Chapter 3.1, Problem 20E

(a)

To determine

The lines that are tangent to the curve at $x=4$ .

Answer to Problem 20E

The required equation of the tangent line is $y=\frac{1}{4}x+1$ .

Explanation of Solution

Given information:

The curve: $y=\sqrt{x}$

Formula used:

Point-slope form of the equation of a line:

  $\left(y-y_1\right)=m\left(x-x_1\right)$

Slope of a curve is:

  $f^{\prime }\left(x\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$

The given curve is $y=\sqrt{x}$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{\sqrt{x}-\sqrt{a}}{x-a}$

Substitute 4 for a in the above limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(4\right)=\underset{x\to 4}{\lim }\frac{\sqrt{x}-\sqrt{4}}{x-4}\\ =\underset{x\to 4}{\lim }\frac{\cancel{\left(\sqrt{x}-2\right)}}{\left(\sqrt{x}+2\right)\cancel{\left(\sqrt{x}-2\right)}}\\ =\underset{x\to 4}{\lim }\frac{1}{\left(\sqrt{x}+2\right)}\end{array}$

Substitute 4 for x in the above limit and simplify.

  $\begin{array}{c}\underset{x\to 4}{\lim }\frac{1}{\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{4}+2}\\ =\frac{1}{2+2}\\ =\frac{1}{4}\end{array}$

Also, substitute 4 for x in the curve $y=\sqrt{x}$ .

  $\begin{array}{c}y=\sqrt{4}\\ =2\end{array}$

So, the slope of the tangent line is $\frac{1}{4}$ and the point through which the line passes is $\left(4,2\right)$ .

To find the equation of the tangent, use point-slope formula of an equation of a line $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Substitute $\frac{1}{4}$ for m , 2 for $y_1$ and 4 for $x_1$ in the above equation and simplify.

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

Hence, the required equation of the tangent line is $y=\frac{1}{4}x+1$ .

(b)

To determine

The lines that are normal to the curve at $x=4$ .

Answer to Problem 20E

The required equation of the tangent line is $y=-4x+18$ .

Explanation of Solution

Given information:

The curve: $y=\sqrt{x}$

Formula used:

Point-slope form of the equation of a line:

  $\left(y-y_1\right)=m\left(x-x_1\right)$

Slope of a curve is:

  $f^{\prime }\left(x\right)=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$

The given curve is $y=\sqrt{x}$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{\sqrt{x}-\sqrt{a}}{x-a}$

Substitute 4 for a in the above limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(4\right)=\underset{x\to 4}{\lim }\frac{\sqrt{x}-\sqrt{4}}{x-4}\\ =\underset{x\to 4}{\lim }\frac{\cancel{\left(\sqrt{x}-2\right)}}{\left(\sqrt{x}+2\right)\cancel{\left(\sqrt{x}-2\right)}}\\ =\underset{x\to 4}{\lim }\frac{1}{\left(\sqrt{x}+2\right)}\end{array}$

Substitute 4 for x in the above limit and simplify.

  $\begin{array}{c}\underset{x\to 4}{\lim }\frac{1}{\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{4}+2}\\ =\frac{1}{2+2}\\ =\frac{1}{4}\end{array}$

Also, substitute 4 for x in the curve $y=\sqrt{x}$ .

  $\begin{array}{c}y=\sqrt{4}\\ =2\end{array}$

So, the slope of the tangent line is $\frac{1}{4}$ and the point through which the line passes is $\left(4,2\right)$ .

It is known that the slope of a normal line is the negative reciprocal to the slope of the tangent line.

So, the slope of the normal line is $\begin{array}{c}m=-\frac{1}{\frac{1}{4}}\\ =\left(-1\right)\times \frac{4}{1}\\ =-4\end{array}$

To find the equation of the tangent, use point-slope formula of an equation of a line $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Substitute $-4$ for m , 2 for $y_1$ and 4 for $x_1$ in the above equation and simplify.

  $\begin{array}{c}\left(y-2\right)=-4\left(x-4\right)\\ y-2=-4x+16\\ y-\cancel{2}+\cancel{2}=-4x+16+2\\ y=-4x+18\end{array}$

Hence, the required equation of the tangent line is $y=-4x+18$ .