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anonymous

  • one year ago

Given the functions x[n] = 1.3n + Sin[n] y[n] = 2(1 + Cos[n/2]) Assuming that these two functions are used to create a parametric plot where x = x[n] and y=y[n] Ah.. but lets say that x[n] is.. x[Pi/2] = (1.3(Pi))/2 + 1 = 3.0420352248333655​ How would you combine these two functions to find the derivative of y= f[x]? so you could work out the slope at that particular point?

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  1. anonymous
    • one year ago
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    Why I am seeing this post as yellow colored? Or anybody else seeing the same?

  2. anonymous
    • one year ago
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    changed it a bit..had it wrong at first. I dont think finding an f[x] is very doable

  3. anonymous
    • one year ago
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    its a paid question to the qualified helpers.. they go yellow

  4. ganeshie8
    • one year ago
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    You may work the derivative directly w/o eliminating the parameter : \[\large \dfrac{dy}{dx} = \dfrac{~\dfrac{dy}{dn}~}{~\dfrac{dx}{dn}~}\] provided ofcourse \(\dfrac{dx}{dn}\ne 0\)

  5. anonymous
    • one year ago
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    thnx, I can probably work this out from here..

  6. anonymous
    • one year ago
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    still not intuitive to me yet though.. even though I've done a dozen problems like this already

  7. ganeshie8
    • one year ago
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    guess it takes some practice, derivation is pretty straightforward though : \[y = f((x)\] plugin the parameterization and get \[2(1+\cos(n/2)) = f\left(x\right)\] differentiating with respect to \(n\) both sides gives \[-\sin(n/2) = f'\left(x\right)*x' = f'(x)*(1.3+\cos(n))\] \[\implies f'\left(x\right) = \dfrac{-\sin(n/2)}{1.3+\cos(n)}\] which is same as what you would get using the earlier formula

  8. anonymous
    • one year ago
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    ahh, I had got this far.. had it upside down dx/dy = 1.3 + Cos[n] / -Sin[n/2]

  9. ganeshie8
    • one year ago
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    Looks good! simply flip both sides to get dy/dx you can play with derivatives just as fractions in single variable calculus

  10. anonymous
    • one year ago
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    ah gotcha, thank you lots, I couldn't see how to approach this problem at all

  11. anonymous
    • one year ago
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    Ive worked out lots of single function derivatives, but this is my first 2 function derivative

  12. ganeshie8
    • one year ago
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    You will see more of these in polar coordinates... If you take calcIII, you get lot of practice while parameterizing for line integrals and surfaces etc. Just want you know that parametric equations are very common and important!

  13. anonymous
    • one year ago
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    I am not a qualified helper..

  14. anonymous
    • one year ago
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    Have you tagged me somewhere writing this question?

  15. anonymous
    • one year ago
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    thank you, ganshie, I'll get this down..found a chapter on predator prey models, looks like it will clear it up for me, well and good..

  16. anonymous
    • one year ago
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    waterineyes.. not tagged mate.. it shows yellow / orange color to all members..

  17. anonymous
    • one year ago
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    The question in which you need a Qualified Helper, that question goes Yellow for all the users?

  18. UsukiDoll
    • one year ago
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    @waterineyes yes. OpenStudy has just added Qualified Helpers and OwlBucks which is their currency used to pay for immediate assistance. You can buy OwlBucks with PayPal or any major credit cards starting at $1.99 for 10 OwlBucks

  19. UsukiDoll
    • one year ago
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    Do you see the advertisement under "ask a question". The blue owl is advertising "Get an Explanation. Ask a Qualified Helper."

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