#$&* course Mth 277 3pm 12/3/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: z_x = 2 x / (x^2 + y^2). At (e, 0, 2) we have z_x = 2 * e / (e^2 + 0^2) = 2 / e (very approximately .75) A unit vector tangent to the plane, in the x z plane, is therefore i + (2/e) k. z_y = 2 y / (x^2 + y^2). At (e, 0, 2) we have z_y = 2 * 0 / (e^2 + 0^2) = 0. The j vector is therefore tangent to the plane, in the yz plane. The cross product of two tangent vectors is a normal vector. The cross product of the two tangent vectors obtained above is k + (2 / e) * (-i) = k - (2 / e) * i. The point (x, y, z) lies on the tangent plane if (x - e) i + (y - 0) j + (z - 2) k is perpendicular to the normal vector. Setting the dot product of the two vectors equal to zero we get the equation -(2/e) * (x - e) + (z - 2) = 0, which we simplify to -(2/e) x + z = 0 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q002. Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x = 2zy^3*cos(xy)sin(z) + [-sin(xy)sin(z)2xzy^2] = 2zy^3sin(z)[cos(xy) - xsin(xy)] f_y = -2xzy^3sin(xy)sin(z) + cos(xy)sin(z)6xzy^3 = 2xzy^2sin(z)[-ysin(xy) + 3cos(xy)] f_z = 2xzy^3cos(xy)cos(z) + sin(z)cos(xy)2zy^3 = 2zy^3cos(xy)[xcos(z) + sin(z)] Total differential f_x + f_y + f_z = 2zy^3sin(z)[cos(xy) - xsin(xy)] + 2xzy^2sin(z)[-ysin(xy) + 3cos(xy)] + 2zy^3cos(xy)[xcos(z) + sin(z)] = 2zy^2 {ysin(z)[cos(xy) - xsin(xy)] + xsin(z)[-ysin(xy) + 3cos(xy)] + ycos(xy)[xcos(z) + sin(z)]} confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The total differential is f_x ds + f_y dy + f_z dz. f_x = ( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z) f_y = (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z) f_z = (2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z) The total differential is therefore ( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z) * f_x + (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z) * f_y + ( 2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z) ) * dz The expression would be simplified by factoring 2 z y^3 out of the coefficient of f_x , 2 x z y^2 out of the coefficient of f_y, and 2 x y^3 out of the coefficient of dz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I did it right. The given solution is confusing. lots of variables. Basically differentiate in terms of x, then y and then z, then add the differentiations together to get the total differentiation and then substitute out the common factors. ------------------------------------------------ Self-critique rating:3
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Given Solution: We approximate by first finding f(x, y) at the point (sqrt(pi), sqrt(pi)). Then we apply the differential with dx = .01 and dy = -.01. f(sqrt(pi), sqrt(pi)) = cos(sqrt(pi) * sqrt(pi)) = cos(pi) = -1. df = f_x dx + f_y dy indicates the change in f due to a given change dx in the value of x, and dy in the value of y. df = -y sin(xy) dx - x sin(xy) * dy. At the point (sqrt(pi), sqrt(pi) ) we have -y sin(x y) = - pi * sin(sqrt(pi) * sqrt(pi)) = -pi * sin(pi) = -pi * 0 = 0. In a similar manner we have -x sin(x y) = 0. That is, both f_x and f_y are zero at this point. Using our values of f_x and f_y at the original point, along with dx = .01 and dy = -.01 we get df = 0 * .01 + 0 * (-.01) = 0. The same procedure would apply to approximate the function at, say, the point (sqrt(pi) + .01, sqrt(pi) / 3 - .02): At the point (x0, y0) = (sqrt(pi) , sqrt(pi) / 3 ) we have f(x, y) = cos(sqrt(pi) * sqrt(pi / 3) ) = cos ( pi / 3) = 1/2 or .5. The differential of our function is still df = -y sin(xy) dx - x sin(xy) * dy. At the point (x0, y0) = (sqrt(pi), sqrt(pi) / 3 ) we have f_x = -y sin(x y) = - pi / 3 * sin(sqrt(pi) * sqrt(pi) / 3) = -pi/3 * sin(pi / 3) = -pi / 3 sqrt(3) / 2, approximately -0.9. f_y = In a similar manner we have -x sin(x y) = -pi sin(pi/3) = -pi sqrt(3) / 2 = -2.7. Using our values of f_x and f_y at our (x0, y0) point, along with dx = .01 and dy = -.02 we get df = -0.9 * .01 + (-2.7) * (-.02) = -.063. Our approximation to the value of f at the given point is therefore .5 + (-.063) = .437. That is f(x, y) = f(sqrt(pi) + .01, sqrt(pi) / 3 -.02) = f(x0, y0) + df = .5 - .063 = .437. You can assess the accuracy of this approximation by evaluating cos(x y) at the given point. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I solve for cos(x,y) and I get -1.0001. Don't understand where you're getting at 0.437 and other numbers. ------------------------------------------------ Self-critique rating: 3
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Given Solution: The tangent plane will be horizontal only if all tangent lines are horizontal. That is, all tangent lines have to have slope zero. Thus all the derivatives need to be zero. This will be the case, for example, at a point where the x and y partial derivatives are both zero. For this function z_x = -2 x + 6 and z_y = -2 y. Thus our conditions z_x = 0 and z_y = 0 give us the two equations -2 x + 6 = 0 -2 y = 0. Each equation has only one solution. We get x = 3 and y = 0. Thus the point (3, 0) is a critical point. We need to check to be sure that our critical point isn't a saddle point. Our second derivatives z_xx and z_yy are both negative, so (3, 0) is a candidate for a relative maximum. So far so good. We also need to test that the graph doesn't go off into a saddle point when we move at some nonzero angle to the x and y axes. The test for this is that z_xx * z_yy - z_xy ^ 2 must be positive. In this case z_xy = 0. We get z_xx * z_yy - z_xy ^ 2 = -2 * -2 + 0^2 = 4, which is > 0, so we don't have a saddle point We conclude that our point (3, 0) does indeed give us a relative maximum. The relative maximum therefore occurs at (3, 0, f(3, 0) ) = (3, 0, 13). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why do we have to make sure its not a saddle point???????? and all the other information? I understnad how to get the saddle points, but don't understnad why we need it. You asked what the equation is for a horizontal tangent plane. Therefore, the only points on the plane can be 0 and 3, so the equation for the plane would be z = (x-3) + (y-0) ------------------------------------------------ Self-critique rating:3"