#$&* course Mth 173 Question: `q001. Using the differential and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.
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Given Solution: `aThe differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx. At x = 3 the differential is `dy = 5 * 3^4 * `dx, or `dy = 405 `dx. Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5. Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx.. Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The differential will be 1/x * `dx in this case. 'dx = .082, same as the last qa. ln(e) = 1. ln(2.8) = 1.0296. 1/e * .082 = .030 We can add that to the original y value to get 1.030 for the second data point. These actual value is, for all intents and purposes, the same as the differential value once we round. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: As shown before, the derivative of the sqrt(x) is 1/(2 'sqrt(x)). In this case, however, since we are starting at 1, the derivative will just be 1/2. So, the differential in this case is 'dy = .5`dx. Now, using common sense, we can see that any change in x will result in exactly half of that change in y. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore `dy = 1 / 2 * `dx. This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The differential for x^2 is `dy = 2x * `dx So, starting at 1 we have `dy = 2`dx. Again, as the previous question, we can easily see that any change x will result in twice that change in y. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: The differential for this function is -.02(-250 e^(-.02t)), which simplifies to `dy = 5e^(-.02t) * `dt. Then we just calculate the differential for t = 50, and `dt = 2. 5e^(-.02 * 50) * 2 1.84 * 2 = 3.68 will be the change over the next two weeks. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential is `dL = L ' (t) * `dt = -.02 ( -250 e^(-.02 t) ) `dt, so `dL = 5 e^(-.02 t) `dt. At t = 50 we thus have `dL = 5 e^(-.02 * 50) `dt, or `dL = 1.84 `dt. The change over the next `dt = 2 weeks would therefore be approximately `dL = 1.84 * 2 = 3.68. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of this function is -2k / r^3, which makes the differential -2 k / r^3 * `dr. We can see `dy with the function using r = 10 and `dr = .3. -2k / 10^3 * .3 = -.0006k is roughly how much the illumination will change. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ????Why is the derivative of this function -2k / r^3 rather than k / 2r like most other derivatives of exponents?????
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: The function to get the volume of a sphere is (4/3)`pi * r^3 The derivative of this function is 3(4/3) * `pi * r ^ 2. Which simplifies to `dy = 4`pi * r^2. At 20 `dy = 4`pi * 400 = 1600 * `pi. With the radius increasing by .3 each day we can do 1600`pi * .3 = 480`pi cm^3 is the estimated increase in volume each 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): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!