course Mth 271 Is there 1 more test to take on Chapter 3 or is there a Chapter 3 test and a final exam?
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22:18:25 Query 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.
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RESPONSE --> y=3sqrt6x^2 y=(6x^2)^1/3 dy=1/3(6x^2)^-2/3 confidence assessment: 2
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22:19:19 dy is the differential; `dy means 'delta-y' and is the exact change. y = (6x^2)^(1/3) y' = dy/dx = 1/3(6x^2)^(-2/3)(12x) y' = dy/dx = 4x(6x^2)^(-2/3) y' = dy/dx = 4x / (6x^2)^(2/3). So dy = (4x / (6x^2)^(2/3)) dx **
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RESPONSE --> I got the correct answer, however I did not simplify it as much as you did. self critique assessment: 2
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22:21:25 ** Query 3.8.12 compare dy and `dy for y = 1 - 2x^2, x=0, `dx = -.1 **** What is the differential estimate dy for the given function and interval, and what is the actual change `dy?
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RESPONSE --> y=1-2x^2 y'=dy/dx = -4xdx -4(0)(-0.1)=0 dy=f(x + dx) - f(x) f(0-0.1) - f(0) =1-2(-0.1)^2 - (1-0) = -.02 self critique assessment: 2
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22:21:45 y ' = dy /dx = - 4 x so dy = -4x dx. The differential estimate is dy = -4 * 0 * (-.1) = 0. Actual change is y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02. **
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RESPONSE --> I got this answer correct. self critique assessment: 2
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22:28:48 Query Extra Problem: Give the equation of the tangent line to y= 2 * x^(1/3) - 1 at (8,3); tan line prediction and actual fn value at `dx = -.01 and .01. **** What is the equation of the tangent line and how did you obtain it?
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RESPONSE --> 2x^1/3 - 1 2/3x^-2/3 2/3(8)^-2/3 =1/6 y-3=1/6(x-8) y-3=1/6x-8/6 y=1/6x+10/6 or 5/3 Filling in for -.01 & .01 y=1/6x + 5/3 y= 1/6(-.01) + 5/3 = -16.66 + 5/3 = -15 y=1/6(.01) + 5/3 = 18.3333 confidence assessment: 2
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22:30:26 f(x) = 2x^(1/3) - 1 f' (x) = 2/3x ^(-2/3) f ' (8) = 2/3(8)^(-2/3) f ' (8) = 1/6 y - 3 = 1/6(x - 8) y - 3 = 1/6x - 8/6 y = 1/6x + 10/6 y = (1/6)x + (5/3) after simplification. Using `dx = .01 we get x + `dx = 8.01. The tangent-line approximation is thus y = 1/6 * 8.01 + 5/3 = 3.001666666. The actual function value is 2 * 8.01^(1/3) - 1 = 3.001665972. The difference is .0000007, approx. A similar difference is found approximating the function for `dx = -.01, i.e., at 7.99. We see that for this short interval `dx the tangent-line approximation is very good, predicting the change in the y value (approximately .001666) accurate to 4 significant figures. COMMON ERROR: Students often round off to 3.0017, or even 3.002, which doesn't show any discrepancy between the tangent-line approximation and the accurate value. Taken to enough decimal places the values of the function and the tangent-line approximation do not coincide; this must be the case because the function isn't linear, whereas the tangent line approximation is. You should in general use enough significant figures to show the difference between two approximations. The difference should be no greater than -.02 * .01^2 = -.000002 (based on a Taylor's Theorem estimate I did in my head so don't hold me responsible for its accuracy, and you aren't responsible for Taylor's Theorem at this point of the course); the discrepancy might therefore appear in the 6th decimal place, almost certainly not later than the 7th. **
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RESPONSE --> I didn't multiply the .01 by 8. I did the rest of it correctly. self critique assessment: 2
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22:34:10 **** Query 3.8.42 drug concentration C = 3t / (27+t^3) **** When t changes from t = 1 to t = 1.5 what is the approximate change in C, as calculated by a differential approximation? Explain your work.
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RESPONSE --> Filling in for the quotient rule you get the following: 27 + t^3(3) - 3t(3t^2)/(27+t^3)^2 81+3t^3 - 9t^3 = 81-6t^3/(27+t^3)^2(dx) Filling in for the ts you get: (81-6(1)^3/(27+1^3)^2 (1.5) = 75-784(1.5) = 1.43494mg per milliliter confidence assessment: 2
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22:35:23 By the Quotient Rule we have C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2, or C' = (81 - 6t^3) / (27 + t^3)^2. The differential is therefore dC =( (81 - 6t^3) / (27 + t^3)^2) dx. Evaluating for t = 1 and `dt = .5 we get dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5) dC = (75 / 784) (.5) dC = .0478 mg/ml **
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RESPONSE --> I think I did this correctly. For some reason you multiplied by .5 instead of 1.5. self critique assessment: 2
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