#$&* course Mth 277 Question: `q001. Give the standard form equation for the tangent plane to the surface z(x,y) = ln(x^2 + y^2) at the point P_0 = (e,0,2).YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: z_x = 2 x / (x^2 + y^2). At (e, 0, 2) we have z_x = 2 * e / (e^2 + 0^2) = 2 / e (very approximately .75) A unit vector tangent to the plane, in the x z plane, is therefore i + (2/e) k. z_y = 2 y / (x^2 + y^2). At (e, 0, 2) we have z_y = 2 * 0 / (e^2 + 0^2) = 0. The j vector is therefore tangent to the plane, in the yz plane. The cross product of two tangent vectors is a normal vector. The cross product of the two tangent vectors obtained above is k + (2 / e) * (-i) = k - (2 / e) * i. The point (x, y, z) lies on the tangent plane if (x - e) i + (y - 0) j + (z - 2) k is perpendicular to the normal vector. Setting the dot product of the two vectors equal to zero we get the equation -(2/e) * (x - e) + (z - 2) = 0, which we simplify to -(2/e) x + z = 0 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x= 2zy^3*cos(xy)*sin(z)-2xzy^4*sin(z)*sin(xy) f_y=6xzy^2*cos(xy)*sin(z)-2zx^2y^3*sin(z)*sin(xy) f_z=2xy^3*cos(xy)*sin(z)+2xzy^3*cos(z)*cos(xy) df=[2zy^3*cos(xy)*sin(z)-2xzy^4*sin(z)*sin(xy)]dx+[6xzy^2*cos(xy)*sin(z)-2zx^2y^3*sin(z)*sin(xy)]dy+[2xy^3*cos(xy)*sin(z)+2xzy^3*cos(z)*cos(xy)]dz df=2zy^3[cos(xy)*sin(z)-xy*sin(z)*sin(xy)]dx+2zx^2y^2[3*cos(xy)*sin(z)-y*sin(z)*sin(xy)]dy+2xy^3[cos(xy)*sin(z)+z*cos(z)*cos(xy)]dz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 believe this is as far as you can simplify it. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Use an incremental approximation to estimate f(sqrt(pi) + .01, sqrt(pi) - .01), where f(x,y) = cos(xy) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(sqrt(pi)+.01,sqrt(pi)-.01) f(x,y)=cos(xy) cos(sqrt(pi)sqrt(pi))=cos(pi)=1 df =-ysin(xy)dx-xsin(xy)*dy sin(sqrt(pi)*sqrt(pi))=sin(pi)=0 I got somewhat lost after doing this and Im not sure how to continue even after glancing down at the solution. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.
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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): That is as much as I could do without looking at the solution. I see where I needed to do another test to ensure that it was a relative minimum. ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!