# Thread: solve explicitly for y and diff to get dy/dx in terms of x

1. ## solve explicitly for y and diff to get dy/dx in terms of x

Hi. Can you tell me what to do next.
I got that dy/dx=-72x/50y

Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

36x2 + 25y2 = 900

I think next it would be
$\displaystyle 25y^2=900-36x^2$
then square root everything and
$\displaystyle 5y=300-6x$
and
$\displaystyle y=60-6/5x$
what's next? did i do something wrong?
thanks

2. Originally Posted by kimjb
Hi. Can you tell me what to do next.
I got that dy/dx=-72x/50y

Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

36x2 + 25y2 = 900

I think next it would be
$\displaystyle 25y^2=900-36x^2$
then square root everything and
$\displaystyle 5y=300-6x$
and
$\displaystyle y=60-6/5x$
what's next? did i do something wrong?
thanks
yes, you did something wrong. in general, $\displaystyle \displaystyle \sqrt{a \pm b} \ne \sqrt a \pm \sqrt b$, yet that is what you did with the right side of the equation when you took the square root. Also note that there should be a $\displaystyle \displaystyle \pm$ sign on the right hand side. since if $\displaystyle \displaystyle x^2 = a$, then $\displaystyle \displaystyle x = \pm \sqrt a$

EDIT: the "first and second quadrants" part of the problem means you should only consider the positive square root.

3. Originally Posted by kimjb
Hi. Can you tell me what to do next.
I got that dy/dx=-72x/50y

Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

36x2 + 25y2 = 900

I think next it would be
$\displaystyle 25y^2=900-36x^2$
then square root everything and
$\displaystyle 5y=300-6x$
and
$\displaystyle y=60-6/5x$
what's next? did i do something wrong?
thanks
$\displaystyle \sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$!!!

So what you have to do from $\displaystyle 25y^2=900-36x^2$ is take square roots of both sides to get

$\displaystyle 5y=\sqrt{900-36x^2}\implies 5y=6\sqrt{25-x^2}\implies y=\frac{6}{5}\sqrt{25-x^2}$.

Thus, your $\displaystyle \dfrac{dy}{dx}=-\dfrac{72x}{50\left(\frac{6}{5}\sqrt{25-x^2}\right)}=-\dfrac{72x}{60\sqrt{25-x^2}}=-\dfrac{6x}{5\sqrt{25-x^2}}$ which should be the same as differentiating $\displaystyle y=\frac{6}{5}\sqrt{25-x^2}$ wrt x.

Does this make sense?

4. Originally Posted by kimjb
Hi. Can you tell me what to do next.
I got that dy/dx=-72x/50y

Solve the equation explicitly for y and differentiate to get dy / dx in terms of x.(Consider only the first and second quadrants for this part.)

36x2 + 25y2 = 900

I think next it would be
$\displaystyle 25y^2=900-36x^2$
then square root everything and
$\displaystyle 5y=300-6x$ *
and
$\displaystyle y=60-6/5x$
what's next? did i do something wrong?
thanks
The DE is seperable:

$\displaystyle \displaystyle 50 \int y \, dy = -72 \int x \, dx \Rightarrow 25 y^2 = -36 x^2 + C$. You obviously know something we don't to get that C = 900.

By the way, according to the line with * next to it, you seem to think that if $\displaystyle A^2 - B^2 = C^2$ then $\displaystyle A - B = C$. An instructor would expect better from a student studying differential equations ....

5. Thanks that did make a lot more sense Chris.
Mr fantastic, I'm actually in a calculus class and I put this in the calculus forum but it got moved, but I do understand about the line with the star. I'm not sure why I put that- that's kind of embarassing.
Thanks for helping