Question

One more question for today...
Find the equation of the tangent line in Cartesian coordinates of the curve given polar coordinates by r=3sin(theta) at (theta)=pi/3

Answer 1

oh wait i should read sorry

Answer 2

that was totally wrong. i deleted it.

Answer 3

can't you take the derivative, then convert to cartesian?

Answer 4

graph is a circle with center at (1.5,0) and radius 1.5 yes?

Answer 5

Lets take the pts where tangent intersect the circle (x,y)

Now eq turns out be x^2 + y^2 >= (radius)^2

Here, radius = 3/2

Answer 6

i'm with you so far but this circle is
\[x^2+(y-\frac{3}{2})^2=\frac{9}{4}\] i htink

Answer 7

center (0,3/2) radius 3/2

Answer 8

bt how that y-3/2 comes up?

Answer 9

\[\sqrt{x ^{2}+ y ^{2}} = R\]

Answer 10

I did tht wrong sry.........3.sin pi/3 should be 3.(3)^1/2
----
2

Answer 11

(radius)^2 should be 27/4

Answer 12

this graph is in polar form yes? so when you graph
\[r=\sin(\theta) \] it traces out a circle starting at (0,0) going up to (1,0) and back down to (0,0)

Answer 13

bt its centre shoul be 0,0......am i right? satellite

Answer 14

the graph is in polar form, not rectangular. the rectangular form of this circle, unless i really messed up somewhere, is
\[x^2+(y-\frac{3}{2})^2=\frac{9}{4}\] a circle with radius [\frac{3}{2}\] and center at
\[(0,\frac{3}{2})\] i am fairly certain of this but let me get a book

Answer 15

a circle with center (0,0) and radius 1 would be
\[r=1\]

Answer 16

okay i might be wrong i haven't done question of polar with circle..... :)

Answer 17

ok i have my book open. the idea is not to switch to rectangular. the idea is that
\[\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]

Answer 18

which book is tht ?

Answer 19

which turns out to be
\[\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(x)-r\sin(\theta)}\]

Answer 20

that is what we need to compute

Answer 21

garden variety calc book. and i made a typo above. that
\[\cos(x)\] should be
\[\cos(\theta)\]

Answer 22

now just plug in the numbers and get the answer

Answer 23

but y do we differentiate everythng is constnt

Answer 24

\[\frac{dr}{d\theta}=3\cos(\theta)\] and if \[\theta=\frac{\pi}{3}\] then
\[r=\frac{\sqrt{3}}{2}\]

Answer 25

damn another typo. this is annoying to write

Answer 26

\[r=\frac{3\sqrt{3}}{2}\]

Answer 27

so we get a numerator of
\[3\cos(\theta)3\sin(\theta)+\frac{3\sqrt{3}}{2}\cos(\theta)\]

Answer 28

and a denominator of
\[3\cos^2(\theta) -\frac{3\sqrt{3}}{2}\sin(\theta)\]

Answer 29

now put
\[\theta =\frac{\pi}{3}\] to get the answer