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anonymous

  • one year ago

I have a question about question 2A-11 from unit 2 Applications of Differentiation-Approximation and Curve Sketching. It is attached as a word document. Any help would be appreciated. Thanks!

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  1. anonymous
    • one year ago
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  2. phi
    • one year ago
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    Can you make your file a pdf or png? Meanwhile, I assume the attached is the problem? The "generic formula" for a 2nd order approximation is \[ f(x) \approx f(x_0) + f'(x_0) (x-x_0)+\frac{f''(x_0)}{2} (x-x_0)^2\] this tells us how to approximate f(x) when x is "near" x0. notice we only need values for f(x0) (and its derivatives). Generally we pick x0 where f(x0) is easy to find, and then use this formula to approximate f(x) nearby.

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  3. phi
    • one year ago
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    In this problem, p is a function of v : \(p(v)= c v^{-k} \) v plays the role of x. \( v= v_0 + \Delta v\) and thus \( \Delta v= (v- v_0) \) we will need \(\frac{dp}{dv} \) and \( \frac{d^2 p}{dv^2} \) \[ p= c v^{-k} \\ \frac{dp}{dv} = c \frac{d}{dv} v^{-k} = -k c v^{-(k+1)} \] and \[ \frac{d^2 p}{dv^2}= k(k+1) c v^{-(k+2)} \]

  4. phi
    • one year ago
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    now plug these expressions into \[ f(x) \approx f(x_0) + f'(x_0) (x-x_0)+\frac{f''(x_0)}{2} (x-x_0)^2 \\ p(v) \approx c\ v_0^{-k} -k c v_0^{-(k+1)} \Delta v +\frac{k(k+1)}{2} v_0^{-(k+2)}\Delta v^2 \] we can factor out the leading term, and write this as \[ p(v_0+\Delta v) \approx \frac{c}{v_0^k} \left( 1-k \frac{\Delta v}{v_0}+\frac{k(k+1)}{2} \left( \frac{\Delta v}{v_0}\right)^2\right) \]

  5. anonymous
    • one year ago
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    I've attached the pdf. Let me know if you think your previous response answers my question. Thanks.

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  6. phi
    • one year ago
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    Your solution is OK. if you divide (1+ax + bx^2) into 1 (where a and b are the coeffs) you get 1-ax +(a^2-b)x^2 plus higher order terms of x which we drop with a= k, b= k(k-1)/2 this becomes 1 -k x + k(k+1)/2 x^2 where x is \( \frac{\Delta v}{v_0} \)

  7. anonymous
    • one year ago
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    One final question. Can you explain the division to me: 1/(1+ax+bx^2). I tried to do it using the polynomial division algorithm but wasn't able to reach the solution. Thanks.

  8. phi
    • one year ago
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    |dw:1437347137318:dw|

  9. phi
    • one year ago
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    the last term should have x^2 in there... but hopefully you get the idea...

  10. phi
    • one year ago
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    |dw:1437347464949:dw|

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