Answer to Question #280822 in Differential Equations for Shell

Question

Find the equation of the curve at every point that passes through the point (0, −1) and which the normal line at any point (x, y) has a slope of 1/(3x + 2).

Answer 1

Let the equation of the curve be $y=f(x).$ Then slope of the nornal to the curve at any point $(x, y)$ will be

$slope=m=-\dfrac{1}{f'(x)}$

Given $slope=m=\dfrac{1}{3x+2}.$

Then

$f'(x)=-\dfrac{1}{m}=-(3x+2)$

Integrate

$f(x)=-\int(3x+2)dx=-\dfrac{3}{2}x^2-2x+C$

The curve passes through the point $(0, −1)$

$-1=-\dfrac{3}{2}(0)^2-2(0)+C=>C=-1$

The equation of the curve at every point is

$y=-\dfrac{3}{2}x^2-2x-1$