Need help finding the intersection of these two vector valued functions: r1(t)= and r2(s)=<4t+6,4t^2,7-t>. Thanks!
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Given r1(t) = and r2(s) = <4s + 6, 4s^2, 7 - s>:
Equate corresponding entries to obtain
t = 4s + 6
t^2 = 4s^2
t^3 = 7 - s.
The second equation implies that t = 2s or t = -2s.
(i) If t = 2s, then substituting this into the first equation yields
2s = 4s + 6 ==> s = -3 (and t = -6).
Double check in third equation: -216 = 10, which is false (so ignore this!).
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(ii) If t = -2s, then substituting this into the first equation yields
-2s = 4s + 6 ==> s = -1 (and t = 2).
Double check in third equation: 8 = 8, which is true.
So, the point of intersection is r1(2) = r2(-1) = <2, 4, 8>.
I hope this helps!
Equate corresponding entries to obtain
t = 4s + 6
t^2 = 4s^2
t^3 = 7 - s.
The second equation implies that t = 2s or t = -2s.
(i) If t = 2s, then substituting this into the first equation yields
2s = 4s + 6 ==> s = -3 (and t = -6).
Double check in third equation: -216 = 10, which is false (so ignore this!).
-----------
(ii) If t = -2s, then substituting this into the first equation yields
-2s = 4s + 6 ==> s = -1 (and t = 2).
Double check in third equation: 8 = 8, which is true.
So, the point of intersection is r1(2) = r2(-1) = <2, 4, 8>.
I hope this helps!