course Mth 173 3/4 9 Question: `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.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(2.8) = 1.03 1 / 2.8 = 0.357 * ‘dx Ln(e) = 1/e * ‘dx = 0.368 * ‘dx = 0.03 Difference of 1. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Since ln(e) = 1, I have to add 1 to equation. This would make the difference from the actual value almost nothing. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q003. Using the differential verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the differential approximation for the function f(x) = `sqrt(x) at an appropriate point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dy = 1 / 2 * ‘dx This equation shows that y only goes half as far as x. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ‘dy = 1 / (2 ‘sqrt(x)) * ‘dx, OK ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dy = 2 * ‘dx confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q005. The lifting strength of an athlete in training changes according to the function L(t) = 400 - 250 e^(-.02 t), where t is the time in weeks since training began. What is the differential of this function? At t = 50, what approximate change in strength would be predicted by the differential for the next two weeks? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L’(t) = -.02 * -250 e^(-.02(50)) * ‘dx 1.839 * ‘dx Two weeks = 2 1.839 * 2 = 3.678 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’(r) = (-2k)/(r^3) * ‘dx (-2k)/(1000) * .3 = -6k / 1000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): -6k / 1000 = [-.0006k] ------------------------------------------------ Self-critique rating #$&*: 2 ********************************************* Question: `q007. A certain crystal grows between two glass plates by adding layers at its edges. The crystal maintains a rectangular shape with its length double its width. Its width changes by .1 cm every hour. At a certain instant its width is 5 cm. Use a differential approximation to determine its approximate area 1 hour later. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not sure how to set it up. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2x^2 4x = 20 20 * .1 = 2 + 50 = 52 Difference of only .02 from actual value. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V = 4/3’pi (r^3) V’ = 4’pi (r^2) 5026.548 * .3 = 1507.964 cm/day confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. Self-Critique: 1600’pi(cm^3) [480 pi = 1507.964] "