# Question 3e542

Aug 30, 2016

The Curves intersect at two pts., $O \left(0 , 0\right) , \mathmr{and} , A \left(1 , 5\right)$

At $O$, the $X$-axis is their common tgt, &, hence, the $\angle$ btwn them is ${0}^{\circ}$.

At $A$, the $\angle$ btwn them is ${1.8965}^{\circ}$.

#### Explanation:

Let the curves be ${C}_{1} : y = 5 {x}^{3} , \mathmr{and} , {C}_{2} : y = 5 {x}^{2}$.

To find their pt. of intersection, we have to solve their eqns.

$5 {x}^{3} = y = 5 {x}^{2} \Rightarrow 5 {x}^{2} \left(x - 1\right) = 0 \Rightarrow x = 0 , x = 1$.

$y = 5 {x}^{2} , x = 0 , x = 1 \Rightarrow y = 0 , y = 5$. Thus, we have,

${C}_{1} \cap {C}_{2} = \left\{O \left(0 , 0\right) , A \left(1 , 5\right)\right\}$.

To find the slopes of tgts. to the curves ${C}_{1} , \mathmr{and} , {C}_{2}$, at the pts. $O , \mathmr{and} , A$, we will find $\frac{\mathrm{dy}}{\mathrm{dx}}$ as it denotes the reqd. slopes.

For C_1, dy/dx=15x^2, &, "for" C_2, dy/dx=10x

$\therefore f \mathmr{and} {C}_{1} , \mathmr{and} , {C}_{2} , {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{O} = 0.$

This means that, both tgts. to C_1,&,C_2# have the same slope, namely $0$, and, both pass thro. the same pt. $O \left(0 , 0\right)$.

Clearly, (1) the $X$-axis is the common tgt. to ${C}_{1} \mathmr{and} {C}_{2}$ touching at the Origin. (2) The angle btwn. them, at $O$ is ${0}^{\circ}$.

$\text{At the pt." A(1,5), dy/dx=15, "for" C_1, and, "for} {C}_{2} , \frac{\mathrm{dy}}{\mathrm{dx}} = 10$

:. ${m}_{1} = 15 , \mathmr{and} {m}_{2} = 10 ,$ are the respective slopes of tgts.

$\therefore \alpha = t h e \angle \left({C}_{1} , {C}_{2}\right) \Rightarrow \tan \alpha = | \frac{{m}_{1} - {m}_{2}}{1 + {m}_{1} {m}_{2}} |$

$= | \frac{15 - 10}{1 + 150} | = \frac{5}{151} \approx 0.0331$

$\therefore \alpha = a r c \tan 0.0331 = {1.8965}^{\circ}$.

Enjoy Maths.!