At what point do the curves r1(t) = t, 5 − t, 48 t2 and r2(s) = 8 − s, s − 3, s2 intersect?

Set components equal:
[tex]t = 8-s \\
5-t = s-3 \\
48 t^2 = s^2[/tex]

From this we find:
[tex]t = 8-s \\
s = 4 \sqrt{3} t[/tex]

Substitute and solve system:
[tex]t = \frac{8}{1+4 \sqrt{3}} = 1.009 \\ \\
s = 8-t = 8 - 1.009 = 6.991[/tex]

The point where they intersect is:
[tex](1.009, 5-1.009, 48(1.009)^2)[/tex]

Answer Link

joaobezerra joaobezerra · Answer 2 · 2021-09-17T19:45:43+02:00

The parametric curves intersect at point (4, 1, 32).

-------------------------

Using the parametric functions for both r1 and r2 as functions of t, we have that:

Curve 1:

[tex]x_1(t) = t, y_1(t) = 5 - t, z_1(t) = 48 - t^2[/tex]

Curve 2:

[tex]x_2(t) = 8 - t, y_2(t) = t - 3, z_2(t) = t^2[/tex]

-------------------------

They intersect where they are equal, thus:

[tex]x_1(t) = x_2(t)[/tex]

[tex]t = 8 - t[/tex]

[tex]2t = 8[/tex]

[tex]t = \frac{8}{2}[/tex]

[tex]t = 4[/tex]

-------------------------

Using curve 1:

[tex](x(t), y(t), z(t)) = (t, 5-t, 48 - t^2)[/tex]

Then

[tex](x(4), y(4), z(4)) = (4, 5-4, 48 - 4^2) = (4, 1, 32)[/tex]

They intersect at point (4, 1, 32).

A similar problem is given at https://brainly.com/question/17426048