Set the equations equal to each other 3x2−2x−1=x3−1 ⇒x3−3x2+2x=0 ⇒x(x−1)(x−2)=0 ⇒x=0,x=1 and x=2 For x=0,y=03−1=−1, Point : (0,−1) For x=1,y=13−1=0, Point : (1,0) For x=2,y=23−1=7, Point : (2,7) So, the point of intersection in the first quadrant is (2,7). For, y1=3x2−2x−1 So, m1=
dy1
dx
=6x−2 At (2,7),m1=6(2)−2=12−2=10 For y2=x3−1,m2=
dy2
dx
=3x2 At (2,7),m2=3(22)=12 ∴tanθ=|
m1−m2
1+m1m2
| ⇒|
10−12
1+(10)(12)
| ⇒|
−2
1+120
|⇒|
−2
121
| ⇒tanθ=
2
121
⇒θ=tan−1(
2
121
) Thus, the acute angle between the curves at their point of intersection in the first quadrant is tan−1(
