calculus help anyone?

[hdeck]

this should be a fun challenge while helping me finish my homework at the same time

ok the problem :

find an equation (in terms of x and y) for the line tangent to the polar curve... (lets say t=theta, for my own laziness)

*easier on the eyes* r = 4 / [5 -cos( t )] at t = pi / 6

i know you are supposed to plug t into the equation and then convert to parametric form by finding x(t) and y(t), but after that i just get lost in the math. the answer is supposed to be x-5y+4=0, but i just can't get there.

 

[oLLie]

calculus help anyone?

No thanks I'm good.

 

[hdeck]

wow, nice way to up your posts...

 

[GoodToGo]

umm Dude I cant understand the equation of the polar curve, u are going to have to express it better.

 

[Crusty]

I think he meant:
r = 4/(5-cos(t))
t = pi/2

 

[hdeck]

r = 4/[5 -cos(t)] at t = pi/6

*edit!!!! its supposed to be at pi/6, not pi/2*

 

[Legendary]

this may help.

I forgot how to convert to parametric - maybe
x = rcosT
y = rsinT

so get that to look like that somehow and you got x or y and then use the polar derivative? I dunno, I have forgotten my calculus.

 

[TuxDave]

Here's my idea... but my calculations don't match up well with your solution. Whatever. To get a line in x,y coordinates, I guess all we have to figure out is a point and a slope.

To get the point.
x = r(t)*cos(t)
y = r(t)*sin(t)

To get the slope, we want dy/dx:

We calculate dy/dt and dx/dt from the two equations above. Then use dy/dt*dt/dx = dy/dx. There, a point and a slope. That's what I can think up of this late of an hour.

Edit:
I'm pretty sure either your question is wrong or your solution is wrong.

Plugging in for the point, @ t=pi/6, x = 0.84, y = 0.48, which does not fall on the line that you gave. I think it's supposed to be pi/2.