Question #3e542

1 Answer
Aug 30, 2016

The Curves intersect at two pts., O(0,0), and, A(1,5)O(0,0),and,A(1,5)

At OO, the XX-axis is their common tgt, &, hence, the /_ btwn them is 0^@0.

At AA, the /_ btwn them is 1.8965^@1.8965.

Explanation:

Let the curves be C_1 : y=5x^3, and, C_2 : y=5x^2C1:y=5x3,and,C2:y=5x2.

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

5x^3=y=5x^2 rArr 5x^2(x-1)=0 rArr x=0, x=15x3=y=5x25x2(x1)=0x=0,x=1.

y=5x^2, x=0, x=1 rArr y=0, y=5y=5x2,x=0,x=1y=0,y=5. Thus, we have,

C_1nnC_2={O(0,0), A(1,5)}C1C2={O(0,0),A(1,5)}.

To find the slopes of tgts. to the curves C_1, and, C_2C1,and,C2, at the pts. O, and, AO,and,A, we will find dy/dxdydx as it denotes the reqd. slopes.

For C_1, dy/dx=15x^2, &, "for" C_2, dy/dx=10xC1,dydx=15x2,&,forC2,dydx=10x

:. for C_1, and, C_2, [dy/dx]_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(0,0).

Clearly, (1) the X-axis is the common tgt. to C_1 and C_2 touching at the Origin. (2) The angle btwn. them, at O is 0^@.

"At the pt." A(1,5), dy/dx=15, "for" C_1, and, "for" C_2, dy/dx=10

:. m_1=15, and m_2=10, are the respective slopes of tgts.

:. alpha=the /_(C_1,C_2) rArr tan alpha=|(m_1-m_2)/(1+m_1m_2)|

=|(15-10)/(1+150)|=5/151~~0.0331

:. alpha=arc tan 0.0331=1.8965^@.

Enjoy Maths.!