course Mth 272 8/1 12:15 pm 028.
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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: Before finding the derivatives with regards to x, y, and z, we can identify that this expression has an outer function, u^(-1/2). We can first take the derivative of this and apply it to the derivatives of W’x, W’y, and W’z. According to the power rule, the derivative of u^(-1/2) = -1/2u^(-3/2). Now we can find du for x, y, and z. du with regard to x = -2x, du with regard to y = -2y, and du with regard to z = -2z. We can now use our derivative rules to find the whole derivatives of W’x, W’y, and W’z. W’x = -2x(-1/2(1-x^2 – y^2 – z^2)^(3/2)) = x/(1-x^2 – y^2 – z^2)^(3/2)). W’y = -2y(-1/2(1-x^2 – y^2 – z^2)^(3/2)) = y/(1-x^2 – y^2 – z^2)^(3/2)). W’z = -2z/(-1/2(1-x^2 – y^2 – z^2)^(3/2)) = z/(1-x^2 – y^2 – z^2)^(3/2)). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 the derivatives are all at the origin, or (0,0,0), they all have the derivative of 0/1. So they are all equal to 0. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To solve this problem, we must first find F’x and F’y. According to our derivative rules, F’x = 9x^2 – 3y^2 and F’y = -12x + 3y^2. If we set both epressions equal to 0, we get 0 = 9x^2 -3y^2 and 0 = -12x + 3y^2. Now we can use numeric substitution to solve. We find that both are 0 at the point (0,0). confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I did not solve for the second point, but after seeing the given answer, it makes more sense. I will make sure to solve for two points if needed in the future.