#$&* course MTH 173 10/26 22:05
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Given Solution: `aThe differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx. At x = 3 the differential is `dy = 5 * 3^4 * `dx, or `dy = 405 `dx. Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5. Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx.. Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function f(x) = ln(x) The differential function of f(x) will be dy = f’(x)*dx Since f’(x) is 1/x, thus dy = (1/x)*dx The value of the differential function at x = e will be dy = (1/e)*dx The approximate value of e is 2.7182 Between x values 2.8 and 2.7182, dx = 0.0818 Thus dy between this x values, dy = 0.0818 / 2.7182 = 0.03 approximated Value of the function at x = e will be 1. Thus the value of the function at x = 2.8 = f(e) + dy = 1 + 0.03 = 1.03 approx The actual value of the function at x = 2.8 will be ln(2.8) = 1.0296 which differs from the estimated value of the function by 0.0004. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. STUDENT QUESTION I see were the rate is coming from but why this value of one is used?????? INSTRUCTOR RESPONSE The function is very easy to evaluate with pleanty of accurately if x = e. All you need to know is that, to four significant figures, e = 2.718. So we very easily see that ln(e) = 2.718. You can't accurately evaluate ln(2.8). However 2.8 is close to e, so if you know how quickly the function y = ln(x) is changing when x = e, you can easily extrapolate to x = 2.8. Of course you can just plug 2.8 into your calculator and get an accurate answer, but that provides no insight into the behavior of the function, or into the nature of this approximation, and gives you nothing on which to build later understanding. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Given function f(x) = sqrt(x) Thus the differential function of f(x) will be dy = f’(x) dx f’(x) = 1/(2*sqrt(x)), thus the differential function will be dy = 1/(2*sqrt(x))dx The differential value of the function at x = 1 will be dy = (1/2)*dx Thus we can see that the square root of a number close to 1 is twice as close to 1 as the number. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore `dy = 1 / 2 * `dx. This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function is f(x) = x^2 The differential function of f(x) is dy = f’(x)*dx, f’(x) = 2x Thus the differential function is dy = 2x*dx Thus value of the differential at x = 1 is dy = 2*dx Thus we can approximate that the square of a number close to 1 is twice as far from 1 as the number. confidence rating #$&*: 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: OK ********************************************* 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: Given function is L(t) = 400 - 250 e^(-.02t) The differential function thus will be dL = L’(t)*dt Thus L’(t) = -250 e^(-.02t)*(-.02) = 5 e^(-.02t) where t is the time in weeks since training Thus the differential function is dL = 5e^(-.02t)*dt Thus the value of the differential at t = 50 will be dL = 5*e^(-.02*50)dt = 1.83939*dt Thus for the next 2 weeks the approximate change in strength would be dL = 1.83939 * 2 = 3.67878 Thus the approximate change in strength predicted by the differential for the next 2 weeks will be 3.67878. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential is `dL = L ' (t) * `dt = -.02 ( -250 e^(-.02 t) ) `dt, so `dL = 5 e^(-.02 t) `dt. At t = 50 we thus have `dL = 5 e^(-.02 * 50) `dt, or `dL = 1.84 `dt. The change over the next `dt = 2 weeks would therefore be approximately `dL = 1.84 * 2 = 3.68. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function I(r) = k / r^2 where k is a constant number and r is distance in meters from the light. Thus the differential function will be dI = I’(r)*dr And I’(r) = -2k / r^3 Thus the differential function will be dI = -2k / r^3 * dr The value of the differential at r = 10 meters will be dI = -2k / 1000 * dr The change in the illumination as we move from r = 10 meters to r = 10.3 meters, dr will be 0.3 Thus dI = (-2k / 1000) * 0.3 = -0.0006k illumination unit Thus the change in illumination from r = 10 meters to r = 10.3 meters will be -0.0006k illumination unit. confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Let the width be x, thus the length becomes 2x. Thus the area function of the crystal will be A(x) = 2x^2 Given that the width change by .1 cm every hour Thus dx = 0.1 dt , where x is in cm and t is time in hours Thus the differential function of the area will be dA = A’(x) * dx and A’(x) = 4x thus dA = 4x dx, this is the area differential function with respect to width but we also know that dx = .1 dt thus the differential may also be written as dA = 4x * .1dt = .4x * dt thus dA = .4x*dt is the differential function of area with respect to time in hours The value of the differential at x = 5 cm will be dA = 2*dt Thus to approximate the change in area after 1 hour, dt will be 1 hour Thus dA = 2*1 = 2 cm^2 When width is 5 cm the area of crystal is 2*5*5 cm^2 = 50 cm^2 Thus the area of the crystal after 1 hour approximated is = 50 cm^2 + 2 cm^2 = 52 cm^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5. f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20. The rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width. A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area. Thus the approximate area should be 50 + 2 = 52. This can easily be compared with the accurate value of the area which is 52.02. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given that the volume of the sphere is V(r) = (4/3)*(pi)*(r^3) Given that the radius of the sphere in cm increases at the rate of .3 cm per day, Thus dr = .3 dt, where t is the number of days Thus to find the change in volume per day we use the derivative of the volume function dV = V’(r)*dr and V’(r) = (4/3)*(pi)*(3r^2) thus the differential function of volume with respect to radius is dV = (4/3)*(pi)*(3r^2) dr since dr = .3 dt, thus the differential function with respect to days will be dV = (4/3)*(pi)*(3r^2)*(.3dt) = .4*(pi)*(3r^2)*dt thus the value of the differentiation when r = 20 cm dV = .4*(pi)*(3*20*20)*dt = 1507.9644*dt thus the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm is 480pi cm^3 / day or approximately 1507.9644 cm^3/day. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION Were did the initial formula come from? I think I see how the problem was solved after that but were did V(r) = 4/3 * `pi * r^3 come from? INSTRUCTOR RESPONSE That is the formula for the volume of a sphere. This should be general knowledge. This formula was also used in some of the Orientation exercises. You should know everything that was covered in those exercises. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!