Vector valued functions: finding tangent line

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SUMMARY

The discussion focuses on finding the unit tangent vector T(t) and the parametric equations for the tangent line to the space curve defined by r(t) = <2sin(t), 2cos(t), 4sin²(t)> at the point P(1, √3, 1). The user correctly derived T(t) using the formula T(t) = r'(t)/||r'(t)||, with r'(t) calculated as <2cos(t), -2sin(t), 8sin(t)cos(t)>. However, the user expressed confusion regarding the application of point P in deriving the parametric equations for the tangent line.

PREREQUISITES
  • Understanding of vector calculus, specifically unit tangent vectors.
  • Familiarity with parametric equations and their general forms.
  • Knowledge of derivatives and their applications in vector functions.
  • Basic trigonometric identities and algebraic manipulation.
NEXT STEPS
  • Study the derivation of parametric equations for lines in vector form.
  • Learn about the application of the point of tangency in vector calculus.
  • Explore the properties of unit tangent vectors in three-dimensional space.
  • Review the concepts of derivatives in the context of vector-valued functions.
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Students studying calculus, particularly those focusing on vector-valued functions and their applications in physics and engineering. This discussion is also beneficial for educators teaching these concepts.

bfusco
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Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ 2+8sin(t)cos(t)
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
 
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bfusco said:

Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ sqrt([STRIKE]2[/STRIKE] 4+8sin(t)cos(t))
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
You have left out the square root for ||r' ||, as well as there is another error as noted above.

Do you know the general form of a parametric equation of a line in vector form?
 

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