Find the Point/s on the Curve $y-x^3=0$ where the normal line have a slope of $\frac{-1}{3}$.

Question

I am bit clueless on how to start the problem. The only idea I have is to use derivatives, yet I can't continue on. I have tried researching different problems connected to it as well, but the results are lacking. Detail explanations are welcomed, even with or without the use of derivatives.

Answer 1

If the slope of the normal line is $-\dfrac{1}{3}$, then as @Fred pointed out in the comment above, the slope of the tangent is the negative of the reciprocal of that, which is $3$.

The slope of the tangent is equal to the value of the derivative. The derivative is given by

$ y' = 3 x^2 $

Hence we want

$3 x^2 = 3 $

i.e. $ x^2 = 1$ , whose solutions are $x =\pm 1 $.

Therefore, the points on the graph of $y - x^3 = 0$ that satisfy the given condition are $ (-1, -1)$ and $(1, 1)$.

Answer 2

If a line $l_1$ is perpendicular to another line $l_2$, then the slope of the first line is related to the slope of the second line as $m_1m_2=-1$. If the normal at a point has a slope of $-1/3$, then the tangent at that point will have a slope of $3$. Now, using the fact that the slope of a line tangent to a curve is the derivative of the function at that point, we get, $\frac{dx^3}{dx}=3$. The derivative of $x^3$ with respect to $x$ is $3x^2$. Set $3x^2=3$, and you have your answer.