Question

Find the intersecting angle between the curves $y=x^2$ and $y=x^3$.

Answer 1

To find the angle of intersection, we first find the point of intersection and then find the angle between the tangents at this point.

Hence, the point of intersection of $y=x^2$ and $y=x^3$ can be foud by equating them.

$\therefore x^2=x^3$

$\therefore x^3-x^2=0$

$\therefore x=0$ or $x=1$

Hence, the points of intersection are $(0,0)$ and $(1,1)$.

The slope of tangent of $y=x^2$ is $2x$ and the slope of tangent of $y=x^3$ is $3x^2$ for a given point $(x,y)$ on the curve.

Now, at $(0,0)$,

slope of tangent of $y=x^2$ = 0

slope of tangent of $y=x^3$ = 0

As the slopes are same, the angle of intersection is $0^{\circ }$.

Now, at $(1,1)$,

slope of tangent of $y=x^2$ = 2

slope of tangent of $y=x^3$ = 3

$therefore$ angle of intersection can be found by

$\therefore \tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$\therefore \tan \theta =\left|\frac{3-2}{1+3\times 2}\right|$

$\therefore \tan \theta =\left|\frac{1}{7}\right|$

$\therefore \theta ={\tan }^{-1}\frac{1}{7}$

Using log tables, we have

$\theta ={8.13}^{\circ }$

Hence the angles of intersection between $y=x^2$ and $y=x^3$ are $0^{\circ }$ and ${8.13}^{\circ }$ at $(0,0)$ and $(1,1)$ respectively.