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Math Help - Find the derivative using Leibniz Rule

  1. #1
    Junior Member NowIsForever's Avatar
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    Find the derivative using Leibniz Rule

    I found this question on Yahoo! Answers, and I am stumped (however, I would like to know the answer):

    Let f(x,y) be a continuous function in R2. For t>0 let R_t = [0,t] x [0,2t] = { (x,y) | 0<=x<=t, 0<=y<=2t }
    and define F(t) integral integral R_t of f(x,y)dA
    Use Leibniz Rule to compute the derivative dF/dt.

    Here's the link:

    How do I use Leibniz Rule? - Yahoo! Answers

    Thanks, Charles
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  2. #2
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    A lot, of course, depends on the function f(x,y).

    Define G(x, t)= \int_0^{2t} f(x,y)dy.

    Then \frac{\partial G}{\partial x}= \int_0^{2t}\frac{\partial f}{\partial x} dy.

    Now, F(t)= \int_0^t G(x,t)dx so that \frac{dF}{dt}= G(t,t)+ \int_0^t\frac{\partial G}{\partial x}dx.

    = \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial G}{\partial x}dydx.
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  3. #3
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    Wouldn't it be

    \displaystyle \int_0^{2t} f(t,y)dy + 2\int_0^t f(x,2t)dx
    Last edited by Jester; December 12th 2010 at 03:26 PM.
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  4. #4
    Junior Member NowIsForever's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    = \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial G}{\partial x}dydx.
    Are you sure you didn't intend to write:

    = \int_0^{2t}f(t, y)dy+ \int_0^t\int_0^{2t}\frac{\partial f}{\partial x}dydx

    Anyway, your formula didn't work when I set f(x,y) = x sin y, although perhaps I made a mistake in my calculation. Are you sure your formula is correct?

    Also, I don't understand why you have two terms in your result: G(t,t) and \int_0^t\frac{\partial G}{\partial x}dx
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  5. #5
    Junior Member NowIsForever's Avatar
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    Quote Originally Posted by Danny View Post
    Wouldn't it be

    \displaystyle \int_0^{2t} f(t,y)dy + 2\int_0^t f(x,2t)dx
    After carrying out the calculation, I concur with your result. Furthermore, f(x,y) need only be integrable, and continuity is more than sufficient. The Leibniz Rule was a red herring:

    \displaystyle [\int_0^{t+h} \int_0^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h

    \displaystyle = [\int_0^t \int_0^{2t+2h} f(x,y) dydx + \int_t^{t+h} \int_0^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h


    \displaystyle = [\int_0^t \int_0^{2t} f(x,y) dydx + \int_0^t \int_{2t}^{2t+2h} f(x,y) dydx

    \displaystyle \hspace{10 mm} + \int_t^{t+h} \int_0^{2t} f(x,y) dydx + \int_t^{t+h} \int_{2t}^{2t+2h} f(x,y) dydx - \int_0^t \int_0^{2t} f(x,y) dydx]/h


    \displaystyle = [\int_0^t \int_{2t}^{2t+2h} f(x,y) dydx + \int_t^{t+h} \int_0^{2t} f(x,y) dydx + \int_t^{t+h} \int_{2t}^{2t+2h} f(x,y) dydx]/h

    Approximating the integrals with (single term) Riemann sums, we get:

    \displaystyle \approx [{2h} \int_0^t f(x,2t) dx + h \int_0^{2t} f(t,y) dy + h \cdot 2h \cdot f(t,2t)]/h

    \displaystyle \rightarrow 2 \int_0^t f(x,2t) dx + \int_0^{2t} f(t,y) dy, h \rightarrow 0
    \displaystyle \therefore F'(t) = 2 \int_0^t f(x,2t) dx + \int_0^{2t} f(t,y) dy
    Last edited by NowIsForever; December 30th 2010 at 06:30 PM.
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