#$&* course Mth 271 12/3 530p *********************************************
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Given Solution: `a dy is the differential; `dy means 'delta-y' and is the exact change. y = (6x^2)^(1/3) y' = dy/dx = 1/3(6x^2)^(-2/3)(12x) y' = dy/dx = 4x(6x^2)^(-2/3) y' = dy/dx = 4x / (6x^2)^(2/3). So dy = (4x / (6x^2)^(2/3)) dx ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Query 3b 7th edition 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’ -4x dy = -4x(-1) = 4x dy = 4x 4(0) = 0 ‘dy = f(x+’x) - f(x) (1-2(0)^2) +-1 - (1-2(0)) 1-1-1 = -1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a y ' = dy /dx = - 4 x so dy = -4x dx. The differential estimate is dy = -4 * 0 * (-.1) = 0. Actual change is y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It seems like in some places in f(x+’x) x means just the value of x and in other places x means the whole thing, in this case (1-2(0)^2). That’s what I put in for x when I should have just put in 0. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `qQuery 4 a 7th edition 3.8.22 equation of the tangent line to y=sqrt(52-x^2) at (3, 4); tan line prediction and actual fn value for `dx = -.01 and .01. What is the equation of the tangent line and how did you obtain it? What are your tangent-line approximations for `dx = -.01 and `dx = +.01? What are the corresponding values of the actual function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = (52-x^2)^1/2 f’(x) = ½(52-x^2)^-½(-2x) = (26-(x^2/2))^-½(-2x) 1/sq.rt. of 26-(9/2)*(-2)(3) = 6/2.403 = 2.5 y-4= 2.5(x-3) y = 2.5x-3.5 tangent line equation Actual fn value for ‘dx = -.01 (3+-.01) y = sq. rt. of 52-(2.99)^2 = 6.56 Actual fn value for ‘dx = .01 (3+.01) y = sq. rt. of 52-(3.01)^2 = 6.55 Actual fn value (52-3^2)^½ = sq. rt. of 43 = 6.55 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a f(x) = sqrt(25-x^2) f' (x) = -x / sqrt(25-x^2) so f ' (3) = -3 / sqrt(25 - 3^2) = -3/sqrt(16) f ' (3) = -3/4 so that the tangent line has equation y - 4 = -3/4(x - 3) y - 4 = -3/4 x + 9/4 y = -3/4 x + 25/4. Using `dx = .01 we get x + `dx = 3.01. The tangent-line approximation is thus y = -3/4 * 3.01 + 6.25 = 3.9925. The actual function value is sqrt(25-3.01^2) = 3.992480432. The difference between the actual and approximate values is .00002, approx. A similar difference is found approximating the function for `dx = -.01, i.e., at 2.999. We see that for this short interval `dx the tangent-line approximation is very good, predicting the change in the y value (approximately .00750) accurate to 5 significant figures. COMMON ERROR: Students often round off to 3.99, or even 4.0, 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. “ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I worked out my problem as it was stated: sq. rt. of 52, not 25, but I think since all 3 values were almost the same, it is right. When I have more time while I am studying for the test I will re-work it with sq. rt. of 25 and see if my solution matches. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q **** Query 5a 7th edition 3.8.38 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C’ = [(27+t^3)(3) - (3t)(3t^2)]/(27 + t^3)^2 = 81 +3t^3-9t^3/(27+t^3)^2 = 81-6t^3/(27+t^3)^2 t = 1 dt = 1.5-1 = .5 { [81-6(1)^3]/[27+(1)^3]^2}*.5 = (75/784)*(.5) = .04783 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a By the Quotient Rule we have C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2, or C' = (81 - 6t^3) / (27 + t^3)^2. The differential is therefore dC =( (81 - 6t^3) / (27 + t^3)^2) dx. Evaluating for t = 1 and `dt = .5 we get dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5) dC = (75 / 784) (.5) dC = .0478 mg/ml ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got this one right on my own!!! ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q **** Query 5a 7th edition 3.8.38 drug concentration C = 3t / (27+t^3) **** When t changes from t = 1 to t = 1.5 what is the approximate change in C, as calculated by a differential approximation? Explain your work. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C’ = [(27+t^3)(3) - (3t)(3t^2)]/(27 + t^3)^2 = 81 +3t^3-9t^3/(27+t^3)^2 = 81-6t^3/(27+t^3)^2 t = 1 dt = 1.5-1 = .5 { [81-6(1)^3]/[27+(1)^3]^2}*.5 = (75/784)*(.5) = .04783 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got this one right on my own!!! ------------------------------------------------ Self-critique Rating:3 #*&! ********************************************* Question: `q **** Query 5a 7th edition 3.8.38 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C’ = [(27+t^3)(3) - (3t)(3t^2)]/(27 + t^3)^2 = 81 +3t^3-9t^3/(27+t^3)^2 = 81-6t^3/(27+t^3)^2 t = 1 dt = 1.5-1 = .5 { [81-6(1)^3]/[27+(1)^3]^2}*.5 = (75/784)*(.5) = .04783 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got this one right on my own!!! ------------------------------------------------ Self-critique Rating:3 #*&!#*&!