Chapter 3, Problem 64RE

(a)

To determine

To find: The equation of tangent to the given curve that is parallel to the line $x-4y=1$ .

Explanation of Solution

Given: The given equation curve is $y=e^x$ .

Differentiate the equation of curve.

  $\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(e^x\right)\\ =e^x\end{array}$ ...... (1)

Consider the equation of line.

  $\begin{array}{c}x-4y=1\\ y=\frac{x-1}{4}\\ y^{\prime }=\frac{1}{4}\end{array}$

Substitute $\frac{1}{4}$ for $y^{\prime }$ in equation (1).

  $\begin{array}{c}\frac{1}{4}=e^x\\ \ln \left(1/4\right)=x\\ x=\ln \left(1/4\right)\end{array}$

Since the line are parallel so there slope will be equal.

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

Therefore, the equation of tangent parallel to the given curve is $y=\frac{x}{4}+\frac{1-\ln \left(1/4\right)}{4}$ .

(b)

To determine

To find: The equation of tangent to the given curve that passes through the origin

Explanation of Solution

Given: The given equation curve is $y=e^x$ .

Let assume that the tangent at $\left(a,e^a\right)$ passes through the origin.

Differentiate the equation of curve.

  $\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(e^x\right)\\ =e^x\end{array}$ ...... (1)

Slope at $a$ is,

  $y^{\prime }=e^a$

The equation of tangent is calculated as,

  $\begin{array}{c}\left(y-e^a\right)=e^a\left(x-a\right)\\ y=e^{a}x+e^a-e^{a}a\\ =e^{a}x+e^a\left(1-a\right)\end{array}$

The tangent will pass through the origin ig $y$ intercept is $0$ .

  $\begin{array}{c}{e}^a\left(1-a\right)=0\\ 1-a=0\\ a=1\end{array}$

Thus,

  $\begin{array}{c}y=e^{1}x+e^1\left(1-1\right)\\ =ex\end{array}$

Therefore, the equation of tangentpassing through the origin of the given curve is $y=ex$ .