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Math Help - slope

  1. #1
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    slope

    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)?
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  2. #2
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    Re:

    Hi Jeph!!

    Here is the work:

    Last edited by qbkr21; April 10th 2007 at 01:55 PM.
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    Re:

    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.
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    so does the ^2 just change into a 2 or something and you multiply the 4 by that to get 8?
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    Re:

    Quote Originally Posted by jeph View Post
    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.
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    Quote Originally Posted by qbkr21 View Post
    I Jeph!!

    Here is the work:
    Hi qbkr21 and jeph!

    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
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    Re:

    What do mean what would it be like with Polar? I have never used them before...
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    im getting confused.....is there a formula for calculating cos^2 problems? this differentiating stuff is gonna kill me on the exam
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    Re:

    Relax Jeph... You will do well just remember:


    cos^2(x)= (cos(x))^2

    Always Always Always now can you differentiate it?

    WARNING cos^2(x) IS NOT EQUAL TO cos(x)^2
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Here are the graphs to give you an idea of how different the answers will be if we use the wrong coordinate system
    Attached Thumbnails Attached Thumbnails slope-interp.gif  
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    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
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  12. #12
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    Re:

    I guess polar coordinates are not functions since they do not appear to be one-to-one.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qbkr21 View Post
    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 functions
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    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)?
    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
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  15. #15
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    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...
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