Hi Jeph!!
Here is the work:
i have to find the slope of the tangent line at theta = pi/2 of the graph r=2 + 2cos(theta).
i did the x=rcos(Theta) and y=rsin(theta) thing then the dx/d(theta) dy/d(theta) thing, then dy/dx
i ended up with 4cos^2(theta) + 2cos(theta) - 2 for dy and dx = -2sin(theta) - 4cos(theta)sin(theta).
the slope is -2/-2 = 1 but how do you calculate 4cos^2(pi/2)?
Yes. You leave constants alone. If you were to differentiate 4cos^2(pi/2), you would just be using the chain rule to for cos^2(pi/2). The 2 comes out in front of the constant, 4, that you left alone and makes an 8. Then you rewrite the 8 cos^2(pi/2) then differentiate cos^x(pi/2). Again you would rewrite the problem for the 3rd time and differentiate with respect to pi/2. Then rewrite the problem for the 4th time, thus giving you your final answer.
Polar coordinates is just a different type of coordinate system than the usual cartesian coordinates. I was introduced to it in calc 3, but i think my school starts to do it in calc 2 now. see Polar coordinate system
but don't worry, you should see it soon
just in case.
we can't really draw a line in polar coordinates, so we have to convert to rectangular coordinates before we can find the desired line.
r = 2 + 2cos(theta)
=> r^2 = 2r + 2rcos(theta) ............multiplied both sides by r
now going from polar to rectangular coord:
r^2 = x^2 + y^2
x = rcos(theta)
y = rsin(theta)
so r^2 = 2r + 2rcos(theta)
=> x^2 + y^2 = 2(x^2 + y^2)^(1/2) + 2x
now we differentiate implicitly, we get:
2x + 2y y' = (2x + 2y y')(x^2 + y^2)^(-1/2) + 2
=> 2x + 2y y' - (2x + 2y y')(x^2 + y^2)^(-1/2) = 2 .........factor out the (2x + 2y y'), we get
=> (2x + 2y y')(1 - (x^2 + y^2)^(-1/2)) = 2
=> 2x + 2y y' = 2/[1 - (x^2 + y^2)^(-1/2)]
=> 2y y' = 2/[1 - (x^2 + y^2)^(-1/2)] - 2x
=> y' = 2/{2y[1 - (x^2 + y^2)^(-1/2)]} - x/y
this will give us the slope of the tangent line. but what are the x and y values to plug in? let's find them.
we want theta = pi/2
if theta = pi/2 then r = 2 + 2cos(pi/2) = 2 + 0 = 2
so r = 2
so we have
x = 2cos(theta)
y = 2sin(theta)
so when theta = pi/2
x = 2cos(pi/2) = 0
y = 2sin(pi/2) = 2
so x = 0 and y = 2 at the point we are concerned with. so we get
y' = 2/{2(2)[1 - (0^2 + 2^2)^(-1/2)]} - 0/2
=> y' = 1 ..............the slope of the tangent line when theta = pi/2
so would 4cos^2(theta) + 2cos(theta) -2 be equal to 4cos(theta)cos(theta) + 2cos(theta) -2? 0.o
oh i didnt see there was a second page. we learned polar coordinates in trig and now we are doing things with it in calc 2. i think its the differentiating thats getting me right now. i forget these things way too quickly after not doing it for a while...