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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)?
Hi Jeph!!
Here is the work:
http://item.slide.com/r/1/33/i/fAqxL...qKGmkACPhm2xm/
Ohh...I didn't finish it up that pi/2 comes out. If it makes it any easier for you you and take pi/2 and multiple it by 180/pi and turn it into radians.
so does the ^2 just change into a 2 or something and you multiply the 4 by that to get 8?
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.Quote:
Originally Posted by jeph
Hi qbkr21 and jeph!Quote:
Originally Posted by qbkr21
qbkr21, usually you would be correct, but here i think we are dealing with polar coordinates, so things will be different.
am i right jeph, is this question in polar coordinates? if so, i will type up a solution to the problem
What do mean what would it be like with Polar? I have never used them before...
im getting confused.....is there a formula for calculating cos^2 problems? this differentiating stuff is gonna kill me on the exam :(
Relax Jeph... You will do well just remember:
cos^2(x)= (cos(x))^2
Always Always Always now can you differentiate it?
:eek: WARNING cos^2(x) IS NOT EQUAL TO cos(x)^2:eek:
Here are the graphs to give you an idea of how different the answers will be if we use the wrong coordinate system
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
I guess polar coordinates are not functions since they do not appear to be one-to-one.
correct. they always revolve. the r in a polar equation means radius, so it is based on circles and trig functionsQuote:
Originally Posted by qbkr21
just in case.Quote:
Originally Posted by jeph
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...