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The problem is as follows:

$y^2(2-x)=x^3$

I know now how to solve it. However, my first try is described underneath, and I still do not know why that did not work. Could anyone explain it to me. I hope I did not do some horrible mistake here.

$y^2(2-x)=x^3 \implies y^2=\frac{x^3}{2-x}\implies 2yy'=\frac{6x^2-2x^3}{(2-x)^2} \implies y'=\frac{3x^2-x^3}{(2-x)^2}*\frac{1}{y}$

Akitirija
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1 Answers1

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I am not sure why you think it should be wrong ...

I would not rearrange before differentiating implicitly, though. Instead I would simply have $$ 2yy'(2-x)+y^2\cdot(-1)=3x^2\implies y'=\frac{3x^2+y^2}{2y(2-x)} $$ This is consistent with what you found, since plugging in $y^2=\frac{x^3}{2-x}$ into my expression for $y'$ we get $$ y'=\frac{3x^2+\frac{x^3}{(2-x)}}{2y(2-x)}=\frac{3x^2(2-x)+x^3}{2y(2-x)^2}=\frac{3x^2-x^3}{y(2-x)^2} $$ which is also what you found.

This result is also consistent with the graph of $y=\sqrt{\frac{x^3}{2-x}}$ and it's tangent at the point $(1,1)$ which has slope $a=\frac{3+1}{2(2-1)}=\frac{3-1}{1(2-1)}=2$ and equation $y=2(x-1)+1=x-1$:

enter image description here

Just as an example.

String
  • 18,395
  • Thank you, String. The reason I thought it was wrong was because I tried to plot in different values for x and y, and my solution's values were not the same as the answer's values. I must have done something wrong... I appreciate your explanation! – Akitirija Dec 09 '14 at 13:25
  • @Akitirija: That is great to hear! – String Dec 09 '14 at 13:29