course Mth 271 014. `query 14*********************************************
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Given Solution: `aSTUDENT SOLUTION: The average rate of change over the interval is -1/6. I got this answer by taking the difference of the numbers obtained when you plug both 1 and 4 into the function and then dividing that difference by the difference in 1 and 4. f(b)-f(a)/b-a ------------------------------------------------ Self-critique Rating: I used the equation given in the solution to find the answer. ********************************************* Question: `q **** How does the average compare to the instantaneous rates at the endpoints of the interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rates at the endpoints are: x^-.5 [1,4] x = 1 (1)^-.5 = 1 x = 4 (4)^-.5 = 1/2 The average is -1/6 The instantaneous rate is given by the derivative: x^-.5 y’ = -.5x^-1.5 x = 1 y’ = -.5(1)^-1.5 = -.5 x = 4 y’ -.5(4)^-1.5 = -.0625 The average of these two answers are: - .5 - .0625 / 2 = -.28125 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel fairly confident about this problem.
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Given Solution: `a The average rate of change is change in y / change in x. For x = 1 we have y = 1^-.5 = 1. For x = 4 we have y = 4^-.5 = 1 / (4^.5) = 1 / 2. So `dx = 1/2 - 1 = -1/2 and `dy = 4 - 1 = 3. `dy / `dx = (.5-1) / (4-1) = -.5 / 3 = -1/6 = -.166... . To find rates of change at endpoints we have to use the instantaneous rate of change: The instantaneous rate of change is given by the derivative function y ' = (x^-.5) ' = -.5 x^-1.5. The endpoints are x=1 and x=4. There is a rate of change at each endpoint. The rate of change at x = 1 is y ' = -.5 * 1^-1.5 = -.5. The rate of change at x = 4 is y ' = -.5 * 4^-1.5 = -.0625. The average of the two endpoint rates is (-.5 -.0625) / 2 = -.281 approx, which is not equal to the average rate -.166... . Your graph should show the curve for y = x^-.5 decreasing at a decreasing rate from (1, 1) to (4, .5). The slope at (1, 1) is -.5, the slope at (4, .5) is -.0625. and the average slope is -.166... . The average slope is greater than the left-hand slope and less than the right-hand slope. That is, the graph shows how the average slope between (1,1) and (4,.5), represented by the straight line segment between those points, lies between the steeper negative slope at x=1 and the less steep slope at x = 4. If your graph does not clearly show all of these characteristics you should redraw the graph so that it does. ** ------------------------------------------------ Self-critique Rating: I used the equation from the first problem to find part of the answer. For the other part, I used the given function and plugged in the endpoints, then took the average of those values. ********************************************* Question: `qQuery 2.3.14 H = 33(10`sqrt(v) - v + 10.45): wind chill; find dH/dv, interpret; rod when v=2 and when v=5 What is dH/dv and what is its meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: H = 33(10`sqrt(v) - v + 10.45) dH/dv would be the derivative of the problem. I used the power rule: dH/dv = f’g + fg’ = (10’sqrt(v) – v + 10.45) + 33(1/2(10v)^-.5 * 5v – 1) By looking ahead, I can already tell that this is quite far from the desired answer. What have I done wrong? I can’t understand the given solution. By looking at the next problem, I see how you got your answer, but I don’t understand why you approached the problem in that manner. Why did you not do the product rule? confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I am not very confident about this problem.
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Given Solution: INSTRUCTOR COMMENT: You give the difference quotient, which in the limit will equal the rate of change, i.e., the derivative. The derivative of h with respect to v is dH / dv = 33 * 10 * .5 * v^-.5 + 33 * -1 = 165 v^-.5 - 33. When v = 2, dH / dv is about 85 and when v = 5, dH / dv is about 40. Check my mental approximations to be sure I'm right (plug 2 and 5 into dH/dv = 165 v^-.5 - 33). H is the heat loss and v is the wind velocity. On a graph of H vs. v, the rise measures the change in heat loss and the run measures the change in wind velocity. So the slope measures change in heat loss / change in wind velocity, which is the change in heat loss per unit change in wind velocity. We call this the rate of change of heat loss with respect to wind velocity. dH / dv therefore measures the instantaneous rate of change of heat loss with respect to wind velocity. ** ------------------------------------------------ Self-critique Rating: I attempted to approach the problem using the product rule, but I didn’t get the right answer. I don’t know how to solve this problem. ********************************************* Question: `qQuery 2.3.20 C = 100(9+3`sqrt(x)); marginal cost **** What is the marginal cost for producing x units? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I look ahead at the answer to better understand. I’m not understanding why the derivative is 150/’sqrtx and not: (9 + 3‘sqrtx) + 100(3/2x^-1/2) = (9 + 3’sqrt) + 150x^-1/2 I used the product rule. I’m not understanding why f’g seems to be left out of the answer.
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Given Solution: `aSTUDENT SOLUTION: To get the marginal cost for producing x units, I think you take the first derivative of the cost function. If this is true, the marginal cost for 100(9+3sqrt(x)) is 100 * 3 * .5 x^-.5 = 150 x^-.5. The marginal cost is the rate at which cost changes with respect to the number of units produced. For this problem x is the number produced and C = 100 ( 9 + 3 sqrt(x) ). Marginal cost is therefore dC/dx = 100 * 3 / (2 sqrt(x)) = 150 / sqrt(x). ** STUDENT COMMENT: I got the end of the problem 150 / 'sqrtx, But what I don't understand is where the 900 went and why it is not included in the answer. INSTRUCTOR RESPONSE: 900 is a constant. It doesn't change. So its rate of change with respect to any variable is zero. This is the conceptual answer, which is very important. Rules are also important. In terms of rules, the derivative of a constant is zero. ------------------------------------------------ Self-critique Rating: I don’t understand this problem. " xxxx