Challenging Calculus question

**pluvam** · March 2nd 2010, 01:08 PM

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

**TheEmptySet** · March 2nd 2010, 01:45 PM

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

$y'=3x^2$

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

$m_a=3a^2$

So the equation of the tangent line is

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

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

So we know that

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

So setting the two above equations equal gives

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

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

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

$b=-2a$

Now just plug the points into the derivative to get

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

**pluvam** · March 2nd 2010, 03:07 PM

Thank you so much

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