Tangent slope with polar coordinate

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yecko
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Homework Statement


http://i.imgur.com/4FPnTNS.jpg
Handwritten derivation of the slope of a polar curve r = cos θ − sin 2θ at θ = π/4, with f(θ), f′(θ) and the slope quotient worked through to √2/(2−√2), next to multiple-choice options (a) to (e).


Homework Equations


(Written in above photo)

The Attempt at a Solution


(Written in above photo)

I have tried hard in figuring out what's wong I have done done, but what I finally got is still option d instead of the model answer e.
Are there any bugs and where are they?
Thank you very much.

(Please forgive me for writing out all the thing because of the difficulty in typing all those expressions.)
 
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How much is ##-{1\over 2} -{1\over 2} ## ?
 
BvU said:
How much is ##-{1\over 2} -{1\over 2} ## ?
It is -1 which I forgot to change in the last line of the step, but I have corrected it in the following steps already... It shouldn't be a problem to the answer...

Whats other bug? thank you.
 
yecko said:
It is -1 which I forgot to change in the last line of the step, but I have corrected it in the following steps already... It shouldn't be a problem to the answer...

Whats other bug? thank you.
Try to draw the curve r=f(θ). Is θ=pi/4 in the domain?
 
Attachments
  • Polar plot in black on a grid of concentric circles and radial lines, angles labelled in degrees and radial axis labelled B from 0 to 2.0; the curve has a small loop near 165 to 180 degrees and a larger loop sweeping from 0 to about 315 degrees.
    r(theta).jpg
    54.2 KB · Views: 421
yecko said:
I have corrected it in the following steps already... .
No you haven't.
 
ehild said:
Try to draw the curve r=f(θ). Is θ=pi/4 in the domain?
[EDIT]Typo. I see what you mean -- and what the exercise composer means

Blue polar curve on x-y axes from -0.5 to 2 and -1.5 to 1.5, with a self-intersecting loop pattern and a red dot near (-0.2, -0.2).
 
Attachments
  • Graph of a polar curve on x and y axes from -2 to 0.5 and -1.5 to 1.5, with a closed loop shape having several petals and a red dot marking one point on the lower left petal.
    upload_2017-3-13_16-21-3.png
    4.5 KB · Views: 487
The polar coordinate r can not be negative as it is the distance from the origin.Half of your curve resulted from negative r=f(θ). The problem maker forgot about that.
 
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(Edit)
Unless I blundered, it looks quite tractable if you start by transforming to cartesian co-ordinates.

Edit: I did and it isn't
 
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