Question 1115856
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{-2000x}{\sqrt{25\,-\,x^2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ dy\ =\ \frac{-2000x}{\sqrt{25\,-\,x^2}}\ dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,dy\ =\ \int\,\frac{-2000x}{\sqrt{25\,-\,x^2}}\ dx]


Let *[tex \LARGE u\ =\ 25\ -\ x^2].  Then *[tex \LARGE dx\ =\ -\frac{1}{2x}du]


Substitute


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,dy\ =\ 1000\,\int\,\frac{1}{\sqrt{u}}\ du]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2000\sqrt{u}\ +\ C]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2000\sqrt{25\ -\ x^2}\ +\ C]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1300]


When


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1300\ =\ 2000\sqrt{25\ -\ 9}\ +\ C]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ C\ =\ -6700]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ D(x)\ =\ 2000\sqrt{25\ -\ x^2}\ -\ 6700]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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