we will differentiate implicitly.

$$\begin{align*} \dfrac{d}{dx}\left(9x^2 + 16y^2 \right) &= \dfrac{d}{dx}(144) \\ 18x + 32y\frac{dy}{dx} &= 0 \end{align*}$$

We can rearrange to find the gradient.

$$\begin{align*} \dfrac{dy}{dx} &= -\dfrac{18x}{32y} \\ &= -\dfrac{9x}{16y} \end{align*}$$