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anonymous
 one year ago
I have a question about question 2A11 from unit 2 Applications of DifferentiationApproximation and Curve Sketching. It is attached as a word document. Any help would be appreciated. Thanks!
anonymous
 one year ago
I have a question about question 2A11 from unit 2 Applications of DifferentiationApproximation and Curve Sketching. It is attached as a word document. Any help would be appreciated. Thanks!

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phi
 one year ago
Best ResponseYou've already chosen the best response.1Can 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 \[ f(x) \approx f(x_0) + f'(x_0) (xx_0)+\frac{f''(x_0)}{2} (xx_0)^2\] 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.1In 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 \[ \frac{d^2 p}{dv^2}= k(k+1) c v^{(k+2)} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now plug these expressions into \[ f(x) \approx f(x_0) + f'(x_0) (xx_0)+\frac{f''(x_0)}{2} (xx_0)^2 \\ p(v) \approx c\ v_0^{k} k c v_0^{(k+1)} \Delta v +\frac{k(k+1)}{2} v_0^{(k+2)}\Delta v^2 \] we can factor out the leading term, and write this as \[ p(v_0+\Delta v) \approx \frac{c}{v_0^k} \left( 1k \frac{\Delta v}{v_0}+\frac{k(k+1)}{2} \left( \frac{\Delta v}{v_0}\right)^2\right) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've attached the pdf. Let me know if you think your previous response answers my question. Thanks.

phi
 one year ago
Best ResponseYou've already chosen the best response.1Your solution is OK. if you divide (1+ax + bx^2) into 1 (where a and b are the coeffs) you get 1ax +(a^2b)x^2 plus higher order terms of x which we drop with a= k, b= k(k1)/2 this becomes 1 k x + k(k+1)/2 x^2 where x is \( \frac{\Delta v}{v_0} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.1the last term should have x^2 in there... but hopefully you get the idea...
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