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Math Help - piecewise function

  1. #1
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    Question piecewise function

    Let y be a piecewise function given by

    y(r) =kr^2 − m, if r ≤ R;
    −2kR^3 / r, if r > R


    for some (k,m, r,R > 0) ∈ R. This function represents the gravitational potential of the Earth.
    Find the values of k for which y is continuous at r = R.

    any idea how to do this queation?
    thank you!
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  2. #2
    Super Member Gamma's Avatar
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    Quote Originally Posted by quah13579 View Post
    Let y be a piecewise function given by

    y(r) =kr^2 − m, if r ≤ R;
    −2kR^3 / r, if r > R


    for some (k,m, r,R > 0) ∈ R. This function represents the gravitational potential of the Earth.
    Find the values of k for which y is continuous at r = R.

    any idea how to do this queation?
    thank you!
    You have some typo in your equation which makes it impossible for me to guess what the function is.
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by quah13579 View Post
    Let y be a piecewise function given by

    y(r) =kr^2 − m, if r ≤ R;
    −2kR^3 / r, if r > R


    for some (k,m, r,R > 0) ∈ R. This function represents the gravitational potential of the Earth.
    Find the values of k for which y is continuous at r = R.

    any idea how to do this queation?
    thank you!
    I assume that you mean:

    y(r) = \left\{\begin{array}{lcr}kr^2-m & if & r\leq R\\-\dfrac{2kR^3}{r} & if & r>R\end{array}\right.

    If y_1(r)=kr^2-m ~ if ~ r\leq R and y_2(r)= -\dfrac{2kR^3}{r} ~ if ~ r>R

    Then you have to solve the system of equations for k:

    y_1(r)=y_2(r)~\wedge~y'_1(r) = y'_2(r)

    y'_1(r) = y'_2(r): 2kr=\dfrac{2kR^3}{r^2}~\implies~\boxed{r=R}

    y_1(r)=y_2(r): kR^2-m=-\dfrac{2kR^3}{R}~\implies~3kR^2=m~\implies~ \boxed{k=\dfrac m{3R^2}}

    Actually you have to calculate some limits to be exact. The final result will be the same which I've posted here.
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  4. #4
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    Quote Originally Posted by earboth View Post
    I assume that you mean:

    y(r) = \left\{\begin{array}{lcr}kr^2-m & if & r\leq R\\-\dfrac{2kR^3}{r} & if & r>R\end{array}\right.

    If y_1(r)=kr^2-m ~ if ~ r\leq R and y_2(r)= -\dfrac{2kR^3}{r} ~ if ~ r>R

    Then you have to solve the system of equations for k:

    y_1(r)=y_2(r)~\wedge~y'_1(r) = y'_2(r)

    y'_1(r) = y'_2(r): 2kr=\dfrac{2kR^3}{r^2}~\implies~\boxed{r=R}

    y_1(r)=y_2(r): kR^2-m=-\dfrac{2kR^3}{R}~\implies~3kR^2=m~\implies~ \boxed{k=\dfrac m{3R^2}}

    Actually you have to calculate some limits to be exact. The final result will be the same which I've posted here.

    thanks so much, you express the right queation, the only thing wrong is actually it is not y, I use y to take the place of "that symbol", I cant wrote or copy that symbol, it is o and I together, any idea what is it ?
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  5. #5
    Super Member Gamma's Avatar
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    Quote Originally Posted by quah13579 View Post
    thanks so much, you express the right queation, the only thing wrong is actually it is not y, I use y to take the place of "that symbol", I cant wrote or copy that symbol, it is o and I together, any idea what is it ?
     \Phi ? or \phi ?

    This is the greek letter phi, the first is capital the second is lower case.
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  6. #6
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    Quote Originally Posted by Gamma View Post
     \Phi ? or \phi ?

    This is the greek letter phi, the first is capital the second is lower case.
    yea, it is phi .....thanks
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