#$&* course MTH 271 5/7/126:40pm 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 ** ********************************************* Question: 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: Again, this question is asking for only the top portion of a derivative: y’ = -4x , but the dy differential is equal to dy = -4x(dx). Using the given values: dy = -4(0)(-0.1) = 0 as the estimate. Using the given values with the general differential expression: y(x + `dx) - y(x) → y(0 - 0.1) - y(0) and input the original function: [1 - 2(-0.1)^2] - [1 - 2(0)^2] → ( 1 - 2(0.01) ) - 1 → 0.98 - 1 → -0.02 -0.02 is the actual change. Confidence Rating: 3
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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: 3 ********************************************* Question: `qQuery 4 a 7th edition 3.8.22 equation of the tangent line to y=sqrt(25-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: Find the differential: y = sqrt(25 - x^2) → y’ = (0.5)(-2x)(25 - x^2)^-0.5 → y’ = (-3)/(sqrt(25 - x^2)) When x = 3, y’ = (-3)/(sqrt(25 - 3^2)) → -3/4 So the tangent line is y = -3/4(x) +b → 4 = -3/4(3) + b → 4 + 9/4 = b → b = 25/4 y = -3/4 x + 25/4 When ‘dx = -0.01, ‘y = x + ‘dx, which is 3 + (-0.01) → y’ = 2.090 When ‘dx = 0.01, ‘y = x + ‘dx, which is 3 + (0.01) → y’ = 3.010 Using the tangent line, the equation with the difference would be: y = (-0.75)(3.01) + 6.25 → 3.9925 and y = (-0.75) (2.99) + 6.25 → y = 4.0075 And using the actual function: y = sqrt(25 - (3.01^2)) → 3.992480432 and y = sqrt(25 - (2.99^2)) → 4.007480505 The difference between the actual and estimated values is about 0.00002. 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): OK ------------------------------------------------ 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: The derivative (dC/ dt), according to the quotient rule, is: C = 3t/ (27 + t^3) → dC/ dt = [(27 + t^3)(3) - (3t)(3t^2)] / (27 + t^3)^2 → (-6t^3 + 81) / (27 + t^3)^2 And so the differential (dC) is: dC’ = [(-6t^3 + 81) / (27 + t^3)^2] (dt) and substituting the change between t = 1 and t = 1.5 for (dt) and the value given (t = 1) for t, the approximate change in C is: dC’ = [(-6t^3 + 81) / (27 + t^3)^2] (dt) → dC’ = [(-6(1)^3 + 81) / (27 + (1)^3)^2] (0.5) → [(-6 + 81) / (27 + 1)^2] (0.5) → [(75) / (784)] * (0.5) → (0.09566) * (0.5) = 0.0478mg/ml 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): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!