## anonymous 5 years ago Find $\frac{d^2y}{dx^2}$ at t=2, given: $x=\sqrt{t}\quady=\sqrt{t-1} • This Question is Closed 1. anonymous Sorry, forgot the closing brackets: \[x=\sqrt{t}\quad y=\sqrt{t-1}$

2. anonymous

Does that derivative makes sense? I don't understand if it is second derivative of both of second of y and x squared.

3. anonymous

what do you mean by that? It's just the second derivative of y (with respect to x) and so, since t=x^2, $y=\sqrt{x^2-1}$ $\frac{dy}{dx}=\frac{x}{\sqrt{x^2-1}}$ $\frac{d}{dx}\frac{dy}{dx}=\frac{d^2y}{dx^2}=\frac{\frac{1}{2\sqrt{x^2-1}}-x{\sqrt{x^2-1}}}{x^2-1}=\frac{-2x^3+2x+1}{2(x^2-1)^{3/2}}$ now, since that was kinda messy, I'm not sure if the second derivative is right. But anyway, the idea is find y as a function of x, and differentiate twice.

4. anonymous

oh sorry, i forgot to actually answer the question. plugging 2 in for t, and thus sqrt(2) for x, gives$\frac{-4\sqrt{2}+2\sqrt{2}+1}{2*1^{3/2}}=-\sqrt{2}+\frac{1}{2}$ let me emphasize, I might have made an arithmetic error, but in principle this is right. Also, sorry if the equation came out as unreadable in the last post.

5. anonymous

Maybe I'm missing on the notation, but what is that 2 doing next to the x?

6. anonymous

Thank you, actually it comes out fine, sometimes you need to refresh the page for the MathJax javascript to redraw it properly.

7. anonymous

the idea of the notation is that $\frac{d}{dx}$ is an operator, that takes the derivative of some function of x, with respect to x. so when we take the derivative of the derivative of a function of x, we have $\frac{d}{dx}\frac{dy}{dx}=\frac{d^2y}{dx^2}$ It's just a weird notational thing, since the d and the dx are part of the operator, which is being applied twice, but the y is only there once, and that's why the x is squared and the y isn't. I know that's not a great explanation, but hopefully it makes some sense

8. anonymous

Actually, this is a parametric equation, so isn't the derivative something like: $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ ?

9. anonymous

well I'm pretty sure that will get you the same thing. Let's try it: $\frac{dx}{dt}=\frac{1}{2\sqrt{t}}$ and $\frac{dy}{dt}=\frac{1}{2\sqrt{t-1}}$ thus $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{1}{2\sqrt{t-1}}}{\frac{1}{2\sqrt{t}}}=\frac{\sqrt{t}}{\sqrt{t-1}}$ and if you plug in x^2 for t, this becomes $\frac{x}{\sqrt{x^2-1}}$ which is what you get the other way, as well. sorry for the wait, trouble with the code

10. anonymous

No problem, I am wondering though if it shouldn't be derived twice, like so: $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d^{2}y}{dt^{2}}}{\frac{d^{2}x}{dt^{2}}}=\frac{\frac{d}{dt}\frac{1}{2\sqrt{t}}}{\frac{d}{dt}\frac{1}{2\sqrt{t-1}}}=\frac{\frac{-1}{4t^{\frac{3}{2}}}}{\frac{-1}{4(t-1)^{\frac{3}{2}}}}=\frac{(t-1)^{\frac{3}{2}}}{t^{\frac{3}{2}}}$Then substituting t=2 we get:$\frac{(t-1)^{\frac{3}{2}}}{t^{\frac{3}{2}}}=\frac{(2-1)^{\frac{3}{2}}}{(2)^{\frac{3}{2}}}=\frac{1}{\sqrt{2^{3}}}=\frac{1}{\sqrt{2^{2}2}}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$ Is this correct?