Tangent slope with polar coordinate

Click For Summary
SUMMARY

The discussion focuses on solving a problem related to tangent slopes in polar coordinates, specifically addressing the function r=f(θ). The user initially arrived at option d instead of the correct answer e due to a miscalculation involving the expression -1. Participants pointed out that the polar coordinate r cannot be negative, which affects the curve's representation. A suggestion was made to convert the polar coordinates to Cartesian coordinates for easier analysis.

PREREQUISITES
  • Understanding of polar coordinates and their properties
  • Familiarity with Cartesian coordinate transformations
  • Knowledge of calculus, particularly derivatives and slopes
  • Ability to interpret and manipulate mathematical expressions
NEXT STEPS
  • Study the properties of polar coordinates and their graphical representations
  • Learn how to convert polar equations to Cartesian coordinates
  • Explore the concept of tangent slopes in calculus
  • Practice solving problems involving polar curves and their domains
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and tangent slopes, as well as educators looking for examples of common pitfalls in mathematical problem-solving.

yecko
Gold Member
Messages
275
Reaction score
15

Homework Statement


http://i.imgur.com/4FPnTNS.jpg
4FPnTNS.jpg


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.)
 
Physics news on Phys.org
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

  • r(theta).jpg
    r(theta).jpg
    54.2 KB · Views: 390
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

upload_2017-3-13_16-26-34.png
 

Attachments

  • upload_2017-3-13_16-21-3.png
    upload_2017-3-13_16-21-3.png
    4.5 KB · Views: 459
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.
 
Last edited:
(Edit)
Unless I blundered, it looks quite tractable if you start by transforming to cartesian co-ordinates.

Edit: I did and it isn't
 
Last edited:

Similar threads

Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
16K
Replies
1
Views
2K