Question

Please help. I have a question like this on a upcoming midterm and I can't seem to find the answer.
Given Polar equation of curve C.
C: r= sinθ 0<= θ <= pi
Find the point, on C such that:
Tangent line: a)Horizontal b)Vertical

Answer 1

so far I have x= sinθcosθ, y= sinθsinθ-> sin^2(θ)
dx = cos(2θ) dy = sin(2θ)
\[\frac{ dy }{ dx} =\frac{ dy/dθ }{ dx/dθ}=\frac{ \sin(2θ) }{ \cos(2θ) }\]
a) horizontal :
sin(2θ) = 0

Answer 2

tan?

Answer 3

I don't seem to understand what you mean by "tan"

Answer 4

sin/cos = tan

Answer 5

I don't think it can work that way.

Answer 6

k i just woke up so bear with me

Answer 7

It's cool, no problem. I say it won't because I need to find the horizontal and vertical lines

Answer 8

I feel like your question changed

Answer 9

nvm

Answer 10

How about L'hopitals?

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx

Answer 12

we know that
x = r cos theta,
but r is a function of theta itself
x = r(theta) * cos theta
y = r(theta) * sin theta
. ok im going to use t instead of theta.
x = r(t) * cos(t)
dx/dt = r'(t) * cos(t) + r(t) * (-sin(t))
dy/dt = r'(t)*sin(t) + r(t) * cos(t)

Answer 13

So product rule...ooooooh figures... Thanks!

Answer 14

but your approach is interesting too

Answer 15

Yours makes more sense haha. Thanks for the link!

Answer 16

i am trying to figure out what she was doing and looked it up but can't make sense

Answer 17

she's turning it into polar coordinates is what I gather

Answer 18

she did, since r= sin theta (that is given in the directions.)
then the conversion equation is
x = r cos theta, ---> x = sin theta * cos theta

Answer 19

then identity: sin (2*theta) = 2 sin theta cos theta
so it should be
x = sin (2theta) / 2

Answer 20

k now I see

Answer 21

so dx/dtheta = cos (2theta). that part is correct

Answer 22

and dy/dtheta = 2 sin(theta) * cos(theta) = sin (2theta) . that is correct too

Answer 23

now you could say , since tan x = sin x / cos x

then dy/dx = tan (2theta) . but its easier if we keep it as sin(2theta)/cos(2theta)
since then you can see that horizontal tangents occur when the numerator = 0, vertical tangents (slope undefined) is when denominator equal to zero.

Answer 24

yes

Answer 25

so either way, you can use paul's general formula or you can do it her way. should come out the same

Answer 26

The answers that are shown here are Horizontal: (0,0) and (1, pi/2)
Vertical: (sqrt(2)/2, pi/4) and (sqrt(2)/2, 3pi/4)

Answer 27

I cannot seem to know how to get that...

Answer 28

right, so you did sin (2theta) = 0

Answer 29

a) sin (2theta) = 0
sin^-1 (0) = 2*theta

Answer 30

by tan(x), it helps to tackle the problem intuitively because there are common graphs

Answer 31

right, they share zero points and undefined points

Answer 32

sin^-1(0) = 0, pi, 2pi, 3pi, ...
so
sin^-1(0) = 2x
gives us
0,pi,2pi,3pi,...= 2x
divide both sides by 2
0, pi/2, pi, 3pi/2, 2pi, 5pi/2, ...

Answer 33

i knew this polar coordinates is going to haunt me.

Answer 34

:)

Answer 35

so theta = 0, pi/2 (and then it repeats pi, 3pi/2, so we don't need those)

Answer 36

Same as vertical line right?

Answer 37

now plug back in to find x and y
x = r cos theta, x = (sin0) cos(0)
y = rsin theta y = sin(0) sin(0)

Answer 38

vertical line is when the denominator is zero , so solve cos(2theta) = 0

Answer 39

or alternatively, when tan(2theta) is undefined

Answer 40

So in other words, we are finding theta when dy/dx = 0 (horizontal slope)
and when dy/dx is undefined (vertical slope). Once we find theta, then we can plug theta into x = r cos theta, y = r sin theta to find the x and y coordinates

Answer 41

http://sites.csn.edu/istewart/mathweb/math127/polar_equ/polar_equ.htm

Answer 42

so theta = 0 , theta = pi/2 gives you horizontal slopes
theta = pi/4 , 3pi/4 gives you vertical

Answer 43

also we found earlier x = sin(2theta)/2 , y = sin^2(theta)

Answer 44

horizontal -> (0,0) (0,1)
vertical -> (1/2, 1/2) , (-1/2, 1/2)

Answer 45

a fun problem is , convert this polar equation to cartesian equation, this one is doable

Answer 46

r = sin theta
multiply both sides by r
r^2 = r sin theta ,
now we know r^2 = x^2 + y^2
and y = r sin (theta)
so we have
x^2 + y^2 = y
x^2 + y^2 - y = 0 , and complete the square you have equation of circle