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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 $tanα = 4/3$ .

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Find the points at which the two curves intersect. At one point, the tangents coincide, at the other they don't. That's the place where you have some work to do. By the way, what system are you solving to get your $x$ values? – Chris Leary Mar 14 '14 at 15:01
tangent of an angle? An angle? – imranfat Mar 14 '14 at 15:02
The functions intersect, not their derivatives! – alex Mar 14 '14 at 15:02
up vote 3 down vote accepted

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

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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.

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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.

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