Finding the Slope of a Polar Curve at a Given Point

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SUMMARY

The discussion focuses on finding the slope of the tangent line to the polar curve defined by the equation r² = 9 sin(3θ) at the point (3, π/6). The relevant formula for calculating the slope is dy/dx = (r cos θ + sin θ dr/dθ) / (-r sin θ + cos θ dr/dθ). The user successfully derived r = 3√(sin(3θ)) but encountered challenges with the derivative due to the complexity of the square root of a trigonometric function. The discussion highlights the importance of considering both positive and negative values of r in polar coordinates, which can lead to mirrored functions.

PREREQUISITES
  • Understanding of polar coordinates and curves
  • Familiarity with trigonometric functions and their derivatives
  • Knowledge of implicit differentiation techniques
  • Experience with polar equations and their graphical representations
NEXT STEPS
  • Study the derivation of polar coordinates and their tangent lines
  • Learn about the implications of negative values in polar equations
  • Explore graphing polar curves using software like Desmos or GeoGebra
  • Investigate advanced differentiation techniques for trigonometric functions
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Students studying calculus, particularly those focusing on polar coordinates and derivatives, as well as educators looking for examples of polar curve analysis.

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Homework Statement


Find the slope to the tangent line to the polar curve r^2 = 9 sin (3θ) at the point (3, π/6)


Homework Equations



dy/dx = (r cos θ + sin θ dr/dθ)/(-r sin θ + cos θ dr/dθ)

The Attempt at a Solution



So I have no issues with taking r^2 = 9 sin (3θ) and taking the root to get r = 3√(sin 3θ)
and then subbing that into the equation.

My problem is that it's only one case, and the derivative of a square root of a trig equation is messy. I think I'm missing something - any assistant would be appreciated.
 
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(i.e. what about the case where r = -3sqrt(sin 3theta)?
 
I think this is one of those things where only one of the answers you get will make sense with the situation:

If r can equal +- (stuff) that means the function is mirrored about the center of your polar function. I didn't graph it but I think this is like a clover thing, right? Lemme draw it.

It's not very well drawn, but I think it gets the point across. Polar functions are weird lol
 

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You were right. I followed it through, and because the two were opposites of each other, I had one case where it was a negative divided by a positive, and the other was a positive divided by a negative, getting the same answer.

Thanks!
 
No problem
 

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