Calculus III Parametric Equations

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SUMMARY

This discussion focuses on solving problems related to parametric equations in Calculus III, specifically addressing homework questions 4, 5, and 6. The participant has successfully demonstrated that the curve passes through the points (3, 0) for t = ±√3 and has derived the tangent line for t = √3 as y = √3x - 3√3. However, they have not completed the analysis for t = -√3. The discussion also highlights the conditions for horizontal and vertical tangents, which occur when y' = 3t² - 1 = 0 and x' = 2t = 0, respectively.

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of derivatives and their applications
  • Familiarity with tangent lines in calculus
  • Ability to solve cubic equations
NEXT STEPS
  • Review the concept of parametric equations in depth
  • Learn how to find horizontal and vertical tangents in parametric curves
  • Study the process of deriving tangent lines for parametric equations
  • Explore cubic equation solutions and their graphical interpretations
USEFUL FOR

Students studying Calculus III, particularly those working on parametric equations and tangent line analysis, as well as educators seeking resources for teaching these concepts.

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Homework Statement


I've uploaded a scan of the questions. Questions 4, 5, and 6 are given in the 3 files uploaded. They all come from the given information from the first scan of the problem.


Homework Equations





The Attempt at a Solution


I've worked everything I could on paper; what am I missing. How should I go about finishing the problem and am I correct so far?

Any good links for parametric equations also?

Thanks for the help!
 

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For part 3, you have correctly shown that the curve passes through (3, 0) for t=\sqrt{3} and for t= -\sqrt{3}. I confess that I don't understand the rather complicated way you do that! It is sufficient to show that y= t^3- 3t= 0 has roots 0, \sqrt{3} and -\sqrt{3} and that only the last two also satisfy x= t^2= 3.

You have shown that the tangent line at t= \sqrt{3} is y= \sqrt{3}x- 3\sqrt{3} but you don't seem to have completed the problem for t= -\sqrt{3}.

For (4), asking where the tangent lines are horizontal or vertical, again, you have again written a complicated calculation starting with writing x^2+ y^2 in terms of t. I can see no reason for that. Horizontal tangents occur where y&#039;= 3t<sup>2<sup>-1= 0</sup></sup> and vertical tangents occur where x&#039;= 2t= 0.
 
Thanks HallsofIvy, once again, for the great help. These forums are great to get the wheels turning!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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