1. ## Challenging Calculus question

Suppose that the tangent at a point P on the curve y=x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

I have no idea where to start except to take the derivative of y=x^3

2. Originally Posted by pluvam
Suppose that the tangent at a point P on the curve y=x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

I have no idea where to start except to take the derivative of y=x^3
I like to draw a picture to help me see what is going on.

So you know the derivative is

$\displaystyle y'=3x^2$

So the slope at the point $\displaystyle (a,a^3)$ is

$\displaystyle m_a=3a^2$

So the equation of the tangent line is

$\displaystyle y-a^3=3a^2(x-a) \iff y=3a^2x-2a^3$

but we know that the tangent line intersects the curse again at some point $\displaystyle (b,b^3)$

So we know that

$\displaystyle y=3a^2b-2a^3$ and that $\displaystyle y=b^3$

So setting the two above equations equal gives

$\displaystyle b^3=3a^2b-2a^3 \iff b^3-3a^2b+2a^3=0$

This is cubic in b which can make it difficult to solve but know that this is trivially true when $\displaystyle a=b$ so $\displaystyle b-a$ must be a factor of the equation by long division we get
$\displaystyle (b-a)(b^2+ab-2a^2)$ this can be factored to gives

$\displaystyle (b-a)(b-a)(b+2a)$ so the only non trivial root is

$\displaystyle b=-2a$

Now just plug the points into the derivative to get

$\displaystyle y'(a)=\textcolor{red}{3a^2};y'(b)=y'(-2a)=3(-2a)^2=12a^2=4\textcolor{red}{(3a^2)}$

3. Thank you so much