Chapter 3.1, Problem 74E

To determine

To sketch: The parabolas $y=x^2$ and $y=x^2-2x+2$ and check whether there is a tangent line to both the curves or not.

Explanation of Solution

Derivative rules:

(1) Power Rule: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

(2) Constant multiple rule: $\frac{d}{dx}\left(cf\right)=c\frac{d}{dx}\left(f\right)$

(3) Sum rule: $\frac{d}{dx}\left(f+g\right)=\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$

(4) Difference rule: $\frac{d}{dx}\left(f-g\right)=\frac{d}{dx}\left(f\right)+-\frac{d}{dx}\left(g\right)$

Result used:

The equation of the tangent line at $\left(x_1,y_1\right)$ is, $y-y_1=m\left(x-x_1\right)$. (1)

where, m is the slope of the tangent line at $\left(x_1,y_1\right)$ and ${m=\frac{dy}{dx}|}_{x=x_1}$.

Graph:

The graph of two parabolas $y=x^2$ and $y=x^2-2x+2$ is shown below in Figure 1.

From Figure 1, it is observed that there may be a line that is tangent to both the parabolas.

It is required to find the equation of the tangent line to the parabolas.

Calculation:

Consider the parabolas $y=x^2$ and $y=x^2-2x+2$.

Choose the point P$\left(a,a^2\right)$ and Q$\left(b,b^2-2b+2\right)$ on the parabolas $y=x^2$ and $y=x^2-2x+2$, respectively.

Suppose the slope of the required tangent line passes through the points P$\left(a,a^2\right)$ and Q$\left(b,b^2-2b+2\right)$, then $m=\frac{b^2-2b+2-a^2}{b-a}$. (2)

The derivative of parabola $y=x^2$ is $\frac{dy}{dx}$, which is obtained as follows.

$\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)$

Apply the power rule (1) and simplify the terms,

$\frac{dy}{dx}=2x$

Thus, the derivative of $y=x^2$ is 2x.

Therefore, the slope of the tangent to $y=x^2$ at $\left(a,a^2\right)$ is 2a. (3)

The derivative of parabola $y=x^2-2x+2$ is $\frac{dy}{dx}$, which is obtained as follows.

$\frac{dy}{dx}=\frac{d}{dx}\left(x^2-2x+2\right)$

Apply the derivative rules (1), (2), (3) and (4),

$\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(2\right)\\ =\frac{d}{dx}\left(x^2\right)-2\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\\ =2x-2\end{array}$

Thus, the derivative of $y=x^2-2x+2$ is $2x-2$.

Therefore, the slope of the tangent to $y=x^2-2x+2$ at $\left(b,b^2-2b+2\right)$ is $2b-2$. (4)

Since the required equation of the tangent is linear from $\left(a,a^2\right)$ to $\left(b,b^2-2b+2\right)$, the slopes of tangent line are equal.

From equations (2) and (3), $2b-2=\frac{b^2-2b+2-a^2}{b-a}$.

From equations (3) and (4), $a=b-1$.

Substitute $a=b-1$ in $2b-2=\frac{b^2-2b+2-a^2}{b-a}$.

$\begin{array}{c}2b-2=\frac{b^2-2b+2-{\left(b-1\right)}^2}{b-\left(b-1\right)}\\ 2b-2=b^2-2b+2-\left(b^2+1-2b\right)\\ 2b-2=b^2-2b+2-b^2-1+2b\\ 2b-2=1\end{array}$

Add 2 on both sides and obtain the value of b.

$\begin{array}{l}2b=1+2\\ 2b=3\\ b=\frac{3}{2}\end{array}$

Substitute the value $b=\frac{3}{2}$ in $a=b-1$,

$\begin{array}{l}a=\frac{3}{2}-1\\ a=\frac{1}{2}\end{array}$

For $a=\frac{1}{2}$, the slope 2a is 1 and the point P$\left(a,a^2\right)$ is $\left(\frac{1}{2},\frac{1}{4}\right)$.

Substitute $\left(\frac{1}{2},\frac{1}{4}\right)$ for $\left(x_1,y_1\right)$ and 1 for m in equation (1),

$\begin{array}{l}y-\frac{1}{4}=1\left(x-\frac{1}{2}\right)\\ y=x-\frac{1}{2}+\frac{1}{4}\\ y=x-\frac{2}{4}+\frac{1}{4}\\ y=x-\frac{1}{4}\end{array}$

Therefore, the equation of the tangent line to the parabolas is $y=x-\frac{1}{4}$.