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Math Help - Vector Fields

  1. #1
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    Vector Fields

    I don't think this problem is too difficult, but I have no idea for what it's asking:

    For the vector field \textbf{F}(x,y)=\textbf{i}+x\textbf{j}, if the parametric flow lines are x=x(t), y=y(t), what differential equations do these functions satisfy? Deduce that \frac{dy}{dx}=x.

    First off, I don't know what x=x(t) and y=y(t) mean, and how t comes into play. Could someone help?
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  2. #2
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    Quote Originally Posted by MathTooHard View Post
    I don't think this problem is too difficult, but I have no idea for what it's asking:

    For the vector field \textbf{F}(x,y)=\textbf{i}+x\textbf{j}, if the parametric flow lines are x=x(t), y=y(t), what differential equations do these functions satisfy? Deduce that \frac{dy}{dx}=x.

    First off, I don't know what x=x(t) and y=y(t) mean, and how t comes into play. Could someone help?

     x = x(t) , y= y(t) is/are the parametric equation(s) which you are looking for .  t is the parameter , not necessary to be the time , we can replace it with a,b,c,d,e,f,g any letter you like .

    Usually , we could find out the parametric equation AFTER we find out the function(family) of the lines  y = f(x) , so don't worry about  x(t) and  y(t)
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  3. #3
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    Are you asked to solve the differential equation ?


    for the flow lines  y = f(x) in a vector field  F(x,y) = P(x,y)i + Q(x,y)j

    we always have  \frac{dy}{dx} = f'(x) = \frac{Q(x,y) }{P(x,y)}
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  4. #4
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    Quote Originally Posted by simplependulum View Post
    Are you asked to solve the differential equation ? Nope.


    for the flow lines  y = f(x) in a vector field  F(x,y) = P(x,y)i + Q(x,y)j

    we always have  \frac{dy}{dx} = f'(x) = \frac{Q(x,y) }{P(x,y)}
    From that, I get  \frac{dy}{dx} = \frac{x}{1}= x . But how should I go about figuring out what differential equations x=x(t) and y=y(t) satisfy? I don't really know what a step of t should mean in such a parametrization. I think x'(t)=1, but what would y'(t) be?
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