#$&* course Mth 277 9/04 6 pm 015. The differential and the tangent line*********************************************
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Given Solution: `aThe differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx. At x = 3 the differential is `dy = 5 * 3^4 * `dx, or `dy = 405 `dx. Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5. Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx.. Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When x = e, the value of the function f(x) = ln(x) is ln(e) which is 1. The derivative of this function is 1/x, so at x = e, the slope of the tangent is 1/e. e is approximately 2.718, so the difference between e and 2.8 is about 0.082. The rise from e to 2.8 should be close to 1/e * 0.082, or 0.0302. Since the y value at e is 1, the y value at 2.8 should be about 1.0302. When calculated, the value of the function at x = 2.8 is 1.0296 approx. There is a difference of 0.0006 between the two, and the actual value is lower. confidence rating #$&*: 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: For the function f(x) = sqrt(x) the y value at x = 1 is 1, and the derivative is 1/(2 * sqrt(x) ). The equation for the tangent line will then be y = 1/2 x + 1/2. Now we can use this equation to see whether or not the statement above is correct. I will use x = 1.1. When I substitute 1.1 into the equation, I get ½ (1.1) + ½ = 1.05. 1.05 is twice as close to 1 as 1.1, so the statement seems correct. I’ll verify by using the original equation. f(1.1) = sqrt(1.1) = 1.049 approx. These two values are very close, so the statement is correct for numbers close to 1. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore `dy = 1 / 2 * `dx. This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the function f(x) = x^2 the y value when x = 1 is 1 and the derivative is 2x. The equation for the tangent line is y = 2x - 1. Like the previous problem, I will estimate using the tangent line and verify using the function. For x = 1.1, the y value on the tangent line is 1.2. this number is twice as far from x as the original, so it supports the statement for numbers close to 1. When I use 1.1 in the function, I get 1.21, which is close to my estimate and supports the statement. confidence rating #$&*: 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: The differential of the function is L’ (t) = 5 e^(-0.02 t). When t = 50, the rate of change is about 1.84 per week, so the approximate change in strength would be 3.68. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe differential is `dL = L ' (t) * `dt = -.02 ( -250 e^(-.02 t) ) `dt, so `dL = 5 e^(-.02 t) `dt. At t = 50 we thus have `dL = 5 e^(-.02 * 50) `dt, or `dL = 1.84 `dt. The change over the next `dt = 2 weeks would therefore be approximately `dL = 1.84 * 2 = 3.68. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The differential for this function is I’ (r) = -2k / r^3. When r = 10, he rate of change is -2k / 10^3, or -k / 500 per meter. To find the estimated change, I take the change in meters and multiply it by the rate of change. 0.3 * -k / 500 = -3k/5000. confidence rating #$&*: 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: f (x) = 2 x^2 f’ (x) = 4x When the width of the crystal is 5 cm, the crystal’s area is 50 cm^2 and its rate of change is 20 cm^2/ cm. One hour later the crystal’s width will be 5.1 cm. Since the crystal’s area is growing (for the moment) at a rate of 20 cm^2 per 1 cm of width, the estimated growth of the crystal will be 2 cm^2, making the total area about 52 cm^2. The actual area is 52.02, so estimate is close. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5. f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20. The rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width. A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area. Thus the approximate area should be 50 + 2 = 52. This can easily be compared with the accurate value of the area which is 52.02. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V(r) = 4/3 pi * r^3 V’ (r) = 4 pi * r^2 When r = 20 cm, the approximate rate at which the volume of the sphere is increasing is: V’ (20) = 4 pi * (20)^2 = 1600 pi = 5026.55 cm^3/ cm approx. confidence rating #$&*: 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): I forgot the radius was changing at a rate of 0.3 cm per day. This means I left the answer in terms of “change in volume”/ “change in radius” instead of converting to “change in volume”/ day. A simple and easily corrected mistake, but my answer is still wrong. ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: