course Mth173 assignment #015015. The differential and the tangent line
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20:11:11 `q001. Using the differential 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 --> x= 3->3.1 f(x)= x^5 f'(x)= 5x^4 Since 'dy = y' * 'dx 'dy= (5(3)^4) (.1) 'dy= 40.5 f(3)= (3)^5 = 243 f(3.1) = 243 + 40.5 = 283.5 When in actuality f(3.1)= 286.3 due to the change in rate from 3 to 3.1, but the answer from the differential is still reasonably close. confidence assessment: 2
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20:12:03 The differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx. At x = 3 the differential is `dy = 5 * 3^4 * `dx, or `dy = 405 `dx. Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5. Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx.. Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3.
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RESPONSE --> okay self critique assessment: 3
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20:17:55 `q002. 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 --> x= e-> 2.8 y=ln(x) y'= 1/x 'dy = (1/x) ('dx) = (1/e) (2.8 - e) =.0301 y=ln(e) =1 1+ .0301= 1.0301 When in actuality ln(2.8) = 1.0296 again a reasonably close estimation using the differential. confidence assessment: 2
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20:18:17 The differential of the function y = f(x) = ln(x) is `dy = f ' (x) `dx or `dy = 1/x `dx. If x = e we have `dy = 1/e * `dx. Between e and 2.8, `dx = 2.8 - e = 2.8 - 2.718 = .082, approx.. Thus `dy = 1/e * .082 = .030, approx.. Since ln(e) = 1, we see that ln(2.8) = 1 + .030 = 1.030, approx..
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RESPONSE --> okay self critique assessment: 3
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20:23:10 `q003. Using the differential verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the differential approximation for the function f(x) = `sqrt(x) at an appropriate point.
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RESPONSE --> y= 'sqrt(x) = x^.5 y'= .5x^(-.5) When evaluated for x=1 we get y' = .5(1)^(-.5) =.5(1) =.5 Since x =1 then y'= .5 which is twice as close to 0 as x as we move away from zero both move away from 0 but x moves away twice as fast as y'. confidence assessment: 2
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20:26:35 The differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore `dy = 1 / 2 * `dx. This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much.
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RESPONSE --> I used went about finding the derivative with a different approach but ultimatly we ended at the same answer of y'= .5 x self critique assessment: 2
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20:30:28 `q004. Using the differential approximation 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 y'=2x 'dy = (2x) ('dx) When evaluated for x=1 'dy = (2(1)) ('dx) 'dy= 2 ('dx) we can see that with x=1 then y'= 2 which is twice as far away from zero as 1 is, and this pattern will continue as we move farther from 1. confidence assessment: 2
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20:31:20 The differential for this function is easily seen to be `dy = 2 x * `dx. At x = 1 the differential is therefore `dy = 2 * `dx. This shows that as we move away from x = 1, the change in y is double the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but by twice as much.
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RESPONSE --> okay self critique assessment: 3
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20:36:24 `q005. The lifting strength of an athlete in training changes according to the function L(t) = 400 - 250 e^(-.02 t), where t is the time in weeks since training began. What is the differential of this function? At t = 50, what approximate change in strength would be predicted by the differential for the next two weeks?
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RESPONSE --> t= 50->52 L(t)= 400 - 250 e^(-.02t) L'(t)= -250 (-.02e^(-.02t)) = 5e^(-.02t) 'dy= (5e^(-.02(50)))(2) = 3.68 A 3.68 change in lifting units would be expected by working out for an additional 2 weeks. (again this is approximate because of the rate change between 50 and 52 hwich isn't taken into account here) confidence assessment: 2
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20:36:41 The differential is `dL = L ' (t) * `dt = -.02 ( -250 e^(-.02 t) ) `dt, so `dL = 5 e^(-.02 t) `dt. At t = 50 we thus have `dL = 5 e^(-.02 * 50) `dt, or `dL = 1.84 `dt. The change over the next `dt = 2 weeks would therefore be approximately `dL = 1.84 * 2 = 3.68.
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RESPONSE --> okay self critique assessment: 3
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20:45:26 `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters.
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RESPONSE --> r= 10 -> 10.3 I(r)= k/r^2 I'(r)= -2k/r^3 'dI= (-2k/r^3) ('dr) =(-2k/(10)^3) (.3) =(-2k/1000) (.3) = (-.002k)(.3) = -6*10^-4 k The change in illumination as you move away from 10 meters to 10.3 meters is -.0006k units (lumens)? confidence assessment: 2
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20:45:40 The differential is `dI = I ' (r) * `dr, where I ' is the derivative of I with respect to r. Since I ' (r) = - 2 k / r^3, we therefore have `dI = -2 k / r^3 * `dr. For the present example we have r = 10 m and `dr = .3 m, so `dI = -2 k / (10^3) * .3 = -.0006 k. This is the approximate change in illumination.
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RESPONSE --> okay self critique assessment: 3
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20:54:10 `q007. A certain crystal grows between two glass plates by adding layers at its edges. The crystal maintains a rectangular shape with its length double its width. Its width changes by .1 cm every hour. At a certain instant its width is 5 cm. Use a differential approximation to determine its approximate area 1 hour later.
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RESPONSE --> The area of the crystal is given by A=lw since l=2w so we get the formula A=2w^2. A'= 4w A(5)= 2(5)^2 A(5)= 50 A'(5)= 20 Since there is a .1cm growth in width/hour 'dx is .1 'dy = (20) (.1) 'dy= 2cm added to the orginal value of 50cm we get 52 cm. confidence assessment: 2
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20:57:24 If the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5. f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20. The rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width. A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area. Thus the approximate area should be 50 + 2 = 52. This can easily be compared with the accurate value of the area which is 52.02.
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RESPONSE --> I forgot to add the actual value to compare against, it is 52.02 rather than 52. But again it's pretty close using the differential equation. self critique assessment: 2
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21:02:56 `q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm.
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RESPONSE --> V= (4/3) 'pi r^2 V'= 4'pi r^2 The rate of change is .3cm of radius /day at 20cm v'(20)= 4'pi(20)^2 = 5026.55cm^3 / cm of radius (.3cm/day) So we get an increase of 1507.97cm^3 / day when the radius is around 20cm. confidence assessment: 2
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21:04:19 The volume of a sphere is V(r) = 4/3 * `pi * r^3. The rate at which the volume changes with respect to the radius is dV / dr = 4 * `pi * r^2. Thus when r = 20 the volume is changing at a rate of 4 * `pi * 20^2 = 1600 `pi cm^3 volume per cm of radius. It follows that if the radius is changing by .3 cm / day, the volume must be changing at 1600 `pi cm^3 / (cm of radius) * .3 cm of radius / day = 480 `pi cm^3 / day. Note that this is the instantaneous rate at the instant r = 20. This rate will increase as r increases.
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RESPONSE --> I didn't use the 'pi notation which could have simplified my answer some but ultimatly we have the same answer. self critique assessment: 2
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