Answer to Question #246836 in Calculus for Akki

Question

Find acute angle between intersecting curves. Write normal line, tangent line for each curves at the intersec tion point.

y = x ^ 3 , x = y ^ 2

Answer 1

Solution:

$y=x^3\ ...(i) \\x=y^2\ ...(ii)$

Put (ii) in (i)

$y=(y^2)^3 \\ \Rightarrow y=y^6 \\ \Rightarrow y-y^6=0 \\ \Rightarrow y(1-y^5)=0 \\ \Rightarrow y=0, y=1$

When $y=0,x=0$

When $y=1,x=1$

So, the points of intersection are (0,0), (1,1).

Now, for (0,0):

$y=x^3 \\ \Rightarrow y'=3x^2 \\ \Rightarrow y'(0,0)=3(0)=0=m_1$

$x=y^2 \\\Rightarrow y^2=x \\\Rightarrow 2y.y'=1 \\\Rightarrow y'=\dfrac1{2y} \\\Rightarrow y'(0,0)=\dfrac10=\infty=m_2$

Now, $\tan \theta=|\dfrac{m_1-m_2}{1+m_1m_2}|=|\dfrac{0-\infty}{1+0.\infty}|=\infty=not\ defined$

Thus, rejecting this case.

Next, for (1,1):

$y=x^3 \\ \Rightarrow y'=3x^2 \\ \Rightarrow y'(1,1)=3(1)^2=3=m_1$

$x=y^2 \\\Rightarrow y^2=x \\\Rightarrow 2y.y'=1 \\\Rightarrow y'=\dfrac1{2y} \\\Rightarrow y'(1,1)=\dfrac1{2(1)}=\dfrac12=m_2$

Now, $\tan \theta=|\dfrac{m_1-m_2}{1+m_1m_2}|=|\dfrac{3-\dfrac12}{1+3\times\dfrac12}|=1$

So, $\theta=45\degree$