Chapter 4.4, Problem 64E

To determine

To prove: Curves in the family $y=-\frac{1}{2}{x}^2+k$ are perpendicular to all curves in the family of $y=\ln x+c$ .

Answer to Problem 64E

  $y=-\frac{1}{2}{x}^2+k\text{ and }y=\ln x+c$ are perpendicular to each other.

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y=-\frac{1}{2}{x}^2+k\text{ and }y=\ln x+c$

Formula Used:

  $\frac{dy}{dx}{x}^n=n{x}^{n-1}$

Calculation: Let us find the slope for first family of curves,

  $\begin{array}{l}\text{Slope of }y=-\frac{1}{2}{\text{x}}^2\text{+k }\\ \text{Differentiating with respect to }x,\\ \frac{dy}{dx}=-\frac{1}{2}\frac{d(x^2)}{dx}\\ \frac{dy}{dx}=-\frac{1}{2}\times 2x\\ \frac{dy}{dx}=-x=m_1\end{array}$

Slope for second family of curves,

  $\begin{array}{l}\text{Slope of }y=\ln x\text{+c }\\ \text{Differentiating with respect to }x,\\ \frac{dy}{dx}=\frac{d(\ln x+c)}{dx}\\ \frac{dy}{dx}=\frac{1}{x}=m_2\\ \text{If the slope of curves are perpendicular to each other then, }\\ \text{first family of curves is perpendicular to second family of curves}\text{.}\\ \text{For that,}\\ {\text{m}}_1\times m_2=-1\\ -x\times \frac{1}{x}=-1\\ -1=-1\\ \text{Therefore, Left Hand side is equals to right hand side}\text{.}\end{array}$

So, $y=-\frac{1}{2}{x}^2+k\text{ and }y=\ln x+c$ are perpendicular to each other.