Tangent slope with polar coordinate

In summary, the conversation involves a student seeking help with a math problem and discussing potential bugs in their solution. The conversation also touches on the concept of polar coordinates and transforming to Cartesian coordinates. The student ultimately realizes their mistake in using a negative value for the polar coordinate, which resulted in half of the curve being incorrect.
  • #1
yecko
Gold Member
279
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
  • #2
How much is ##-{1\over 2} -{1\over 2} ## ?
 
  • #3
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.
 
  • #4
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: 348
  • #5
yecko said:
I have corrected it in the following steps already... .
No you haven't.
 
  • #6
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: 430
  • #7
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:
  • #8
(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:

FAQ: Tangent slope with polar coordinate

What is the equation for finding the slope of a tangent line with polar coordinates?

The equation for finding the slope of a tangent line with polar coordinates is m = (dy/dθ) / (dx/dθ), where dy/dθ and dx/dθ are the derivatives of the polar coordinates.

How is the slope of a tangent line with polar coordinates different from Cartesian coordinates?

The slope of a tangent line with polar coordinates is different from Cartesian coordinates because the x and y values are expressed in terms of angle and radius, rather than horizontal and vertical distances.

Can the slope of a tangent line with polar coordinates be negative?

Yes, the slope of a tangent line with polar coordinates can be negative. This occurs when the tangent line is below the x-axis, meaning the angle is in the third or fourth quadrant.

What is the significance of the slope of a tangent line with polar coordinates?

The slope of a tangent line with polar coordinates represents the rate of change of the radius with respect to the angle. It can also be interpreted as the instantaneous rate of change of the polar coordinates.

How can the slope of a tangent line with polar coordinates be used in real-life applications?

The slope of a tangent line with polar coordinates is used in various fields such as physics, engineering, and navigation to calculate rates of change, velocities, and accelerations. It is also used in computer programming to create smooth and accurate curves in graphics and animation.

Back
Top