So I'm given the following problem:

Find the angle where these two curves intersect

a) $y=x^2$, $x=y^2$

b) $y=\frac{x^2}{2}$, $y=\frac{1}{1+x^2}$

here's what I did for the second one, I don't even know where to start the first one

So I found that they intersect at $(1;1/2)$ and $(-1;1/2)$, then I took the derivatives of the two in order to get the equation of the tangent line at the intersection point and I got

$y'=x$ and $y'=\frac{-2x}{1+x^2}$

so the slopes of the tangent lines at that point have to be $1; -1$ or $-1; 1$

then the angle between those two lines have to be $\tan^{-1}\dfrac{-1 - 1}{1-1}$.

From here I don't know what to do because I clearly did something wrong since the answer is $arctan3$

If you could help me with these problems by giving me hints or solutions or pointing out where i made mistakes that would be great. Thank you in advance.

P.S. We have not studied any series or integrals in class yet (I don't know if there was any need to write this, but just in case). We have just started learning differentiation.

  • 1
    $\begingroup$ You should double-check your derivatives. And a) is just as easy as b). Why do you not know where to start? $\endgroup$ – Ted Shifrin Apr 25 at 22:59
  • $\begingroup$ I object; (a) is harder than (b). One has to make a bit of thinking (admittedly, a very little bit, but still): if $x=y^2$, then what is $y$ in terms of $x$? $\endgroup$ – Ivan Neretin Apr 25 at 23:15
  • $\begingroup$ @TedShifrin oh, thank you I got the mistake $\endgroup$ – Britanica Apr 26 at 9:17
  • $\begingroup$ @IvanNeretin what I struggle to understand is whether I should take the derivative with respect to $x$ or $y$? $\endgroup$ – Britanica Apr 26 at 9:18
  • $\begingroup$ Can you write $y$ in terms of $x$? Will that make the question void? $\endgroup$ – Ivan Neretin Apr 26 at 9:21


Points of intersections: $O(0, 0)$ and $A(1, 1)$

1): at O:

$y=x^2 \Rightarrow y'=2x\Rightarrow m_1=0\Rightarrow \theta_1=0^o$

$y=\sqrt x\Rightarrow y'=\frac1{2\sqrt x}\Rightarrow m_2, x=0\Rightarrow m_2=\infty\Rightarrow \theta_2=90^o$

$\Rightarrow \theta =\theta_2-\theta_1=90-0=90^o$

  1. at A(1, 1):

$m_1=2$, $m_2=\frac1{2\sqrt 1}=\frac12$

$tan(\theta)=\frac {2-\frac12}{1+2\times \frac12}=\frac 34\Rightarrow \theta =tan^{-1}(\frac34)$

Your calculation for b is correct and angle between tw curves is $90^o$. $tan^{-1} 3$ is wrong.

  • $\begingroup$ But I got the textbook answer later. I just didn’t square the denominator when taking the derivative. $\endgroup$ – Britanica Apr 26 at 14:26

I'd suggest calculating the angle from horizontal of each tangent line separately then subtracting the two angles. You're on the right track that the arctan will translate from slope to angle, but it will be trickier if you want to calculate the angle between the tangent lines directly.


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