1. ## Parametric Calc

Given that $\displaystyle y=x^2-x$, $\displaystyle {dx}/{dt}>0$ and that the particle is traveling at a constant $\displaystyle 2\sqrt{10}$, what is dy/dt?

So far I've used a formula for speed as $\displaystyle \sqrt{(dy/dt)^2+(dx/dt)^2}$ and set it equal to (40)^(1/2). Any help would be hot.

Thanks.

2. ## Re: Parametric Calc

perhaps I should learn how to type those math symbols in text.
Sorry for bad handwriting, and I was lazy to simplify the last expression. I hope its right )

3. ## Re: Parametric Calc

Originally Posted by pyromania
perhaps I should learn how to type those math symbols in text.
Sorry for bad handwriting, and I was lazy to simplify the last expression. I hope its right )
It's right. I forgot to mention that the point of interest is the rectangular point (2,2), and a friend of mine mentioned the answer was 6, which is concurrent with what you said. No worries about your handwriting either. Thanks again!

4. ## Re: Parametric Calc

Try this
dy/dt =dy/dx times dx/dt=(2x-1)dx/dt
So (2x-1)^2(dx/dt)^2+(dx/dt)^2=40
Now put x=2

5. ## Re: Parametric Calc

Originally Posted by biffboy
Try this
dy/dt =dy/dx times dx/dt=(2x-1)dx/dt
So (2x-1)^2(dx/dt)^2+(dx/dt)^2=40
Now put x=2
Don't put x= 2 until after you have solved the differential equation!

6. ## Re: Parametric Calc

We are not solving a differential equation we just want dx/dt when x=2. So I think x=2 can be substituted as I said. I get the answer dx/dt=2
So dy/dt=3.2=6