#$&* course Mth 271 Question: `q 2b 7th edition 2.7.16 (was 2.7.10) dy/dx at (2,1) if x^2-y^3=3
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Given Solution: `a The derivative of x^2 with respect to x is 2 x. The derivative of y^3 with respect to x is 3 y^2 dy/dx. You can see this by realizing that since y is implicitly a function of x, y^3 is a composite function: inner function is y(x), outer function f(z) = z^3. So the derivative is y'(x) * 3 * f(y(x)) = dy/dx * 3 * y^3. So the derivative of the equation is 2 x - 3 y^3 dy/dx = 0, giving 3 y^2 dy/dx = 2 x so dy/dx = 2 x / ( 3 y^2). At (2,1), we have x = 2 and y = 1 so dy/dx = 2 * 2 / (3 * 1^2) = 4/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 3b 7th edition 2.7.28 (was 2.7.22) slope of x^2-y^3=0 at (1,1) What is the desired slope and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: d/dx (x^2 - y^3) = 2x - 3y^2y’ y’ = 2x/3y^2 y’ = 2/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of the equation is 2 x - 3 y^2 dy/dx = 0. Solving for dy/dx we get dy/dx = 2x / (3 y^2). At (-1,1) we have x = 1 and y = 1 so at this point dy/dx = 2 * -1 / (3 * 1^2) = -2/3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 6 7th edition 2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x)) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p^2 = (500 - x)/ 2x p^2 * 2x = 500 - x p^2 2x +x - 500 = 0 2p2x + 2x’p^2 + x’ -0 = 0 4 px + 2x’ p^2 + x’ = 0 X’ (2p^2 + 1) = -4px X’ = - 4px/(2p^2 +1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 + dx / dp = 0 (2 p^2 + 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 + 1) ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 6 7th edition 2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x)) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p^2 = (500 - x)/ 2x p^2 * 2x = 500 - x p^2 2x +x - 500 = 0 2p2x + 2x’p^2 + x’ -0 = 0 4 px + 2x’ p^2 + x’ = 0 X’ (2p^2 + 1) = -4px X’ = - 4px/(2p^2 +1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 + dx / dp = 0 (2 p^2 + 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 + 1) ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `q 6 7th edition 2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x)) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p^2 = (500 - x)/ 2x p^2 * 2x = 500 - x p^2 2x +x - 500 = 0 2p2x + 2x’p^2 + x’ -0 = 0 4 px + 2x’ p^2 + x’ = 0 X’ (2p^2 + 1) = -4px X’ = - 4px/(2p^2 +1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 + dx / dp = 0 (2 p^2 + 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 + 1) ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!