Find Tangent Line to Path: x(t)=4costi-3sintj+5tk, t=\pi/3

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SUMMARY

The discussion focuses on finding the tangent line to the path defined by the vector function x(t) = 4cos(t)i - 3sin(t)j + 5tk at t = π/3. The derivative x'(t) is calculated as x'(t) = -4sin(t)i - 3cos(t)j + 5k. Substituting t = π/3 into the derivative yields x'(π/3) = -4(√3/2)i - (3/2)j + 5k, confirming the correctness of the approach. The user expresses confusion regarding Calculus III concepts but receives affirmation that their calculations are accurate.

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  • Understanding of vector functions in three-dimensional space
  • Knowledge of derivatives and their application to parametric equations
  • Familiarity with trigonometric functions and their values at specific angles
  • Basic proficiency in Calculus III concepts
NEXT STEPS
  • Study the concept of tangent vectors in vector calculus
  • Learn how to apply the chain rule to parametric equations
  • Explore the geometric interpretation of derivatives in three dimensions
  • Practice problems involving vector functions and their derivatives
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Students in Calculus III, particularly those struggling with vector functions and derivatives, as well as educators seeking to clarify these concepts for their students.

rgalvan2
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Find an equation for the line tangent to the given path at the indicated value for the parameter.

x(t)=4costi-3sintj+5tk, t=[tex]\pi[/tex]/3

So what I did here was take x'(t) and then plugged in [tex]\pi[/tex]/3 after that to get an equation containing i, j, and k.
x'(t)= -4sinti-3costj+5k
x'([tex]\pi[/tex]/3)=-4[tex]\sqrt{3}[/tex]/2i-3/2j+5k
Am I doing this right? Calc III is confusing me so much. Any help would be appreciated. This is due Thursday. Thanks!
 
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Looks all right so far.
 
Alright sweet thanks a lot!
 

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