Answer:
The angles between the curves at the points of intersection are:
0º, 1.3º
Step-by-step explanation:
The intersections points are found by setting the equations equal to each other and solving the resulting equation:
[tex]7x^2=7x^3\\x^3-x^2=0\\x^2(x-1)=0\\x=0,x=1[/tex]
The angles of the tangent lines can be found by stating their slopes.
To find the slope we differentiate the equations:
[tex]y'_1=14x,y'_2=21x^2[/tex]
Then we plug the x-coordinates of the intersections:
For x=0 we get the slopes are both 0:
[tex]y'_1=14(0)=0,y'_2=21(0)^2=0[/tex]
So the angles of inclination of the lines are the same their difference is 0. Hence the angle between the tangent curves is also 0º at the point of intersection at x=0
For x=1 we get the following slopes:
[tex]y'_1=14(1)=14,y'_2=21(1)^2=21[/tex]
The slopes are the tangents of the angles. Therefore, to get the angle between the lines we do:
[tex]arctan(21)-arctan(14)\approx87.2737\º-85.9144\º\approx1.3\º[/tex]
So, 1.3º is the angle between the curves at the second point of intersection at x=1.
