course Mth 271 029. `query 29
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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 ** `a** 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y ' = dy /dx = - 4 x dy = -4x dx. The differential estimate is dy = - 4 * 0 * (-.1) = 0. The actual change is y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02.
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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): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `a f(x) = sqrt{25-x^2} f' (x) = -x / sqrt{25-x^2} 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. I used `dx = .01 and got x + `dx = 3.01. For the tangent-line approximation I got 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
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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): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q **** Query 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: Using the Quotient Rule I had C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2 simplify C' = (81 - 6t^3) / (27 + t^3)^2. The differential is therefore: dC =[ (81 - 6t^3) / (27 + t^3)^2] dx. Evaluated for t = 1 and `dt = .5 I got dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5) dC = (75 / 784) (.5) dC = .0478 mg/ml
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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): ------------------------------------------------ Self-critique Rating:ok "