Parameterized tangent line to a parameterized curve

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SUMMARY

The discussion centers on constructing a tangent line to a parameterized curve \( a(t) \) in \(\mathbb{R}^3\). The tangent line is defined as \( L = a(t) + s \cdot a'(t) \), where \( a'(t) \) represents the derivative of the curve at point \( t \). The key theorem referenced states that the slope of the tangent line is equal to the derivative of the curve, confirming that \( L(0) = a(t) \) establishes the point of tangency. This formulation is essential for understanding the geometric properties of curves in three-dimensional space.

PREREQUISITES
  • Understanding of parameterized curves in \(\mathbb{R}^3\)
  • Knowledge of derivatives and their geometric interpretation
  • Familiarity with vector operations
  • Basic concepts of calculus
NEXT STEPS
  • Study the properties of parameterized curves in \(\mathbb{R}^3\)
  • Learn about the Fundamental Theorem of Calculus as it relates to curves
  • Explore vector calculus, focusing on derivatives and tangent vectors
  • Investigate applications of tangent lines in physics and engineering
USEFUL FOR

Students of calculus, mathematicians, and anyone studying the geometric properties of curves in three-dimensional space will benefit from this discussion.

Cauchy1789
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Homework Statement



I seem to remember that a parameterized a(t) curve in [tex]\mathbb{R}^3[/tex] that one can construct the tangent from the slope of a'(t) and the curve itself.

such that the tangent line L = a(t) + s * a'(t) to a. This is supposedly a straight line in [tex]\mathbb{R}^3[/tex].
To make a long question. What theorem allows me to construct the tangent in such a way? confused:
 
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Cauchy1789 said:
I seem to remember that a parameterized a(t) curve in [tex]\mathbb{R}^3[/tex] that one can construct the tangent from the slope of a'(t) and the curve itself.

such that the tangent line L = a(t) + s * a'(t) to a. This is supposedly a straight line in [tex]\mathbb{R}^3[/tex].
To make a long question. What theorem allows me to construct the tangent in such a way? confused:

Hi Cauchy1789! :smile:

It's the theorem that says that the slope of the tangent euqals the derivative …

so, for fixed t, L(s) = s * a'(t) + constant …

and the constant has to be a(t) because L(0) = a(t). :wink:
 

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