Question

A Cardiod in a diamond shaped box. r = 1 - Sin(theta)
How do you find the dimensions of the diamond around the cardoid? Area?

Answer 1

A similar problem was posted several weeks back. I don't know that we reached a conclusion. I am posting the link to that particular thread in the event you want to take a look at it. @rmrjr22

http://openstudy.com/study#/updates/511b3fd8e4b03d9dd0c38459

Answer 2

yes, this was me. No conclusion was reached

Answer 3

Can you help me @jim_thompson5910 ?

Answer 4

does it give you where those points of tangency are?

Answer 5

no.... just that.

Answer 6

well you would at least need to know the points of tangency to find the distance between each point, which you could use to find the lengths of the sides

or you would need to know the slopes of each side of the rhombus, which you could then figure out where those points of tangency are (then use the method given above)

Answer 7

seems like something is missing

Answer 8

to find the tangents, would you need to find the derivatives of x and y? x solved ='s horizontal tangent?

Answer 9

yeah you would need to convert to cartesian form too

but you would also need to know the slopes of the lines that make up the diamond

Answer 10

I had a prior question which is up above in the first link. Which is before this question in particular.

Answer 11

ok let me look at that real quick

Answer 12

x = rcos(theta)
y = rsin(theta)
(1-sin(theta)(cos(Theta)) = x
d/dx = (cos(theta)-sin(theta)Cos(theta)

Answer 13

ok that one is easy because the slopes of the lines are known: 0 or undefined

Answer 14

and I got Pi/6....5pi/6,-1/2.....-/2

Answer 15

but this one, the slopes aren't given or can be figured out easily

Answer 16

can't*

Answer 17

which is where I am stuck on...not sure how to approach it

Answer 18

yeah that's what I'm trying to figure out too

which is making me think there's missing info

Answer 19

I promise there is no more info. Others students and I are at a loss. Teacher says, it is solvable...

Answer 20

If anything we approuched it by inserting a square inside the Cardiod to create triangles, in which can be measured into as the radius.

Answer 21

Inside the diamond, i meant

Answer 22

hmm sorry, i'm completely stumped on this one

I still think there's something missing, but there's probably a way to do it without the missing info

Answer 23

if you knew the angles theta1 and theta2 (that correspond to r1 and r2 shown below)

then you can use the law of cosines to find x

Answer 24

Ok... Thank you though. Its been driving me mad!
What if you make a triangle inside the diamond? Its a 90 degree angle that if taken in half = 45 degrees. Arc tan (1) equals PI/4. We use one bc it is the length is 1.
A^2+b^2 = c^2
Which yields 1^2+1^2 = sqrt(2)

Answer 25

how do you know the angles of this diamond are 90 degrees?

Answer 26

its essenstually a square really. In the problem its a "square shipping box"....

Answer 27

this says diamond

the previous problem was a square

Answer 28

so they're the same problem?

Answer 29

previous problem was a bigger square. The way the box is initally set up, the width of the box would be the horizontal distance from the furthest point. If you take the verticle distance of the heart, it is not the same. So dimensions of said box are different

Answer 30

well all I know is that this says diamond, which is a rhombus (not a square)

Answer 31

Ok, if its a sideways square then how would you approach that?

Answer 32

hmm not sure, maybe we can figure out the diagonals somehow

Answer 33

@rmrjr22 I mentioned this problem to my former AP Calc BC teacher. I am posting his reply here in the event it has relevance to your work.
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Polar Arc Length is in BC syllabus and the formula is "Integral from a to b (in this case 0 to 2 pi) of square root of ((r^2) + (der of r)^2) d theta". My TI-89 says "8".

Later, followed up with the following:

Just did the problem by hand and got "8". The formula for Arc Length for Polar Curves is on page 736 of the Anton 8th ed. book that we still use.