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Finding the point of intersection between two curves. (Vectors)
At what point do the curves r1(t) = (t, 4-t, 63+t^2) and r2(s)= (9-s, s-5, s^2) intersect?Answer in the form: (x,y,z) = ____
Find the angle of intersection theta to the nearest degree.
The Attempt at a Solution
i: t=9-sj: 4-t=s-5k: 63+t^2=s^2
i/j: t-9sk: 63+(9-s)^2=s^2″Solving for “s””s=8t=1…I know not what to do from here. 🙁
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SOLVED:At what point do the curves r_1 (t) = \lan – wizard123 wizard123. Set components equal: From this we find: Substitute and solve system: The point where they intersect isHere are two paths r1(t) and r2(t) intersect if there is a point P lying on both curves. We say that r1(t) and r2(t) collide if r1(t0) = r2(t0) at some time t0. If u(t) = (sin t, cos t, t) and v(t) = (t, cos t, sin t), use Formula 4 of Theorem 3 to find View Answer. If a curve has the property that the position vector r(t) is…The curve C1 is through the point P. If r(t) = ⟨x(t) , y(t) , z(t)⟩ is parametrization for the curve C1 with r(t0) = P, then since the points of C1 are on the surface, we have This shows that the vector ∇F(x(t) , y(t) , z(t)) is perpendicular to the vector r′(t) = ⟨x′(t) , y′(t) , z′(t)⟩. Especially, at t = t0. we will have that…