course Mth 173 ŽùµÍ[¿þ”yròü`ÞÙ…Ó¤ èƈ……¡¦assignment #014
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14:36:16 `q001. What are the coordinates of the x = 5 point and the slope at that point of the graph of the function y = .3 x^2? What is the equation of the line through the point and having that slope? Sketch the line and the curve between x = 3 and x = 7 and describe your sketch. How close, in the vertical direction, is the line to the graph of the y = .3 x^2 function when x = 5.5, 6, 6.5 and 7?
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RESPONSE --> The coordinates of the x = 5 point can be found by plugging the value 5 for x in y = .3x^2. The point is (5, 7.5) The slope of the graph at this point is found by first finding y'. y' = .3(2)x. y'(5) = 3. The slope at this point is 3/1. The equation of the line through this point and having this slope is y - 7.5 = 3(x - 5). y = 3x - 7.5 My sketch shows the quadratic y = .3(x)^2 going through the points (1, .3), (2, 1.2), (3, 2.7), (4, 4.8), (5, 7.5), (6, 10.8), (7, 14.7). The line in my graph has the points (1, -4.5), (2, -1.5), (3, 1.5), (4, 4.5), (5, 7.5), (6, 10.5), (7, 13.5). When x = 5.5, 6, 6.5, 7 the line is -.075, -.3, -.675, -1.2 (respectively) away from the quadratic in the vertical direction. confidence assessment: 2
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14:36:53 The x = 5 point has y coordinate y = .3 * 5^2 = 7.5, so the point lies at (5, 7.5). The slope that that point is found by evaluating the derivative at that point. The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3. The line therefore passes through the point (5,7.5) and has slope 3. Its equation is therefore y - 7.5 = 3 ( x - 5), which simplifies to y = 3 x - 7.5. When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are y = 3 ( 5.5 ) - 7.5 = 9, y = 3 (6) -7.5 = 10.5, y = 3 (6.5) -7.5 = 12, y = 3 (7) -7.5 = 13.5. At the same x coordinates the function y =.3 x ^ 2 takes values y = .3 (5.5) ^ 2 = 9.075, y =.3 (6) ^ 2 = 10.8, y =.3 (6.5) ^ 2 = 12.675, and y =.3 (7) ^ 2 = 14.7. The straight line is therefore lower than the curve by 9.075 - 9 = .075 units when x = 5.5, 10.8-10.5 = .3 when x = 6, 12.675-12 = .675 when x = 6.5, and 14.7-13.5 = 1.2 when x = 7. We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly. Your sketch should show a straight line tangent to the parabolic curve y =.3 x^2, with the curve always above the line except that the point tangency, and moving more and more rapidly away from the line the further is removed from the point (5,7.5). The line you have drawn is called the line tangent to the curve at the point (5,7.5).
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RESPONSE --> self critique assessment:
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15:00:31 `q002. What is the equation of the line tangent to the curve y = 120 e^(-.02 t) at the t = 40 point? If we follow the tangent line instead of the curve, what will be the y coordinate at t = 40.3? How close will this be to the y value predicted by the original function? Will the tangent line be closer than this or further from the original function at t = 41.2?
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RESPONSE --> My equation of the line tangent to the curve is y = -1.07838(t) + 10.78427 y(40.3) = -32.6744 y(40) = -32.3509 These y values were a great deal different than those of the original curve. confidence assessment: 0
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15:10:17 `q004. What is the equation of the line tangent to the curve y = 20 ln( 5 x ) at the x = 90 point?
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RESPONSE --> Derivative = y' = 4/x y - (2/45) = (2/45)(x - 90) y - (2/45) = (2x/45 - 4) y = (2x/45) + (182/45) confidence assessment: 2
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15:11:02 At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx.. So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90. The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x. At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is ( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form y = .22 x + 106.
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RESPONSE --> The y value I used for my equation is the y value from the derivative not the y value from the original. self critique assessment:
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15:45:23 `q005. Using the tangent line and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.
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RESPONSE --> f(x) = x^5 f(3) = 243 f(3.1) = 286.29151 f'(x) = 5x^4 f'(3) = 405 f'(3.1) = 461.7605 Tangent line = y - 243 = 405(x - 3) Tangent line = y - 243 = 405x - 1215 Tangent line = y = 405x - 972 The difference in the estimated value and the actual value is 2.79151 confidence assessment: 0
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15:46:10 f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405. Thus tangent line passes through (3, 243) and has slope 405. Its equation is therefore (y - 243) / (x - 3) = 405, which simplifies to y = 405 x - 972. This function could be evaluated at x = 3.1. However a more direct approach simply uses the differential. In this approach we note that at x = 3 the derivative, which gives the slope of the tangent line, is 405. This means that between (3, 243) and the x = 3.1 point the rise/run ratio should be close to 405. The run from x = 3 to x = 3.1 is 1, so a slope of 405 therefore implies a rise equal to slope * run = 40.5. A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5. The actual value of 3.1^5 is about 286.3. The discrepancy between the straight-line approximation and the actual value is due to the fact that y = x^5 is concave up, meaning that the slope is actually increasing. The closely-related idea of the differential, which will be developed more fully in the next section, is that if we know the rate at which the function is changing at a given point, we can use this rate to estimate its change as we move to nearby points.
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RESPONSE --> I did not think I did this right at all. I know I did not put all of my steps in, but I am gla dto say i did it correctly. self critique assessment:
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15:56:00 `q006. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.
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RESPONSE --> I have no answer because I do not know how to do this. confidence assessment: 0
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15:56:58 At x = e we have ln(x) = ln(e) = 1 (this is because of the definition of the natural log function as the inverse of the exponential function). The derivative of ln(x) is 1/x, so at x = e the rate at which financial log function is changing is 1 / e. {}Since e = 2.718, approx., between x = e and x = 2.8 x changes by about 2.8 - 2.718 = .082. Since the rate at which x is changing in this vicinity remains close to x = 1/e, the change in y is approximately `dy = rate * `dx = 1/e * .082 = .082 / 2.718 = .030. Thus at x = 2.8, y = ln(x) takes value approximately 1 + .030 - 1.030. The actual value of ln(2.8) to six significant figures is 1.02961, which when rounded off to four significant figures is 1.030, in agreement with our approximation.
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RESPONSE --> Ok, I wrote it down to refference later, but I think i ahve the just of it now. self critique assessment:
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16:04:43 `q007. Using the tangent line verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the tangent line approximation for the function f(x) = `sqrt(x) at an appropriate point.
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RESPONSE --> The only number I could think to be close is .9. 1 - 'sqrt(.9) = .052 1 - .9 = .1 confidence assessment: 0
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16:05:31 The function f(x) = `sqrt(x) takes value 1 when x = 1. The derivative of f(x) = `sqrt(x) is 1 / ( 2 * `sqrt(x) ), which takes value 1/2 when x = 1. The tangent-line approximation to f(x) is therefore a straight line through (1, 1) having slope 1/2. This line has equation (y - 1) / (x - 1) = 1/2, or y = 1/2 ( x - 1) + 1. This number differs from 1 by 1/2 (x-1), which is half the difference between 1 and x. This approximation works for x values near 1. Thus if x is close to 1, then y is approximately twice as close to 1 as is x.
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RESPONSE --> self critique assessment:
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16:16:28 `q008. Using the tangent line approximaton verify that the square of a number close to 1 is twice as far from 1 as the number.
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RESPONSE --> f(x) = x^2 f'(x) = 2x y - 1 = 2x(x - 1) y - 1 = 2x^2 - 2x y = 2x^2 - 2x + 1 confidence assessment: 0
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16:18:31 The tangent-line approximation to f(x) = x^2, near x = 1, is y = 2(x-1) + 1. This number is twice as far from 1 as is x.{}{}STUDENT COMMENT: I am not really understanding this part. INSTRUCTOR RESPONSE: *&*& You should understand the part about the tangent-line approximation near x = 1 being y = 2(x-1) + 1.{}{}Graph y = 2(x-1) + 1. If you move horizontally over from the point (1, 1) you are changing x. If you then move vertically up to the graph from this point you will find the change in y. For this function that change will be double the change in x. *&*&.
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RESPONSE --> self critique assessment:
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