course Mth 271 ²¹Ñ„ÄyÚÁü—¶ÌwIÿ¸Ëƒ´ûËßÁü©¹assignment #018
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16:54:39 ** Query problem 2.5.48 der of 3/(x^3-4) **** What is your result?
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RESPONSE --> f(x) = 3/(x^3-4)^2 first I rewrite the function as f(x) = 3(x^3-4)^-2 then I use the general power rule f'(x) = -6(x^3 - 4)^-3 (3x^2) f'(x) = -18x^2 / (x^3 - 4)^3 self critique assessment: 3
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16:57:43 This function can be expressed as f(g(x)) for g(x) = x^3-4 and f(z) = 3 / z. The 'inner' function is x^3 - 4, the 'outer' function is 1 / z. So f'(z) = -3 / z^2 and g'(x) = 3x^2. Thus f'(g(x)) = -3/(x^3-4)^2 so the derivative of the whole function is [3 / (x^3 - 4) ] ' = g'(x) * f'(g(x)) = 3x^2 * (-3/(x^3-4)^2) = -9 x^2 / (x^3 - 4)^2. DER**
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RESPONSE --> I believe I got the question right because in the answer given here, the derivative of the f function does not multiply the constant by the power of z and the power of z stays -2 instead of becoming -3. self critique assessment: 2
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18:16:46 **** Query problem 2.5.66 tan line to 1/`sqrt(x^2-3x+4) at (3,1/2) **** What is the equation of the tangent line?
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RESPONSE --> f(x) = 1/`sqrt(x^2-3x+4) at (3,1/2) I rewrite it as f(x) = (x^2-3x+4)^(-1/2) Now I use the general power rule f'(x) = -.5(x^2-3x+4)^(-3/2) (2x-3) f'(x) = (-x+1.5) / (x^2-3x+4) Now I substitute 3 for x to find the slope. f'(3) = (-3 + 1.5) / (3^2 - 3(3) + 4) = -.19 approx. Finally, I use the point slope form to find the equation of the tangent line. y - 1/2 = -.19(x - 3) y = -.19x + 1.07 confidence assessment: 2
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18:18:55 The derivative is (2x - 3) * -1/2 * (x^2 - 3x + 4) ^(-3/2) . At (3, 1/2) we get -1/2 (2*3-3)(3^2- 3*3 + 4)^(-3/2) = -1/2 * 3 (4)^-(3/2) = -3/16. The equation is thus ( y - 1/2) = -3/16 * (x - 3), or y = -3/16 x + 9/16 + 1/2, or y = -3/16 x + 17/16. DER**
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RESPONSE --> I worked everything right. I probably should have left my answer as a fraction. That way I would have gotten an exact answer instead of a rounded answer. self critique assessment: 2
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18:35:31 **** Query problem 2.5.72 rate of change of pollution P = .25 `sqrt(.5n^2+5n+25) when pop n in thousands is 12 **** At what rate is the pollution changing at the given population level?
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RESPONSE --> P = .25 `sqrt(.5n^2+5n+25) P' = .125(.5n^2+5n+25)^(-1/2) * (n+5) P' = (.125n + .625) / `sqrt(.5n^2+5n+25) When the population is 12,000 P'(12) = [.125(12) + .625] / `sqrt[.5(12)^2 + 5(12) + 25] P'(12) = (1.5 + .625) / `sqrt(72 + 60 + 25) P'(12) = 2.125 / `sqrt(157) = .170 approx. So at a population of 12,000, the pollution is increasing at a rate of .170 parts per million / thousand of population. confidence assessment: 2
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