Question

Find \[\frac{d^2y}{dx^2}\] at t=2, given: \[x=\sqrt{t}\quady=\sqrt{t-1}

Answer 1

Sorry, forgot the closing brackets:

\[x=\sqrt{t}\quad y=\sqrt{t-1}\]

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Actually, this is a parametric equation, so isn't the derivative something like:

\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
?

Answer 9

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

Answer 10

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?