course Mth 272 August 13, 2009 at 9:23 AM Question: `qQuery Problem 7.4.8 fy for xy / (x^2+y^2)
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Given Solution: `a You have to use the quotient rule. The derivative is taken with respect to y, so the ' stands for the derivative with respect to y. You get [ (xy)' (x^2 + y^2) - xy ( x^2 + y^2)' ] / (x^2 + y^2) ^ 2. Remembering that ' represents derivative with respect to y we get [ x ( x^2 + y^2) - xy ( 2y ) ] / (x^2 + y^2 ) ^ 2 or [ x^3 + x y^2 - 2 x y^2 ] / (x^2 + y^2) ^ 2, which simplifies to [ x^3 - x y^2 ] / (x^2 + y^2) ^ 2 or x [ x^2 - y^2 ] / (x^2 + y^2) ^ 2 . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.4.32 wx, wy, wz at origin for w = 1/sqrt(1-x^2-y^2-z^2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: inner derivatives= -2x, -2y,-2z outer derivative= z^-½ with that, Wx= x/(1-x^2-y^2-z^2)^3/2 Wy= y/(1-x^2-y^2-z^2)^3/2 Wz=z/(1-x^2-y^2-z^2)^3/2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a `a Every partial derivative involves the chain rule with inner function 1 = x^2 - y^2 - z^2 and outer function 1/ sqrt(z) = z^(-1/2). The partial derivatives with respect to x, y and z of the 'inner function' (1 - x^2 - y^2 - z^2) are respectively -2x, -2y and -2z. The derivative of the 'outer function' z^-(1/2) is -1/2 z^(-3/2). So the partial derivatives of the entire function are wx = -2x * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = x / (1 - x^2 - y^2 - z^2)^3/2 wy = -2y * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = y / (1 - x^2 - y^2 - z^2)^3/2 wz = -2z * -1/2 ( 1 - z^2 - y^2 - z^2)^(-3/2) = z / (1 - x^2 - y^2 - z^2)^3/2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat are the values of the three requested partial derivatives at the specified point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: since x,y,and z are all equal to ZERO, the answer is going to come out to be zero. Each equation is x/y/z DIVIDED by a number, which results in a value of ZERO. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a At the origin we have x = y = z = 0 so that the three partial derivatives are all of form 0 / 1 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.4.40 (was 7.4.36) fx = fy = 0 for 3x^3-12xy+y^3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fx= 9x^2-12y Fy= -12x+3y^2 substitution method..... 9x^2=12y 9x^2/12=Y SIMPLIFY: (3x^2)/4=Y PLUG in Y to second equation -12x+3((3x^2)/4)^2 =-12x+(27x^4)/16 SIMPLIFY: -3x(4-9x^3 / 16) I am not sure what there is to do from here........ confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Geometric solution: fx = 0 when y = 3/4 x^2, which describes a parabola with vertex at the origin, axis of symmetry conciding with the y axis, opening upward and passing thru the points (1, 3/4) and (-1, 3/4). fy = 0 when x = 1/4 y^2, which describes a parabola with vertex at the origin, axis of symmetry conciding with the x axis, opening to the right and passing thru the points (0, 1/4) and (0, -1/4). If you sketch these parabolas it should be clear that they coincide at one point in the first quadrant. You should estimate the coordinates of these points. Then proceed to solve these equations simultaneously to find the accurate coordinates: ** The partial derivatives are fx = 9x^2 - 12y fy = -12x + 3y^2. If the two partial derivatives are zero we get the equations 9x^2 - 12 y = 0 -12x + 3 y^2 = 0. Solving the first equation for y we get y = 3x^2 / 4. Substituting this expression for y in the second we have -12 x + 3 ( 3x^2 / 4)^2 = 0 so -12 x + 27 x^4 / 16 = 0. Factoring out -3x: -3x ( 4 - 9 x^3 / 16) = 0. This is so if -3x = 0 or 4 - 9 x^3 / 16 = 0. -3x = 0 gives solution x = 0. 4 - 9 x^3 / 16 = 0 if x^3 = 4 * 16 / 9, which happens when x = 64^(1/3) / 9^(1/3) = 4 * 3^(-2/3), which is expressed in standard form as 4 * 3 ^(1/3) / 3. If x = 0 then since 9 x^2 - 12 y = 0 we have y = 0. If x = 4 * 3^(1/3)/3 then 9 x^2 - 12 y = 0 gives us 9 [ 4 * 3^(1/3)/3 ] ^2 - 12 y = 0 so y = 9 [ 4 * 3^(1/3)/3 ] ^2 / 12 = 3/4 * 16 * 3^(2/3)/9 = 4 * 3^(2/3) / 3. So the two partials are both zero at (0,0) and at ( 4 * 3^(1/3)/3, 4 * 3^(2/3)/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: