Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 102

Illustrate the parabolas $y=x^2$ and $y = x^2 - 2x + 2$ and state if there is a line that is tangent to both curves.
If so, find its equation. If not, why not?




For $\displaystyle y = x^2$,
$\displaystyle\qquad \frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x \quad \Longleftarrow \text{Equation 1}$
For $\displaystyle y = x^2-2x +2$,
$\displaystyle\qquad \frac{dy}{dx} = \frac{d}{dx}(x^2) - 2 \frac{dy}{dx}(x) = \frac{d}{dx}(2) = 2x - 2 \quad \Longleftarrow \text{Equation 2}$

Let $y = ax + b$ be the equation of the line that is tangent to $y = x^2$ at $(x_1,x_1^2)$
and $y = x^3 - 2x + 2$ at $(x_2, x_2^2 - 2x_2 + 2)$.

By using the equation of the line and Equations 1 and 2 we get

$\begin{equation} \begin{aligned} ax+1 + b &= x_1^2 && \Longleftarrow \text{Equation 3}\\ \\ ax_2 + b &= x_2^2 - 2x_2 + 2 && \Longleftarrow \text{Equation 4} \end{aligned} \end{equation}$

Also, recall that the slope $a$ is equal to the derivative of the curve. So,

$\begin{equation} \begin{aligned} a & = 2x_1 &;&& x_1 &= \frac{a}{2}\\ \\ a &= 2x_2-2 &;&& x_2 &= \frac{a}{2} +1 \end{aligned} \end{equation}$


Using these equations together with Equations 3 and 4 we get
From Equation 3:

$\begin{equation} \begin{aligned} a \left(\frac{a}{2}\right) + b & = \left( \frac{a}{2} \right)^2\\ \\ \frac{a^2}{2}+b &= \frac{a^2}{4}\\ \\ b & = \frac{a^2}{4} - \frac{a^2}{2}\\ \\ b &= - \frac{a^2}{4} \end{aligned} \end{equation}$

From Equation 4:

$\begin{equation} \begin{aligned} a\left(\frac{a}{2}+1\right) + b & = \left( \frac{a}{2} + 1 \right)^2 - 2 \left( \frac{a}{2} +1 \right) +1\\ \\ \frac{a^2}{2} + a + b &= \frac{a^2}{4} + \cancel{a} + 1 - \cancel{a} - \cancel{2} + \cancel{2}\\ \\ b &= \frac{a^2}{4} - \frac{a^2}{2} - a + 1\\ \\ b &= \frac{-a^2}{4} - a +1 \end{aligned} \end{equation}$


Solving for $a$

$\begin{equation} \begin{aligned} \cancel{\frac{-a^2}{4}} &= \cancel{\frac{-a^2}{4}} - a + 1\\ \\ a & = 1 \end{aligned} \end{equation}$


Solving for $b$

$\begin{equation} \begin{aligned} b &= - \frac{a^2}{4} = -\frac{(1)^2}{4}\\ b &= - \frac{1}{4} \end{aligned} \end{equation}$

Plugging the values of $a$ and $b$ to the equation of the line we get

$\begin{equation} \begin{aligned} y &= ax + b\\ y &= (1)x + \left( - \frac{1}{4} \right)\\ y &= x - \frac{1}{4} \end{aligned} \end{equation}$

Therefore, the equation of the line that is tangent to both curve is $\displaystyle y = x - \frac{1}{4}$