calculus of vector-valued fun

2011-01-14 10:49 pm

At what point do the curves r1(t) = (t,1-t,3+t^2) and r2(t) = (3 -t ,t-2,t^2) intersect?

Find their angle of intersection correct to the nearest degree.

I do not what I need to differentiate ?

Can you give me a whole solution~~THX~

回答 (1)

myisland8132

2011-01-14 11:24 pm

✔ 最佳答案

If r1(t) = (t,1-t,3+t^2) and r2(s) = (3 -s ,s-2,s^2) has a intersection point, then

t = 3 - s.....(1)

1 - t = s - 2....(2)

3 + t^2 = s^2...(3)

Sub. (1) into (3)

3 + (3 - s)^2 = s^2

s^2 - 6s + 12 = s^2

s = 2 and then t = 1

So the intersection point is (1, 0, 4)

The derivative of r1(t) and r2(t) are (1, -1, 2t) and (-1, 1 ,2s) respectively. Sub. (1, 0, 4) into them, we see that the tangent vector of r1(t) and r2(s) at the point (1, 0, 4) are a(1, -1, 2) and b(-1, 1, 4).

The angle of intersection θ can be obtained by the formula

cos θ = (a.b)/||a||||b||

cos θ = 6/√108

cos θ = 1/√3

θ = 55 (correct to the nearest degree)

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