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Math Help - Parametric Equation

  1. #1
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    Parametric Equation

    I'm reviewing for a test and can't find anything in my notes or book to explain this review problem, wondering if anyone can help.

    Given x = 2t, y = 2t + 3t
    a.)find: dy/dx, when t = 2
    b.)dy/dx at the point (2,5)

    I understand that in a.) I have to substitute 2 for t, just not sure when/where to substitute it... do you calculate dy/dx and then plug 2 in for t? and for b.) I know that t = 1 for point (2,5) but no clue what it means to find dy/dx at that point. Any help will be appreciated and a brief explanation if possible, the answer won't do me any good if I don't know how to do it =)
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  2. #2
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    I forgot to mention in my first post that I'm assuming a.) goes like this

    dy/dx = 4t+3 / 2 and then just plug in 2 for t to get 11/2 as the answer but we never did anything like that so I'm not sure, just makes the most sense, and then b I'm not sure what to do, I was gonna remove the parameter and then see if I could come up with something but I got lost.
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  3. #3
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    Quote Originally Posted by skimanner View Post
    I'm reviewing for a test and can't find anything in my notes or book to explain this review problem, wondering if anyone can help.

    Given x = 2t, y = 2t + 3t
    a.)find: dy/dx, when t = 2
    b.)dy/dx at the point (2,5)
    \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    \frac{dy}{dx} = \frac{4t+3}{2}

    dy/dx = 4t+3 / 2 and then just plug in 2 for t to get 11/2

    correct
    at x = 2, y = 5 ... t = 1 ... \frac{dy}{dx} = \frac{7}{2}
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    thx skeeter I was very silly to not realize to just plug in 1 for t in part b.). Quick follow up question. I'm now trying to calculate the arc length from t =1 to t=2 and the surface area when the curve is revolved around the x-axis.

    I got half way through each of them but since I dont know how to use all the symbols in here to type it out it might be easier if I just don't since it will look like a total mess, but its pretty much the point where SQRT(4 + (4t+3) dt) that i'm stuck at. Any help would be appreciated and thx in advance
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  5. #5
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    Quote Originally Posted by skimanner View Post
    thx skeeter I was very silly to not realize to just plug in 1 for t in part b.). Quick follow up question. I'm now trying to calculate the arc length from t =1 to t=2 and the surface area when the curve is revolved around the x-axis.

    I got half way through each of them but since I dont know how to use all the symbols in here to type it out it might be easier if I just don't since it will look like a total mess, but its pretty much the point where SQRT(4 + (4t+3) dt) that i'm stuck at. Any help would be appreciated and thx in advance
    the arc length integral will not be easy ... looks to me like a trig substitution is involved.

    you sure the calculations cannot be done using a calculator alone?
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    Well I wont be able to use a calculator on the final to get answers, allthough I think he may of said we only had to setup the problem for the arc length, I know we have to solve the x-axis curve though
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    Actually I think he may of screwed something up because finding the curve would be complex also. So nevermind I'm just gonna wait and talk to the teacher about that.

    I am a little confused about dy/dx for the same problem of x = 2t, y=2t+3t

    dy/dx = d/dt (4t+3/2) / 2, would that come out to be 4/2 = 2 or would it end up as 0/2 = 0.
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  8. #8
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    Quote Originally Posted by skimanner View Post
    Actually I think he may of screwed something up because finding the curve would be complex also. So nevermind I'm just gonna wait and talk to the teacher about that.

    I am a little confused about dy/dx for the same problem of x = 2t, y=2t+3t

    dy/dx = d/dt (4t+3/2) / 2, would that come out to be 4/2 = 2 or would it end up as 0/2 = 0.
    \frac{d^2y}{dx^2} = \textcolor{red}{\frac{d}{dx}}\left(\frac{4t+3}{2}\  right)
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  9. #9
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    Quote Originally Posted by skeeter View Post
    \frac{d^2y}{dx^2} = \textcolor{red}{\frac{d}{dx}}\left(\frac{4t+3}{2}\  right)
    I think we might be on 2 different pages here, the formula I have to use to solve this problem is:

    dy/dx = d/dt(dy/dx) / dx/dt <-- not sure how clear that is but what I'm trying to illistrate is that the d/dt(dy/dx) would be the top part of the fraction bar and dx/dt underneath the "thick" fraction bar as a complex fraction.

    and thus what needs to be solved is:

    dy/dx = d/dt (4t+3/2) / 2 = ?

    since the top part which is d/dt (4t+3/2) needs to be derived with respect to t I know I would get 4 on top but is it 0 on the bottom so you end up with 4/0? which is undefined and then the bottom of the fraction would just be 2 since it dont get derived. thus the answer would be undefined so I think I'm doing something wrong because that just don't make sense.
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