# At what point do the curves r1 and r2 intersect?

At what point do the curves r1 = < t, 3 - t, 15 + t^2 >

and r2 = < 5 - s, s - 2, s^2 > intersect?

( ____, ____, ____)

Find their angle of intersection, θ correct to the nearest degree.

θ = ____°

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)

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To find angle of intersection, we find gradient vectors at point (1, 2, 16)

Angle between curves at intersection = angle between gradient vectors.

r? = < t, 3-t, 15+t2 >

dr?/dt = < 1, -1, 2t >

u = dr?/dt at t = 1

u = < 1, -1, 2 >

r? = < 5-s, s-2, s2 >

dr?/dx = < -1, 1, 2s >

v = dr?/dx at s = 4

v = < -1, 1, 8 >

We use dot product to find θ

cos θ = u.v / (||u|| ||v||)

u.v = < 1, -1, 2 > . < -1, 1, 8 > = -1 - 1 + 16 = 14

||u|| = √(1+1+4) = √6

||v|| = √(1+1+64) = √66

cos θ = 14 / (√6√66)

cos θ = 14 / (6 √11)

cos θ = 7 / (3 √11)

θ = arccos(7/(3√11)) = 45.289377545 = 45° (to the nearest degree)

-- Ματπmφm --

Sep 15 at 21:42