Finding the angle between two tangent vectors

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To find the angle between the tangent vectors of the curves r1(t) and r2(s) at the intersection point (1, 0, 2), the first step is to determine the parameters t and s that yield this point, resulting in t=1 and s=1. The tangent vectors are found by differentiating the curves, leading to r1'(1) = <0.841, 0, 0> and r2'(1) = <1, 0, 4>. The magnitudes of these vectors are calculated, with |r1'(1)| = 0.841 and |r2'(1)| = √17. Finally, the cosine of the angle between the two tangent vectors can be computed using the dot product and the magnitudes of the vectors.
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Homework Statement


Consider the two space curves
r1(t) = <cos(t − 1), t^2 − 1, 2t^4>
r2(s) = <1 + ln s, s^2 − 2s + 1, 2s^2>,
where t and s are two independent real parameters.
Find the cosine of the angle between the tangent vectors of the two curves at the intersection point
(1, 0, 2).

Please show me steps..thank you!


Homework Equations





The Attempt at a Solution


I set cos(t-1)=1 and got t=1.
In the same manner, I got s=1.
But I'm not sure how to get r'(1)...I'd appreciate any help on this!
 
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You find a tangent vector to a curve by differentiating the curve, don't you?
 
Yeah, I differentiated it and got r'(1)=<0.841,0,0> and |r'(1)|=0.841, which seems like an odd number...I wanted to confirm that I did it right.
I also got r2'(1)=<1,0,4> and |r2'(1)|=sq.root17.
Is this right? And I just set them over each other to get the tangent vector right?
 
0.841=sin(1). Seems ok so far. You set them 'over each other' to get two unit tangent vectors. Then what?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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