Chapter 4.4, Problem 30E

To determine

To calculate: The point at which the tangent to the curve is perpendicular to given line.

Answer to Problem 30E

The point is $x=\ln \left(\frac{1}{6}\right),y=\frac{-2}{3}$ .

Explanation of Solution

Given information: The equation of curve is,

  $y=2e^x-1$

The equation of line is,

  $y=-3x+2$

The slope of two perpendicular lines is equal to -1.

Formula used: $\left(slope\right):m_a=\frac{d{y}_a}{dx}$

Calculation: Slope of curve,

  $\begin{array}{l}{m}_1=\frac{dy}{dx}\\ m_1=\frac{d\left(2e^x-1\right)}{dx}\\ m_1=2e^x\\ \text{Calculate}\text{slope}\text{of}\text{line,}\\ m_2=\frac{dy}{dx}\\ m_2=\frac{d\left(-3x+2\right)}{dx}\\ m_2=-3\\ \text{Equate}\text{both}\text{slopes,}\\ m_1\times m_2=-1\\ 2e^x\times -3=-1\\ x=\ln \left(\frac{1}{6}\right)\\ \text{Calculate}y,\\ y=\frac{-2}{3}\end{array}$

The point at which the tangent to the curve is perpendicular to given line is $x=\ln \left(\frac{1}{6}\right),y=\frac{-2}{3}$ .