I do not have direct access to my webs this week, which is causing delays in my responses. I do have access to my email. If you want quicker feedback, you may use email; however continue to also submit your work by the form. The work included here will be posted by the first of next week. 016. Implicit differentiation. 03-12-2007
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09:01:27 `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^2 * y ' 2x * y + x^2 * y ' confidence assessment: 2
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09:01:31 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 --> self critique assessment: 3
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09:06:23 `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 --> y^3 ' y ' * 3 y ^ 2 confidence assessment: 3
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09:06:27 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 --> self critique assessment: 3
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09:08:09 `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 --> (x^2 * y^3) ' 2x * y^3 + 3 x^2 * y^2 * y' confidence assessment: 2
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09:08:13 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 --> self critique assessment: 3
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09:09:40 `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?
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RESPONSE --> 2 x^2 * y + 7x = 9 2 x^2 * y = 9 - 7x y = 9 - 7x / 2 x^2 confidence assessment: 2
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09:09:44 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 --> self critique assessment: 3
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09:16:18 `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 = 9 / (2 x^2) - 7 / (2 x) y' = 9 / 2 * 1/x^2 - 7 / 2 * 1/x y' = 9 / 2 * -2/x^3 - 7 / 2 * -1/x^2 y' = -9 / x^3 + 7 / 2 x^2 y = 9 / 2(1^2) - 7 / 2(1) y = 9 / 2 - 7 / 2 y = 2/2 y = 1 y' = -9 / 1^3 + 7 / 2(1^2) y' = -9 / 1 + 7 / 2 y' = -2 / 2 confidence assessment: 2
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09:20:12 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 --> y' = -9 + 7 = -2 I don't understand where -11/2 came from. self critique assessment: 1
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09:27:34 `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 ( (x^2) ' y + x^2 y ' ) + 7 = 0 2x * y + x^2 * y ' + 7/2 = 0 x^2 * y ' + 7/2 = -2x * y x^2 * y ' = -2x * y - 7/2 y' = (-2x * y - 7/2) / x^2 y' = (-2 * y/x - 7) / 2x^2
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09:27:54 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 --> I still don't understand where the -11 is coming from.
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09:40:26 `q007. Follow the procedure of the preceding problem to determine the value of y ' when x = 1 and y = 2, for the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0. Also validate the fact that x = 1, y = 2 is a solution to the equation, because if this isn't a solution it makes no sense to ask a question about the equation for these values of x and y.
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RESPONSE --> 2 x^2 y^3 - 3 x y^2 - 4 = 0 (2 x^2 * y^3) - (3 x * y^2) - 4 = 0 (2 x^2) * y^3 + 2 x^2 * (y^3) - (3x) * y^2 - 3x * (y^2) = 0 4x * y^3 + 6x^2 * y^2 * y'- 3y^2 - 6xy * y' = 0 6x^2 * y^2 * y' - 6xy * y' = -4x * y^3 + 3y^2 y'(6x^2 * y^2 - 6xy) = -4x*y^3 + 3y^2 y' = (-4x*y^3 + 3y^2)/(6x^2 * y^2 - 6xy) y' = (-4x * y^2 + 3y)/(6x^2 * y - 6x) 2 x^2 y^3 - 3 x y^2 - 4 = 0 2 * 1^2 * 2^3 - 3(1)(2^2) - 4 = 0 2 * 1 * 8 - 12 - 4 = 0 16 - 12 - 4 = 0 y' = (-4x * y^2 + 3y)/(6x^2 * y - 6x) y' = (-4(1) * (2^2) + 3(2)) / (6(1^2) * 2 - 6(1)) y' = (-4 * 4 + 6) / 6 * 2 - 6 y' = -16 + 6 / 12 - 6 y' = -10 / 6 y' = -1.66667 confidence assessment: 3
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09:40:29 `qNote that this time we do not need to solve the equation explicitly for y. This is a good thing because this equation is cubic in y, and while there is a formula (rather a set of formulas) to do this it is a lengthy and messy process. The derivative of the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 is (2 x^2 y^3) ' - (3 x y^2) ' - (4)' = 0. Noting that 4 is a constant, we see that (4)' = 0. The derivative of the equation therefore becomes (2 x^2) ' * y^3 + 2 x^2 * ( y^3) ' - (3 x ) ' * y^2 - 3 x * (y ^2 ) ' = 0, or
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RESPONSE --> confidence assessment: 3
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09:40:33 4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0. Subtracting from both sides the terms which do not contain y ' we get
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RESPONSE --> confidence assessment: 3
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09:40:36 6 x^2 y^2 y ' - 6 x y y ' = - 4 x * y^3 + 3 y^2. Factoring out the y ' on the left-hand side we have
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RESPONSE --> confidence assessment: 3
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09:40:39 y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ':
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RESPONSE --> confidence assessment: 3
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09:40:42 y ' = (- 4 x * y^3 + 3 y^2) / ( 6 x^2 y^2 - 6 x y ) . The numerator and denominator have common factor y so we end up with
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RESPONSE --> confidence assessment: 3
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09:40:45 y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ). Now we see that if x = 1 and y = 2 the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 gives us 2 * 1^2 * 2^3 - 3 * 1 * 2^2 - 4 = 0, or 16 - 12 - 4 = 0, which is true. Substituting x = 1 and y = 2 into the equation y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ) we get y ' = (- 4 * 1 * 2^2 + 3 * 2) / ( 6 * 1^2 * 2 - 6 * 1 ) = (-16 + 6) / (12 - 6) = -10 / 6 = -5/3 or -1.66... .
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RESPONSE --> confidence assessment: 3
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09:48:45 `q008. (Mth 173 only; Mth 271 doesn't require the use of sine and cosine functions). Follow the procedure of the preceding problem to determine the value of y ' when x = 3 and y = `pi, for the equation x^2 sin (y) - sin(xy) = 0.
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RESPONSE --> x^2 sin (y) - sin(xy) = 0 x^2 ' sin(y) + x^2 * sin(y) ' - sin(xy) = 0 2 x * sin(y) + x^2 (y' cos(y)) - (y + xy ')cos(xy) = 0 2 x * sin(y) + x^2 (cos(y)) y ' - y cos(xy) - x cos(xy) y ' = 0 (x^2 * cos(y) - x * cos(xy)) y ' = y * cos(xy) - 2x * sin(y) y' = y * cos(xy) - 2x * sin(y) / x^2 * cos(y) - x * cos(xy) y' = 'pi * cos(3('pi)) - 2 * 3(sin('pi)) / 3^2 * cos('pi) - 3 * cos (3*'pi) y' = 'pi * -1 - 2 * 3 * 0 / 9 * -1 - 3 * -1 y' = 'pi / -6 confidence assessment: 2
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09:48:55 Taking the derivative of both sides of the equation we obtain (x^2) ' sin(y) + x^2 (sin(y) ) ' - ( sin (xy) ) '. By the Chain Rule (sin(y)) ' = y ' cos(y) and (sin(xy)) ' = (xy) ' cos(xy) = ( x ' y + x y ' ) cos(xy) = (y + x y ' ) cos(xy). So the derivative of the equation becomes 2 x sin(y) + x^2 ( y ' cos(y)) - ( y + x y ' ) cos(xy) = 0. Expanding all terms we get 2 x sin(y) + x^2 cos(y) y ' - y cos(xy) - x cos(xy) y ' = 0. Leaving the y ' terms on the left and factoring ou y ' gives us [ x^2 cos(y) - x cos(xy) ] y ' = y cos(xy) - 2x sin(y), so that y ' = [ y cos(xy) - 2x sin(y) ] / [ x^2 cos(y) - x cos(xy) ]. Now we can substitute x = 3 and y = `pi to get y ' = [ `pi cos( 3 * `pi) - 2 * 3 sin(`pi) ] / [ 3^2 cos(`pi) - 3 cos(3 * `pi) ] = [ `pi * -1 - 2 * 3 * 0 ] / [ 9 * -1 - 3 * -1 ] = -`pi / 6.
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RESPONSE --> self critique assessment: 3
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13:47:06 5.4.2 fn piecewise linear from (0,1)-(1,1)-(3,-1)-(5,-1)-(6,1). F ' = f, F(0) = 0. Find F(b) for b = 1, 2, 3, 4, 5, 6.What are your values of F(b) for b = 1, 2, 3, 4, 5, 6 and how did you obtain them.
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RESPONSE --> 1st: the area is: 1 F(1) = 1 2nd: the area is: .5 F(2) = 1 + .5 = 1.5 3rd: the area is: -.5 F(3) = 1.5 + -.5 = 1.0 4th: the area is: -1 F(4) = 1.0 + -1.0 = 0 5th: the area is: -1 F(5) = 0 + -1 = -1 6th: the area is: 0 F(6) = -1 + 0 = -1
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13:47:09 ** f(x) is the 'rate function'. The change in the 'amount function' F(x) is represented as the accumulated area under the graph of f(x). Since F(0) = 0, F(x) will just the equal to the accumulated area. We can represent the graph of f(x) as a trapezoidal approximation graph, with uniform width 1. The first trapezoid extends from x = 0 to x = 1 and has altitude y = 1 at both ends. Therefore its average altitude is 1 and its what is 1, giving area 1. So the accumulated area through the first trapezoid is 1. Thus F(1) = 1. The second trapezoid has area .5, since its altitudes 1 and 0 average to .5 and imply area .5. The cumulative area through the second trapezoid is therefore still 1 + .5 = 1.5. Thus F(2) = 1.5. The third trapezoid has altitudes 0 and -1 so its average altitude is -.5 and width 1, giving it 'area' -.5. The accumulated area through the third trapezoid is therefore 1 + .5 - .5 = 1. Thus F(3) = 1. The fourth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fourth trapezoid is therefore 1 + .5 - .5 - 1 = 0. Thus F(4) = 0. The fifth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fifth trapezoid is therefore 1 + .5 - .5 - 1 - 1 = -1. Thus F(5) = -1. The sixth trapezoid has altitudes 1 and -1 so its average altitude, and therefore it's area, is 0. Accumulated area through the sixth trapezoid is therefore 1 + .5 - .5 - 1 - 1 + 0 = -1. Thus F(6) = -1. **
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RESPONSE -->
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13:49:25 If you made a trapezoidal graph of this function what would be the area of teach trapezoid, and what would be the accumulated area from x = 0 to x = b for b = 1, 2, 3, 4, 5, 6? What does your answer have to do with this question?
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13:53:51 Query 5.4.12. integral of e^(x^2) from -1 to 1. How do you know that the integral of this function from 0 to 1 lies between 0 and 3?
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RESPONSE --> Average value * length of interval = integral You know that the integral of this function lies less than 3 because the highest value of the function is e, which is 2.71... This means that the average value cannot be over 3. The length of interval is below 3 because the curve is below 3 so this makes the interval below 3.
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13:53:52 ** The max value of the function is e, about 2.71828, which occurs at x = 1. Thus e^(x^2) is always less than 3. On the interval from 0 to 1 the curve therefore lies above the x axis and below the line y = 3. The area of this interval below the y = 3 line is 3. So the integral is less than 3. Alternatively e^(x^2) is never as great as 3, so its average value is less than 3. Therefore its integral, which is average value * length of interval, is less that 3 * 1 = 3. **
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13:53:57 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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