Question

At how many points do the space curves $r_1(t) = \langle t^2, 1 - t^2, t + 1 \rangle$ and $r_2(s) = \langle 1 - s^2, s, s \rangle$ intersect?

          At how many points do the space curves $r_1(t) = \langle t^2, 1 - t^2, t + 1 \rangle$ and $r_2(s) = \langle 1 - s^2, s, s \rangle$ intersect?
        
At how many points do the space curves r1(t) = ⟨ t^2, 1 - t^2, t + 1 ⟩ and r2(s) = ⟨ 1 - s^2, s, s ⟩ intersect?

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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At how many points do the space curves r(t) = t^2, 1 - t, t + 1 and r2(s) = 1 - s^3 intersect?
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Transcript

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00:01 In this question here we recall that given the two vectors a space curve r1t and r1s, and to find the intersection between them, we need to let the r1t equal to r2s, there's not to be r2 here, then we will need to serve for the t and the s here.
00:31 Now in this question we're given the rt l1 t equal to the t square t plus 1 and we have the r2 s equal to the square root of the s s and then s minus 1 so to find the intersection we need to let the r1 t equal to the r2s so it means that we will have the 3 system equations now when x equal to the x y equal to the y and the z equal to the z so the t equal to the square root to the s t squared equal to the s t plus one equal to the s minus one now when we serve for this one we get now it was from the first equation here it is square and up so t here must equal to the t square must equal to the s now and it could be the same as the second equation.
01:37 And if you click into the third one here, so we have the t plus 1 equal to this s here equal to the t squared minus 1.
01:47 Now we need to solve this equation now.
01:51 I bring everything to one side of the equation, so t squared minus t minus 2 equal to 0...
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