Finding the Tangent of a point on a curve (problem driving me crazy )

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SUMMARY

The discussion centers on finding the tangent vector at a specific point on a curve defined by the parametric equations r(t) = (1 + 4√t, t^5 - t, t^5 + t). The user is attempting to compute the derivative r'(t) = (2/√t, 5t^4 - 1, 5t^4 + 1) but encounters an undefined value when substituting t = 0. The correct parameter value to find the tangent at the point (5, 0, 2) is t = 1, not t = 0, which resolves the confusion regarding the evaluation of the tangent vector.

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Homework Statement


I've done this problem about 5 times, several different ways following many different examples but I can't seem to find the right answer or even figure out how they got one of the answers (y answer is given)

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The Attempt at a Solution



r(t) = (1+4*sqrt(t), t5-t, t5+t)

s(t) = P0 + tv (P and v are vectors)
s(t) = r(0) + tr1(0)

r1(t) = (2/sqrt(t), 5t4-1, 5t4+1)
r1(0) = ?
and this is where i get stuck because trying to substitute 0 in x equation will give you an undefined number...so I'm doing something wrong obviously...please help!
 
Physics news on Phys.org
You want to find t such that (x,y,z)=(5,0,2). That's t=1, right? Why do you say t=0?
 

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