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

• Oct 9th 2010, 06:01 PM
kimjb
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
$25y^2=900-36x^2$
then square root everything and
$5y=300-6x$
and
$y=60-6/5x$
what's next? did i do something wrong?
thanks
• Oct 9th 2010, 06:11 PM
Jhevon
Quote:

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

yes, you did something wrong. in general, $\displaystyle \sqrt{a \pm b} \ne \sqrt a \pm \sqrt b$, 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 $\displaystyle \pm$ sign on the right hand side. since if $\displaystyle x^2 = a$, then $\displaystyle x = \pm \sqrt a$

EDIT: the "first and second quadrants" part of the problem means you should only consider the positive square root.
• Oct 9th 2010, 06:11 PM
Chris L T521
Quote:

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

$\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$!!!

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

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

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

Does this make sense?
• Oct 9th 2010, 06:13 PM
mr fantastic
Quote:

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
$25y^2=900-36x^2$
then square root everything and
$5y=300-6x$ *
and
$y=60-6/5x$
what's next? did i do something wrong?
thanks

The DE is seperable:

$\displaystyle 50 \int y \, dy = -72 \int x \, dx \Rightarrow 25 y^2 = -36 x^2 + C$. 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 $A^2 - B^2 = C^2$ then $A - B = C$. An instructor would expect better from a student studying differential equations ....
• Oct 10th 2010, 06:15 PM
kimjb
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