Is this correct. Tangent lines.

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the vector function r(t) = sin(t) i + cos(t) j + t k at the point (0, 1, 0). By substituting t = 0 into r(t), the point is confirmed as (0, 1, 0). The derivative r'(t) is calculated as , leading to r'(0) = <1, 0, 1>. The resulting parametric equations for the tangent line are x = t, y = 1, and z = t, confirming the correctness of the calculations.

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Homework Statement



r(t) = sin t (i) + cos t (j) + t k


Find the equation of the line tangent to r(t) at the point (0,1,0)

If you plug in 0 into to for r(t), you get (0,1,0). Thus t must equal o.

To find the vector of r' (t) or the derivative of r(t)

this equals = < cos t, - sin t, 1 > which also equals cost (i) - sin t (j) + k


So I find r'(0) = < 1,0,1>


Then my equations are

x= 1 t + 0

y = 0t + 1

z = 1t + 0


x= 1t

y = 1

z = 1t
 
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Yes, that is correct.
 

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