Can we find a tangent to $y=1/(1+x^2)$?

Can we find a tangent to $y=1/(1+x^2)$?

Question

Give a sketch of the curve $y=\dfrac{1}{1+x^2}$, for $x \geq 0$.

Find the equation of the line that intersects the curve at $x=0$ and is tangent to the curve at some point with $x>0$. Prove that there are no further intersections between the line and the curve. Draw the line on your sketch.

By considering the area under the curve for $0\le x\le 1$, show that $\pi>3$.

Show also, by considering the volume formed by rotating the curve about the $y$ axis, that $\ln 2>2/3$.

$\left[\text{Note:} \displaystyle\int_0^1\dfrac{1}{1+x^2} \:dx = \dfrac{\pi}{4}.\right]$

Solution

In order to sketch the curve, we notice:

1. We have that $y\ge0$ for all $x\ge 0$.

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1. At $x=0$, we have $y=1$, so it passes through the point $(0,1)$.

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1. As $x$ gets large, $y$ gets small. This means that $y\to 0$ as $x\to \infty$.

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1. Taking the derivative, we find that $\frac{dy}{dx}=-\frac{2x}{(1+x^2)^2},$ so we have $\dfrac{dy}{dx}=0$ at $x=0$, so $(0,1)$ is a stationary point of the curve.

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1. For $x>0$, we have that $\dfrac{dy}{dx}<0$, so the curve is decreasing as $x$ increases.

Using these facts, we can now sketch the curve.

Let us draw the sketch of the curve with the tangent immediately.

The straight line we want intersects the curve at $(0,1)$, that is, its $y$-intercept is $1$. So it has an equation of the form $y=mx+1$.

Solving $y=mx+1$ simultaneously with $y=\dfrac{1}{1+x^2}$ gives a cubic equation, so there can be at most three crossing points (one of which will be repeated if the line is tangent to the curve).

Approach 1: Using the discriminant

Solving these equations simultaneously, we have

$$mx+1=\frac{1}{1+x^2},$$

so

$$(mx+1)(1+x^2)=1.$$

Expanding the brackets gives

$$mx^3+x^2+mx+1=1$$

so

$$\begin{equation} x(mx^2+x+m)=0. \end{equation}$$

This has the root $x=0$ (as the line intersects the curve at $x=0$) and the quadratic $mx^2+x+m=0$ has a double root if the discriminant of the quadratic is zero, so $1-4m^2=0,$ yielding $m=\pm\dfrac{1}{2}$. We must have $m=-\dfrac{1}{2}$ and so $x=1$ at the point of tangency. The line therefore has equation $y=1-\dfrac{1}{2}x$.

Approach 2: Using calculus

The tangent line will be of the form $y = mx +1$. Let the point where the line touches the curve be $(p,q)$.

We know therefore that

1. $(p,q)$ is on the curve

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1. $(p,q)$ is on the line

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1. the gradient of the curve equals that of the line at $(p,q)$.

This tells us that $q = \dfrac{1}{1+p^2}$, $q=mp+1$, and $\dfrac{-2p}{(1+p^2)^2} = m$ (using the gradient function we derived earlier).

Thus $\dfrac{1}{1+p^2}=\dfrac{-2p}{(1+p^2)^2}p+1,$ which yields $p=1$ (since $p > 0$), and hence $q = \dfrac{1}{2}, m = -\dfrac{1}{2}$.

The line therefore has equation $y=1-\dfrac{1}{2}x$.

By considering the area under the curve for $0\le x\le 1$, show that $\pi>3$.

$\left[\text{Note:} \displaystyle\int_0^1\dfrac{1}{1+x^2} \:dx = \dfrac{\pi}{4}.\right]$

The area under the curve for $0\le x\le 1$ is strictly greater than that of the trapezium bounded by the lines $x=0$, $x=1$, $y=0$ and $y=1-\dfrac{1}{2}x$.

So $\int_0^1 \frac{1}{1+x^2} dx> \frac{1}{2}\left(1+\frac{1}{2}\right).$ By the note given in the question, we therefore know $\frac{\pi}{4}>\frac{3}{4},$ that is, $\pi>3.$

The equation of the curve $y = \dfrac{1}{1+x^2}$ can be rearranged to the form $x^2=\dfrac{1}{y}-1$.

When $x=0$, $y=1$ and when $x=1$, $y=1/2$.

The volume of revolution of the curve about the

$y$

axis is therefore given by

$$\begin{align*} \pi\int x^2 dy &= \pi\int_\frac{1}{2}^1 \frac{1}{y}-1 dy\\ &= \pi \left[\ln y-y\right]^1_{1/2}\\ &= \pi \left[(\ln 1-1)-\left(\ln \frac{1}{2}-\frac{1}{2}\right)\right]\\ &= \pi (\ln 2-\frac{1}{2}) \end{align*}$$

The equation of the line $y=1-\dfrac{x}{2}$ can be rearranged to the form $x=2-2y$.

The volume of the shape given by rotating the line

$y=1-\dfrac{x}{2}$

about the

$y$

axis is

$$\begin{align*} \pi\int x^2 dy &= \pi\int_\frac{1}{2}^1 (2-2y)^2 dy\\ &= \pi\int_\frac{1}{2}^1 4-8y+4y^2 dy\\ &= \pi \left[4y-4y^2+\frac{4}{3}y^3\right]^1_\frac{1}{2}\\ &= \pi \left[\left(4-4+\frac{4}{3}\right)-\left(2-1+\frac{1}{6}\right)\right]\\ &= \frac{1}{6}\pi \end{align*}$$

Since volume of the shape of the rotated curve is greater than that of the shape of the rotated line, we have $\pi\left(\ln 2 - \frac{1}{2}\right) > \frac{1}{6}\pi,$ that is, $\ln 2>\frac{2}{3}.$