Chapter 3.1, Problem 53E

To determine

To find:

The equation of the tangent and parallel to the line

Answer to Problem 53E

The equation of the both line is $y-9=12\left(x-2\right),y+7=12\left(x+2\right)$

Explanation of Solution

Given:

The tangent to the curve

  $y=1+x^3$

Parallel to the line

  $\begin{array}{l}12x-y=1\\ y=12x-1\end{array}$

Concept used:

The equation is in slope −intercept form, $y=mx+c$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=1+x^3\mathrm{.......................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}=m\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}y=1+x^3\\ y^{\prime }=3x^2\mathrm{........................}\left(2\right)\end{array}$

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

  $\begin{array}{l}y=12x-1\\ y^{\prime }=\mathrm{12.............................}\left(3\right)\end{array}$

The equation is in slope −intercept form

The slope of the line is $=12$

From equation (2) and equation (3)

  $\begin{array}{l}3x^2=12\\ x^2=4\\ x=\pm 2\end{array}$

The coordinates of points on each line are $\left(2,9\right)\text{and}\left(-2,-7\right)$

An equation for the line through the point $\left(x_1,y_1\right)$ with slope m is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-9=12\left(x-2\right)\\ y+7=12\left(x+2\right)\end{array}$