course mth173 assignment #014014. Tangent Lines and Tangent Line Approximations
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19:44:40 `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 --> We have the functions y=.3x^2 y'= .3(.2x^2-1) y'=.6x 7.5= 3(5) +b 7.5 = 15+b Ytan(5)= 3x - 7.5 x y y' Ytan 5 7.5 3 5.5 9.075 9 6 10.8 10.5 6.5 12.675 12 7 14.7 13.5 average descrepency of terms of y and Ytan(5): .56 As we move away from point (5, 7.5) the tangent y function moves away at an increaing rate from the tangent line confidence assessment: 2
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19:46:15 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 --> okay self critique assessment: 3
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19:58:13 `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 --> y= 120 e^(-.02t) y'= 120 (-.02e^(-.02t)) y'= -2.4e^(-.02t) Ytan(40)= (-1.08) x + 97.12 53.92= -1.08 * 40 + b 97.12 = b x y y' Ytan(40) 40 53.92 -1.08 40.3 53.60 53.60 41.2 52.64 52.62 The tangent line of point 40 is closer to the point at x= 40.3 the two equations show the same answer however they differntiate at a higher sig-sig value. But we can easily see that at point x= 41.2 differs more from the tangent line by -.02 which is more that at the x=40.3 point so as x moves farther away from 40 the function moves away with increasing rate from the tangent line. confidence assessment: 2
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20:01:04 The tangent line at the x = 40 point will have y coordinate y = 120 e^(-.02 * 40) = 53.9194, accurate to six significant figures. The derivative of the function is y ' = -2.4 e^(-.02 t) and at x = 40 has value y ' = -2.4 e^(-.02 * 40) = -1.07838, accurate to six significant figures. The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will therefore be y - 53.9194 = -1.07838 ( x - 40), which we can solve for y to obtain y = -1.07838 x + 97.0546. The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408. The y values corresponding to the tangent line are 53.5958 and 52.6253. We see that the tangent line at x = 40.3 is .0011 units lower than the curve of the function, and at x = 41.2 the tangent line is .0155 units lower--about 14 times as far from the curve as at the x = 40.3 point.
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RESPONSE --> I didn't go to as many points behind the decimal as you but I think I still got the idea that as we move farther from the tangent line point the function moves away from tangent line with an increasing rate. self critique assessment: 2
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20:06:57 `q003. Sketch a graph of this curve and the t = 40 tangent line and describe how the closeness of the tangent line to the curve changes as we move away from the t = 40 point.
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RESPONSE --> The curve forms a parabolic curve in quadrant 1 and the tangent touches the near tip of the curve for an instant at t=40. The tangent line, like the name implies, is a line! It holds the value of slope that the curve has at the t=40 instant, as the slope of the curve changes it moves away from the tangent line at an increasing rate. confidence assessment: 3
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20:07:25 Your sketch should show the gradually curving exponential function very close to the tangent line. The tangent line is straight while the slope of the exponential function is always increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the tangent line, moving away very gradually at first and then with increasing rapidity.
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RESPONSE --> okay self critique assessment: 3
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20:15:30 `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 --> y= 20 ln(5x) using the chain rule we derive y'=20(5 (1/x) =20 (5/5x) =20(1/x) y'= 20/x 122.18 = (.22)(90) + b 122.18 = 19.8 + b 102.38 = b Ytan= .22 x + 102.38 x y y' 90 122.18 .22 confidence assessment: 2
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20:20:55 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 --> our calculations are a bit different but I think it is because I used a calculator where you aproximated the value of ln(450) but otherwise the answers are similiar. self critique assessment: 2
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20:31:07 `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 --> y= x^5 y'= 5x^4 243 = 405 (3) +b 243 = 1215 +b -972= b ytan= 405 x -972 x y y' ytan 3 243 405 3.1 286.29 283.59 We can see that the the tangent of the curve at point x=3 for x=3.1 is 283.59 when the actual point for the function y=x^5 is 286.29 a substanial distance at such a small 'dx. confidence assessment: 2
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20:32:18 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 --> okay self critique assessment: 3
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20:49:39 `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 --> ln(x) when x=e , ln(e) = 1 because the two are inverse equations. The dirivative of ln(x) is 1/(x) so we have 1/e so if y=ln(x) = 1 and y'= 1/e= .368 The differance of 2.8 and e is .0817 'dy = (1/e) (.0817) = .0300 y + 'dy = 1.0300 x y y' e 1 .368 the estimation of ln(2.8) using the differntial was 1.0300 and the actual value of ln(2.8) is : 1.0296 both are rounded to 1.03. confidence assessment: 3
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20:50:07 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 --> okay self critique assessment: 3
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21:08:29 `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 --> Lets let x= 4 y= 'sqrt(x) = x^.5 y'= .5 x ^(-.5) 2= .25 (4) +b 2= 1 +b 1=b Ytan= .25x + 1 x y y' Ytan 4 2 .25 16 4 5 We can see by the tangent line that even though it only touches at 1 point on the 'sqrt funtion it remains close the the curve of the function for some time. Also by looking at some perfect squares we can see that the square root is closer to 1 than the number. confidence assessment: 2
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21:15:49 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 --> My tangent line was in y=mx +b form where yours is y=(1/2) (x-1) +1. I can see the logic behind this and but how did you determine (y - 1) / (x - 1) = 1/2 ? self critique assessment: 2
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21:24:36 `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 --> y= x^2 and y' = 2x 1= 2(1) +b b=1 Ytan= 2x +1 x y y' Ytan 1 1 2 2 4 2 5 4 16 8 9 We can see that as x increase it is more than twice as far away from 1 as the original number, as the Ytan is 2 times the number +1 and this becomes apparent for any number greater than 1. confidence assessment: 1
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21:26:02 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 --> again I don't understand the (x-1) where I have just x. self critique assessment: 2
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