# Polar Coordinates Tangent line

• PsychonautQQ
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. ehilddy/dθ = 6cos(π/12)cos(π/2) + sin(π/2)cos(π/12)dx/dθ = 1/2(5cos(5π/12) + 7cos(

## Homework Statement

I don't know how to make theta so
∅ = theta.
find the slope of the tangent line at
r = sin(6∅) when ∅ = pi/12

## Homework Equations

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

## 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 :)

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.

CAF123 said:
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.

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

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

Precisely.

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

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

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'.

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CAF123 said:
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'.

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

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

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

#### Attachments

• polarcurve.jpg
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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)

CAF123 said:
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)

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

PsychonautQQ said:
slope = dy/dx = (dy/dθ) / (dx/dθ)

y = rsinθ
x = rcosθ

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

• 1 person
ehild said:
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

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.

ehild said:
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

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 ;-(

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

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

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