At what point do the curves r1(t) (t, 5 t, 63 t2) and r20s) (9 s, S 4, s2) in (x, y, z) Find their angle of intersection, 0, correct to the nearest degree.

Both the curves r₁ and r₂ are in the 3 dimentional (x,y,z) plane

All Curves may not meet. To verify whether they meet at a unique point, we should equate the x, y and z components of a point

Equating the x components of points for both the curves

t=9-s  s+t =9.............(1)

Equating the y components of the points for both curves

5-t = s-4   s+t =9 ..............(2) (the same expression previously obtained)

Equating the z components

63+ t² = s^{2...............................(3)}

From equation (2) we can write s = 9-t

putting the value of s in equation (3), we get 63+ t² =81 -18t + t²

Cancelling the t² on both sides of the above equation, we get 63 = 81-18t

Hence the value of t can be found to be 1

From equation (1), the value of s = 8 when t = 1

By using these values of t and s, we get the point of intersection to be (1,4,64)

To find the angle between the curves, we have to find the dot products of slope of the tangent. Hence we can find cosine of angle between them.

The differenciation of r₁ wrt t and r₂ wrt s gives <1,-1,2t> and <-1,1,2s> as the component of slopes of the tangent

 <1,-1,2> and <-1,1,16> respectively

Dot product of 2 quantities u and v is u.v =|u||v cos = 1*(-1) + -1*1+ 2*16 = 30

Hence cos = u.v/|u||v| = 

 0.7624

Cos^-1 (0.7624) = 40.3

Hence the angle of intersection correct to nearest degree is 40