1. ## 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!

2. Originally Posted by quah13579
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.

3. Originally Posted by quah13579
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.

4. Originally Posted by earboth
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 ?

5. Originally Posted by quah13579
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.

6. Originally Posted by Gamma
$\Phi$? or $\phi$?

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