# Polar Coordinates Tangent line

1. Sep 6, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
I don't know how to make theta so
∅ = theta.
find the slope of the tangent line at
r = sin(6∅) when ∅ = pi/12

2. Relevant equations
y=rsin(6∅)
x=rcos(6∅)
r=sin(6∅)
tangent line equation
y-y' = m(x-x')
m = dy/dx

3. The attempt at a solution
when ∅ = pi/12 then sin(6∅) = 1
so
r=1
x=rcos(6∅) = 0
y = rsin(6∅) = 1

so If I did the previous parts correctly, then I'm now attempting to find the slope (m).
I do not know how to find it.

m = dy/dx =EITHER= (dy/dr) / (dx/dr) or (dy/d∅) / (dx/d∅). Either way I wouldn't know how to take the derivatives if I did know which one to choose.

Am I doing this correctly? Am I on the right track? Thanks :)

2. Sep 6, 2013

### CAF123

You do not need calculus for this question. As you observed, when ∅=pi/12, sin(6∅)=1. Now recall the sin graph and this should give you the answer.

Alternatively, you know r=r(∅) and you want r'(∅=pi/12).

theta should also be in quick symbols - look for θ.

3. Sep 6, 2013

### PsychonautQQ

so when r=sin6θ=1 that is when the sin function is at a max. therefore the slope is zero?

4. Sep 6, 2013

Precisely.

5. Sep 6, 2013

### PsychonautQQ

why when I type zero into my online homework for the slope of the tangent curve if says i'm wrong ;-(

6. Sep 7, 2013

### CAF123

Not sure, try '0' instead of typing zero. Have you tried it again using calculus?

If you parametrize the curve r, writing expressions for x and y, then your method in the OP is also good.
$(dy/dx) = (dy/d\theta) (d\theta/dx)$, but it is probably simpler to just find r'.

Last edited: Sep 7, 2013
7. Sep 7, 2013

### PsychonautQQ

it's not accepting 0 as an answer either ;-( booo

8. Sep 7, 2013

### ehild

r is maximum as function of theta, which does not mean that the slope of the curve is zero. See picture.

The equation of the curve in polar coordinates is r=sin(6θ). How would you write the Cartesian coordinates x,y as functions of θ?

ehild

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Last edited: Sep 7, 2013
9. Sep 7, 2013

### CAF123

Wow, somehow I completely ignored that ∅ was the polar angle.
What I suggested in my opening post is wrong because you are not plotting r versus ∅. r and ∅ are polar representation.

Likewise, evaluating r'(∅=pi/15) here is incorrect because we are not plotting r versus theta.

Do as I said in my last post and parametrise the curve to get eqns for x and y. The find dy/dx.

(Thanks ehild for letting me know of my mistakes)

10. Sep 9, 2013

### PsychonautQQ

slope = dy/dx = (dy/dθ) / (dx/dθ)
I went in and saw my teacher at this part but I lost the sheet she wrote on because I'm a noob ;-(. But it went something like dy/dr(some sin's and cosines) / dx/dr(some sin's and cosines)
but I have no idea what she did now or how she did it or why it made sense at all. Any help appreciated.

y = rsinθ
x = rcosθ

11. Sep 9, 2013

### ehild

It is given that r=sin(6θ). So the Cartesian coordinates are x=sin(6θ)cos(θ) and y=sin(6θ)sin(θ).
Derive both of them with respect to θ. The slope is dy/dx=(dy/dθ)/(dx/dθ).

ehild

12. Sep 9, 2013

### PsychonautQQ

you are so beautiful!! I love this website so much. I'll learn LaTex so soon!! 18 credits (4 physics classes + calculus three) is quite demanding and you have no idea how grateful I am I discovered this website and all of you!! You're all so patient with me it's amazing.

13. Sep 10, 2013

### PsychonautQQ

dy/dθ = 6sinθcos(6θ) + sin(6θ)cos(θ)
dx/dθ = 1/2(5cos(5θ) + 7cos(7θ)

(dy/dθ)/(dx/dθ) = (6sinθcos(6θ) + sin(6θ)cos(θ)) / (1/2(5cos(5θ) + 7cos(7θ))
when θ = ∏/12...
(6sin(∏/12)cos(6∏/12) + sin(6∏/12)cos(∏/12)) / (1/2(5cos(5∏/12) + 7cos(7∏/12))
(0 + .9659258) / (.6470476128 + -1.8117333)
= -.8293446126
Which my online thing says is wrong ;-(

14. Sep 10, 2013

### ehild

Are you sure you did it correctly? Try again. Leave it in terms in of 6θ and θ. As 6θ=pi/2, both derivatives become very simple at θ=pi/12.

ehild

Last edited: Sep 10, 2013