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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Sorry, forgot the closing brackets: \[x=\sqrt{t}\quad y=\sqrt{t-1}\]

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Maybe I'm missing on the notation, but what is that 2 doing next to the x?

  6. anonymous
    • 5 years ago
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    Thank you, actually it comes out fine, sometimes you need to refresh the page for the MathJax javascript to redraw it properly.

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Actually, this is a parametric equation, so isn't the derivative something like: \[\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\] ?

  9. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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?

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