Calculus and analytic geometry question

Question

Find the tangent of the angle in which the functions $x^3 $, and $x^2 $ intersect $(x≠0)$ .

I find this question to be quite funny since the intersection point has two tangents going to it, with apparently different slopes...

attempt at a solution: $(1) y' = 3x^2 (2) y' = 2x $. solving the system we get $x=0.666666...$ . hence$ m=y'(0.666666....) = 4/3$, or $tanα = 4/3$ .

Answer 1

Yes there are two tangents. Let the slope of first tangent be $m_1$, and let the slope of second tangent be $m_2$. In this case one is $3$, the other is $2$. It is asking you to find the tangent(the trigonometric function) of the angle between these tangents. (The wording is very unclear, took me a while).

So we need to find, $\tan(\alpha-\beta)$ where $tan(\alpha)=3$ and $\tan(\beta)=2$.

Use the formula $$\tan(\alpha-\beta) = \frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$$

Answer 2

Your attempt is completely wrong. Solving that system of equations gives you an $x$ where the two slopes are the same. You want to start by solving $x^2 = x^3$ to find a point where the two curves intersect.

Answer 3

Well, they intersect at two points, the Origin and $(1,1)$ In the Origin, their tangents perfectly coincide and so the angle would be zero. At $(1,1)$ you find for the angle the difference of $arctan3$ and $arctan2$ which amounts to approx $8.13°$ No need to solve a system of equations here.