Find the slope of the tangent at the given angle theta

  • Thread starter Calpalned
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  • #1
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Homework Statement


Find the slope of the tangent line to the give polar curve at the point specified by the value of theta
R = 1/θ, θ = π

Homework Equations


Slope of a polar equation is (dR/dθsinθ+Rcosθ/dR/dθcosθ-Rsinθ)

The Attempt at a Solution


Using my calculator, I plugged π for θ, -.101 for DR/Dtheta (after differentiating R) and R = 1/π
I got π/12 as my answer, but the correct answer is -π. Why am I off by a factor of 12?
 

Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


Find the slope of the tangent line to the give polar curve at the point specified by the value of theta
R = 1/θ, θ = π

Homework Equations


Slope of a polar equation is (dR/dθsinθ+Rcosθ/dR/dθcosθ-Rsinθ)

The Attempt at a Solution


Using my calculator, I plugged π for θ, -.101 for DR/Dtheta (after differentiating R) and R = 1/π
I got π/12 as my answer, but the correct answer is -π. Why am I off by a factor of 12?
You'll need to show the details of how you got π/12. I don't see where the factor of 12 could come from.
 
  • #3
34,687
6,393

Homework Statement


Find the slope of the tangent line to the give polar curve at the point specified by the value of theta
R = 1/θ, θ = π

Homework Equations


Slope of a polar equation is (dR/dθsinθ+Rcosθ/dR/dθcosθ-Rsinθ)
Your formula is pretty much unreadable. I have no idea what this means.
[QUOTE="Calpalned"

The Attempt at a Solution


Using my calculator, I plugged π for θ, -.101 for DR/Dtheta (after differentiating R) and R = 1/π
I got π/12 as my answer, but the correct answer is -π. Why am I off by a factor of 12?[/QUOTE]
In general we have x = rcos(θ) and y = rsin(θ). Differentiate both with respect to θ to get dy/dx. When I do this, I get the answer you're supposed to get.
 
  • #4
297
6
[itex]\frac{(dr/dθ)sinθ + rcosθ}{(dr/dθ)cosθ - rsinθ}
frac{(dr/dθ)sinθ + rcosθ}{(dr/dθ)cosθ - rsinθ}
[itex]\frac{([itex]\frac{dr}{dθ}sinθ + rcosθ}{([itex]\frac{dr}{dθ}cosθ - rsinθ}
[itex]frac(dr)
 
  • #5
34,687
6,393
##\frac{(dr/dθ)sinθ + rcosθ}{(dr/dθ)cosθ - rsinθ}##
Fixed.
What I typed was # #\frac{(dr/dθ)sinθ + rcosθ}{(dr/dθ)cosθ - rsinθ}# # (with no extra spaces between the # pairs).

The LaTeX tags come in pairs, with [ itex ] or [ tex ] at the beginning, and [ /itex ] or [ /tex ] at the end (omit the extra spaces. I prefer to use # # at beginning and end (again without the extra spaces, for inline LaTeX, or $ $ at beginning and end for standalone stuff.

You have r = 1/θ. When you write x = r cos(θ) and y = r sin(θ), replace r in each equation by 1/θ. Then take your derivatives. Alternatively, you could calculate dr/dθ from r = 1/θ.
 

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