course Mth 173 MzfvX̚ψ癙Student Name:
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13:04:17 `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 --> f(3)=(3)^5 = 243 f(3.1) = 286.29
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13:09:21 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 --> ok i understand
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13:12:50 `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 --> I do not understand the function. Is the function just x=e? I don't understand what I'm supposed to plug into where.
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13:17:33 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 --> ok, I understand now
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13:26:15 `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 --> f(1) = sqrt(1) = 1 f(.9) = sqrt(.9) = .9486 f(1-.9) = sqrt(.1) = .316 .316 x 2 = .632 which means .316 is twice as far away as sqrt(.9)
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13:28:29 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 --> ok
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13:58:41 `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 --> 'dy = 2 * 'dx
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14:00:38 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 --> ok
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14:11:08 `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 --> L'(t) = -250(-.02)*e^(-.02t) = 5e^(-.02t) L(50) = 400- 250e^(-.02(50)) = 216.06 L(64) = 260.98 260.98 - 216.06 = 44.92
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14:12:42 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 --> I am confused because I thought f ' (x) represents the rate of change, not the amount of change?
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14:34:53 `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 --> OK, I believe I have figured this out. You are asking for the change/unit, not the total amount of change.
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14:37:48 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 --> ok i understand
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14:46:38 `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 --> A = 2x^2 = 2(5)^2 = 50 A' = 4x dx A'(5) = 4(5) *.1 = 8 A = 58 1 hour later
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14:47:51 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 --> ok
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14:51:13 `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 --> A = 4/3 pi r^2 A' = 8/3 pi r A' = 8/3 pi (20) * .3 = 50.265 cm^3 per day
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14:59:19 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 --> ok
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ดv Student Name: assignment #016 016. Implicit differentiation.
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16:47:57 `q001. Suppose that y is a function of x. Then the derivative of y with respect to x is y '. The derivative of x itself, with respect to x, is by the power function rule 1, x being x^1. What therefore would be the derivative of the expression x^2 y?
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RESPONSE --> x^1y + x^2
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16:52:33 By the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' .
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RESPONSE --> ok
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16:58:48 `q002. Note that if y is a function of x, then y^3 is a composite function, evaluated from a given value of x by the chain of calculations that starts with x, then follows the rule for y to get a value, then cubes this value to get y^3. To find the derivative of y^3 we would then need to apply the chain rule. If we let y' stand for the derivative of y with respect to x, what would be the derivative of the expression y^3 with respect to x?
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RESPONSE --> I am completely confused
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17:07:01 The derivative of y^3 with respect to x can be expressed more explicitly by writing y^3 as y(x)^3. This reinforces the idea that y is a function of x and that we have a composite f(y(x)) of the function f(z) = z^3 with the function y(x). The chain rule tells us that the derivative of this function must be (f ( y(x) ) )' = y ' (x) * f ' (y(x)), in this case with f ' (z) = (z^3) ' = 3 z^2. The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2. In shorthand notation, (y^3) ' = y ' * 3 y^2. This shows how the y ' comes about in implicit differentiation.
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RESPONSE --> So y(x)^3 is like saying G(F(x)). ok
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17:08:16 `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3?
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RESPONSE --> 2x * 2x* 3y^2 = 4x^2 * 3y^2
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17:16:37 The derivative of x^2 y^3, with respect to x, is (x^2 y^3) ' = (x^2)' * y^3 + x^2 * (y^3) ' = 2x * y^3 + x^2 * [ y ' * 3 y^2 ] = 2 x y^3 + 3 x^2 y^2 y '. Note that when the expression is simplified, the y ' is conventionally placed at the end of any term of which is a factor.
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RESPONSE --> ok
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17:19:52 `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?
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RESPONSE --> hmmm...
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17:22:23 Starting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain 2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain y = (9 - 7 x ) / (2 x^2), or if we prefer y = 9 / (2 x^2 ) - 7 / ( 2 x ).
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RESPONSE --> That was simple and I made it difficult. I was looking at the equation as 2x^(2y) + 7x = 9
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17:29:07 `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1?
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RESPONSE --> y' = -22 y = 4 y = 18x^ -2 - 14x^ -1 y' = -26x^ -3 + 14x^ -2
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17:41:02 y ' = 9/2 * ( 1/x^2) ' - 7/2 ( 1 / x) ' = 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) = -9 / x^3 + 7 / (2 x^2). So when x = 1 we have y = 9 / (2 * 1^2 ) - 7 / ( 2 * 1 ) = 9/2 - 7/2 = 1 and y ' = -9 / 1^3 + 7 / (2 * 1^2) = -9 + 7/2 = -11 / 2.
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RESPONSE --> ok
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17:50:42 `q006. The preceding calculation could have been obtained without solving the original equation 2x^2 y + 7 x = 9 for y. We could have taken the derivative of both sides of the equation to get 2 ( (x^2) ' y + x^2 y ' ) + 7 = 0, or 2 ( 2x y + x^2 y ' ) + 7 = 0. Complete the simplification of this equation, then solve for y ' . Note that when x = 1 we have y = 1. Substitute x = 1, y = 1 into the equation for y ' and see what you get for y '.
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RESPONSE --> 2(2(1)(1)) + 1^2 y' +7 = 0 4 + 7 + y' = 0 -11 = y'
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17:51:32 Starting with 2 ( 2x y + x^2 y ' ) + 7 = 0 we divide both sides by 2 to obtain 2x y + x^2 y ' + 7/2 = 0. Subtracting 2 x y + 7 / 2 from both sides we obtain x^2 y' = - 2 x y - 7 / 2. Dividing both sides by x^2 we end up with y ' = (- 2 x y - 7 / 2) / x^2 = -2 y / x - 7 / ( 2 x^2). Substituting x = 1, y = 1 we obtain y ' = -2 * 1 / 1 - 7 / ( 2 * 1^2) = - 11 / 2. Note that this agrees with the result y ' = -11/2 obtained when we solved the original equation and took its derivative. This shows that it is not always necessary to explicitly solve the original equation for y. This is a good thing because it is not always easy, in fact not always possible, to solve a given equation for y.
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RESPONSE --> ok
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