Derivative of parametric function

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SUMMARY

The discussion focuses on finding the tangent line to the parametric function defined by x = cos(t) and y = √3 cos(t) at the point corresponding to t = 2π/3. The slope of the tangent line, calculated as dy/dx, is confirmed to be √3, leading to the tangent equation y = √3x. Additionally, the second derivative d²y/dx² is determined to be 0, as dy/dx is a constant, confirming that the tangent line is horizontal at the specified point.

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of derivatives and the quotient rule
  • Familiarity with trigonometric functions and their derivatives
  • Ability to interpret and manipulate calculus equations
NEXT STEPS
  • Study the application of the quotient rule in calculus
  • Explore the concept of parametric derivatives in more depth
  • Learn about the implications of constant derivatives in tangent lines
  • Investigate the geometric interpretation of second derivatives
USEFUL FOR

Students studying calculus, particularly those focusing on parametric functions and derivatives, as well as educators seeking to clarify concepts related to tangent lines and their properties.

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Homework Statement


Find the line tangent to the point 2pi/3 when x=cost y=sqrt3 cost.
Also find the value of d2y/dx2 at the point given.


Homework Equations


I found dy/dx to be -sqrt3 sint/ -sint. I found that to be just sqrt3. This matched what my calculator told me the slope was. The equation of the tangent is y=sqrt3 x or at least I think.

I am having trouble with the second derivative. I believe that it is 0. When I applied the quotient rule to dy/dx I got sqrt3 sintcost-sqrt3 sintcost/sin2t. I then multiplied by dt/dx. Either way, the numerator is still 0.

Am I right, on the right path, or completely wrong.
 
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I think you're right. d2y/dx2 = 0
 
Yes that is correct. You can get the same answer easily by noting that since dy/dx = constant. Hence d/dx (dy/dx) = 0.
 

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