Find the equation of the normal to the curve with the equation $y=x^3 + 1$ at the point $(1,2)$
I'm having trouble solving this:
Find the equation of the normal to the curve with the equation $y=x^3 + 1$ at the point $(1,2)$.
$\endgroup$ 31 Answer
$\begingroup$The normal line is defined as the line perpendicular to the tangent line of the curve at that point. Knowing this we can evaluate slope of the normal line of this curve (or in general), by
$m_{normal}=-\frac{1}{f'(x)}$
Where $f(x)$ is our function in question. In this case, $y=x^3+1$.
The slope or the derivative at this point is the derivative evaluated at $x=1$. Because $f'(x)=3x^2$, $f'(1)=3$.
The slope of normal line is then $-\frac13$.
Recall that to find the equation of ant line given a point and a slope, we could use the point slope formula. (Really, we can use any form of a linear equation and solve accordingly). The point slope is $$y-y_{point}=m_{slope}(x-x_{point})$$
Since the normal line also intersects the tangent which intersects the curve at $(1,2)$, we can use the point $(1,2)$ and the normal's slope to get the following tangent line:
$$y-2=-\frac13(x-1)$$ Or $$y=-\frac13x+\frac73$$
$\endgroup$ 4
