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Math Help - slope

  1. #16
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    Re:

    Quote Originally Posted by jeph View Post
    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...
    Hi Jeph! Hope you are doing OK.

    No

    4cos^2(theta) + 2cos(theta) -2 does not yield 4cos(theta)cos(theta) + 2cos(theta) -2

    The correct answer is:

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  2. #17
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    oh! i wasnt trying to differentiate that equation

    it started out as r=2+2cos(theta)
    then i did y=rsin(theta)
    dy/d(theta)=2cos(theta) + 2 cos^2(theta) - 2sin^2(theta) simplified down to
    4cos^2(theta) + 2cos(theta) -2

    so dy/dx = slope. i have to replace the thetas in 4cos^2(theta) + 2cos(theta) -2 with pi/2. but i dont know how to calculate 4cos^2(theta)......
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  3. #18
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    Quote Originally Posted by jeph View Post
    oh! i wasnt trying to differentiate that equation

    it started out as r=2+2cos(theta)
    then i did y=rsin(theta)
    dy/d(theta)=2cos(theta) + 2 cos^2(theta) - 2sin^2(theta) simplified down to
    4cos^2(theta) + 2cos(theta) -2

    so dy/dx = slope. i have to replace the thetas in 4cos^2(theta) + 2cos(theta) -2 with pi/2. but i dont know how to calculate 4cos^2(theta)......
    Ok, so this approach is incorrect. i did the question a few posts up, refer to it to see the correct approach to this problem.

    but anyway, to answer your question, you would calculate 4cos^2(theta) as 4(cos(theta))^2

    so 4cos^2(theta)
    = 4(cos(theta))^2

    at theta = pi/2, we get

    = 4(cos(pi/2))^2
    = 4 (0)^2
    = 0

    so anyway, not that this is the right problem but dy/d(theta) = 4cos^2(theta) + 2cos(theta) -2 at theta = pi/2:

    dy/d(theta) = 4cos^2(pi/2) + 2cos(pi/2) -2 = 0 + 0 - 2 = -2

    which is not the answer you're looking for. i believe you said the answer was 1. you want to find dy/dx not dy/d(theta)
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  4. #19
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    oh! so you do cos(theta) first then square it...><

    yeah i know. dy was only the top part. dx on the bottom equals -2 also so -2/-2=1

    thank you!!
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    oh! so you do cos(theta) first then square it...><

    yeah i know. dy was only the top part. dx on the bottom equals -2 also so -2/-2=1

    thank you!!
    what i'm saying is you didn't have dy/dx, you had dy/d(theta), that's two different derivatives. dy/dx is that complicated function i derived

    and even if you did, it wouldn't be dy = -2, it would be dy/dx = -2 so dx could only be -2 if dy = 4
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  6. #21
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    what? but on the answer sheet it says dx/dt is -2sin(theta) - 4cos(theta)sin(theta)... i left out the dx/dt because i just needed help understanding the dy/dt part of it.
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    what? but on the answer sheet it says dx/dt is -2sin(theta) - 4cos(theta)sin(theta)... i left out the dx/dt because i just needed help understanding the dy/dt part of it.
    O ok, i get what you are trying to do now.

    You want to find dy/d(theta) and dx/d(theta)

    then dy/dx = dy/d(theta) * d(theta)/dx ..........since the d(theta)'s would cancel.

    Ok, give me a sec and i'll post the solutions using your method

    i take it you want a step by step walk-through when finding dy/d(theta) ?
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    let t be theta

    r = 2 + 2cost, find dy/dt when t = pi/2
    now in polar coordinates:

    y = rsint

    x = rcost

    let's find dx/dt

    x = rcost
    since r = 2 + 2cost
    => x = (2 + 2cost)cost
    => x = 2cost + 2cos^2(t) .........we have to use the chain rule on cos^2(t)
    => dx/dt = -2sint + 4cos(t)*(-sint)
    => dx/dt = -2sint - 4sintcost
    now when t = pi/2
    dx/dt = -2sin(pi/2) - 4sin(pi/2)cos(pi/2)
    => dx/dt = -2(1) - 4(1)(0)
    => dx/dt = -2 at t = pi/2

    let's find dy/dt

    y = rsint
    since r = 2 + 2cost
    => y = (2 + 2cost)sint
    => y = 2sint + 2costsint
    now we are going to find dy/dt. note that we have to find the derivative of 2costsint by using the product rule, since we have a product of two functions.

    To refresh your memory, the product rule says:
    if x and y are functions, then
    (xy)' = x'y + xy'
    that is we take the derivative of the first function times the second, plus the derivative of the second function times the first (or vice versa)

    so for sintcost we have:

    (sint)' = cost
    (cost)' = -sint

    so (sintcost)' = (sint)'cost + sint(cost)' = costcost + sint(-sint) = cos^2(t) - sin^2(t)

    that was quite a side bar, let's go back to the problem

    we left off at:
    y = 2sint + 2costsint ..............now take the derivative
    => dy/dt = 2cost + 2(costcost - sintsint)
    => dy/dt = 2cost + 2cos^2(t) - 2sin^2(t) ..........now simplifying this any further is unnecessary, since it's not the function itself we are concerned about but it's numerical value, there's no need to waste time making this form look pretty (in case you are wondering though, your simplifications were correct)

    now when t = pi/2

    dy/dt = 2cos(pi/2) + 2(cos(pi/2))^2 - 2(sin(pi/2))^2
    => dy/dt = 0 + 0 - 2(1)^2
    => dy/dt = -2

    now, derivative notations are more than notations, you can actually treat them like fractions. you can multiply them, cancel them out, seprate them, do whatever with them you can do with fractions. In that spirit, we will find dy/dx by the following.

    dy/dx = (dy/dt)/(dx/dt) or you can say dy/dt = dy/dt * dt/dx ....same thing

    dy/dt = -2
    dx/dt = -2

    so dy/dx = -2/-2 = 1 ............The answer

    if you wanted to use dy/dt = dy/dt * dt/dx

    dy/dt = -2
    dt/dx = 1/-2

    => dy/dx = -2 * 1/-2 = 1

    any questions? I'll have to answer them a bit later since i have class now

    I have to adimit, the length of the methods are about the same, but this is easier than the method i originally proposed. my method you could do relatively quickly if you are good with implicit differentiation, but this way is much simpler--which is what math is all about, making life simple
    Last edited by Jhevon; April 11th 2007 at 11:21 AM.
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