Question

How do you find the maximum r value in a polar equation?

Answer 1

I like to recall the ranges of the trig functions
like sin(x) is between -1 and 1
and cos(x) is between -1 and 1
the max value of sin(x) is 1
the max value of cos(x) is 1
the max value of 1-sin(x) is 2 because sin(x) has min -1 and 1-(-1)=2
---
do you have a certain problem you want to look at ?

Answer 2

Is the equation in polar form? You'd maximize it like any other function using critical points.

Answer 3

\[r=6(\sqrt{2}-2\sin \theta) \]
and could we also find the zeroes of r that are between 0 and 2pi?

Answer 4

i would use the min value of sin being -1
since you have subtraction to find the max value of r

Answer 5

How do I find the max value though? I still don't understand.

Answer 6

Well, we can actually do this one without calculus.

Answer 7

In this case, we want to minimize \(\sin\theta\).

Answer 8

Which would be at \(\sin\theta = -1\)

Answer 9

So -1, like freckles said?

Answer 10

Ok, whats next?

Answer 11

plug in

Answer 12

to find max r

Answer 13

plug in as in put in calculator?

Answer 14

no calculator needed

Answer 15

\[r=6(\sqrt{2}-2\sin \theta) \]
the lowest sin(theta) is -1

Answer 16

so replace sin(theta) with -1

Answer 17

Ohhh, ok. So the answer would be \[6\sqrt{2}+12\]?

Answer 18

yah

Answer 19

Thanks

Answer 20

How do you find the zeroes of r though?

Answer 21

you also had another question somewhere up there

Answer 22

there it is

Answer 23

yeah haha

Answer 24

let r be 0

Answer 25

and solve for theta

Answer 26

\[0=6 (\sqrt{2}-2 \sin(\theta))\]

Answer 27

this equation might look easier to you if just go ahead and divide both sides by 6

Answer 28

I got \[\sin \Theta = \sqrt{2}/2\]

Answer 29

sounds good

Answer 30

Ok thank you

Answer 31

one of my favorite equations
(because I can use my handy dandy unit circle to find such thetas)

Answer 32

Yeah i need to memorize my unit circle for our test tomorrow. Any tips?

Answer 33

I remember having trouble with that in trig
and then I was like I know I will just remember the first quadrant
and basically repeat it for all the other quadrants
and remember that first quadrant we have (+,+)
second (-,+)
third (-,-)
fourth (+,-)

Answer 34

I wasn't as good with fractions back then as I am now... but I had to figure out the special angles by doing
\[0 \\ \frac{\pi}{6} \\ \frac{\pi}{4} \\ \frac{\pi}{3} \\ \frac{\pi}{2} \\ \frac{\pi}{2}+\frac{\pi}{6} \\ \frac{\pi}{2}+\frac{\pi}{4} \\ \frac{\pi}{2}+\frac{\pi}{3} \\ \pi \\ \pi+\frac{\pi}{6} \\ \pi+\frac{\pi}{4} \\ \pi+\frac{\pi}{3} \\ \frac{3\pi}{2} \\ \frac{3\pi}{2}+\frac{\pi}{6} \\ \frac{3\pi}{2}+\frac{\pi}{4} \\ \frac{3\pi}{2}+\frac{\pi}{3} \\ 2\pi\]
well that is how I would do it back then in radians

Answer 35

like to figure out what the special angles were

Answer 36

\[(\cos(\theta),\sin(\theta)) \\ (1,0) \\ (\frac{\sqrt{3}}{2},\frac{1}{2}) \\ (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) \\ (\frac{1}{2},\frac{\sqrt{3}}{2}) \\ (0,1) \\\]
then when you get to that point the x's right while the y's fall
except the x's will be negative and the y's will be positive

Answer 37

I see. I have the hardest time when I am to do something like sqrt(2)/2 and then find where those are zeroes, like in the last problem we did

Answer 38

those points that I was going to list after (0,1) was going to be points in the second quadrant

Answer 39

well you know that the y-coordinate is sin(theta)?

Answer 40

ok

Answer 41

do you know how to read the unit circle
it seems like most people have trouble reading it

Answer 42

like you look for when the y-coordinate is sqrt(2)/2
and then grab the angle that corresponds to y values with sqrt(2)/2

Answer 43

Yes i know how to

Answer 44

for your question above

Answer 45

I think so... would the answers be pi/4 and 7pi/4

Answer 46

sin is positive in the 1st and 2nd quadrant

Answer 47

because y is positive in the 1st and 2nd quadrant

Answer 48

7pi/4 won't do because it is in the 4th
y is negative in the 4th

Answer 49

so then a correct choice would be pi/4 and 3pi/4 ?

Answer 50

yep