at what point do the curves r1(t)= and r2(t)=<3-s,s-2,s^2>intersect? find their
ID: 3284439 • Letter: A
Question
at what point do the curves r1(t)= and r2(t)=<3-s,s-2,s^2>intersect? find their angle of intersection correct to the nearest degree.Explanation / Answer
When x, y, and z coordinates are all equal at the same time x-coordinates: t = 5 - s y-coordinates: 3 - t = s - 2 -----> -t = s - 5 ----> t = 5 - s z-coordinates: 15 + t² = s² Notice that equations we get from x- and y-coordinates both give us t = 5 - s So we will replace t with this value in equation we derived from z-coordinate: 15 + (5-s)² = s² 15 + 25 - 10s + s² = s² 40 - 10s = 0 10s = 40 s = 4 t = 5 - s = 1 So we will get point of intersection when t = 1 and s = 4 (t, 3-t, 15+t²) = (1, 3-1, 15+1) = (1, 2, 16) (5-s, s-2, s²) = (5-4, 4-2, 4²) = (1, 2, 16) Point of intersection: (1, 2, 16) To find angle of intersection, we find gradient vectors at point (1, 2, 16) Angle between curves at intersection = angle between gradient vectors. r1 = dr1/dt = < 1, -1, 2t > u = dr1/dt at t = 1 u = < 1, -1, 2 > r2 = < 5-s, s-2, s² > dr2/dx = < -1, 1, 2s > v = dr2/dx at s = 4 v = < -1, 1, 8 > We use dot product to find ? cos ? = u.v / (||u|| ||v||) u.v = < 1, -1, 2 > . < -1, 1, 8 > = -1 - 1 + 16 = 14 ||u|| = v(1+1+4) = v6 ||v|| = v(1+1+64) = v66 cos ? = 14 / (v6v66) cos ? = 14 / (6 v11) cos ? = 7 / (3 v11) ? = arccos(7/(3v11)) = 45.289377545 = 45° (to the nearest degree)Related Questions
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