Hi there,

I'm having difficulty with this problem:

Assume that x and y are differentiable functions of t. Find dy/dt when x^2-xy+y^2=19, dx/dt=3 for x=2, and y>0.

Anyone have any ideas?

Thanks in advanced for the help!

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- Dec 7th 2011, 12:37 AMjuicysharpieDifferentiable functions
Hi there,

I'm having difficulty with this problem:

Assume that x and y are differentiable functions of t. Find dy/dt when x^2-xy+y^2=19, dx/dt=3 for x=2, and y>0.

Anyone have any ideas?

Thanks in advanced for the help! - Dec 7th 2011, 12:45 AMProve ItRe: Differentiable functions
- Dec 7th 2011, 12:51 AMjuicysharpieRe: Differentiable functions
Oh sorry, I wrote it wrong. I'm supposed to find dy/dt. I'll correct it in my question.

- Dec 7th 2011, 12:57 AMProve ItRe: Differentiable functions
$\displaystyle \displaystyle \begin{align*} x^2 - x\,y + y^2 &= 19 \\ \frac{d}{dt}\left(x^2 - x\,y + y^2\right) &= \frac{d}{dt}(19) \\ \frac{d}{dt}\left(x^2\right) - \frac{d}{dt}\left(x\,y\right) + \frac{d}{dt}\left(y^2\right) &= 0 \\ \frac{d}{dx}\left(x^2\right)\frac{dx}{dt} - \left(x\,\frac{dy}{dt} + y\,\frac{dx}{dt}\right) + \frac{d}{dy}\left(y^2\right)\frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} - x\,\frac{dy}{dt} - y\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 0 \end{align*} $

Now substitute your values and solve. - Dec 7th 2011, 01:04 AMjuicysharpieRe: Differentiable functions
Thanks for that part. But I wonder if y(dx/dt) is supposed to be negative for 2x(dx/dt) - [x(dy/dt)+y(dx/dt)] + 2y(dy/dt) because of the subtraction sign in front where you used the product rule? Also I have a hard time solving for the y-value. When I do it, there seems to be imaginary numbers. Am I just solve wrong?

- Dec 7th 2011, 01:07 AMProve ItRe: Differentiable functions
- Dec 7th 2011, 01:17 AMjuicysharpieRe: Differentiable functions
But don't I need y to solve for dy/dt? So, I would need to substitute x, dx/dt, and y. I plugged in x=2 into the original function x^2-xy+y^2=19 in order to find y, but that's when I kept getting imaginary numbers. My result of (2)^2-(2)y+y^2=19 is y^2+2y+15=0, which I can't factor.

- Dec 7th 2011, 01:53 AMjuicysharpieRe: Differentiable functions
Ok I figured it out. I just made mistakes with my signs. (2)^2-(2)y+y^2=19 is y^2+2y-15=0. So, y=5 or -3. But since the problem says y>0, y must equal 5.

Thanks for the help!