#$&* course Mth 277 1055pm 12/3/2011 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: grad(f) = del f = e^(x + y + z) i + e^(x + y + z) j + e^(x + y + z) k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Find the directional derivative of f(x,y) = x^2 + xy at the point (1, -1) in the direction of the vector v = i - j. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x = 2x + y f_y = x grad(f) = (2 x + y) i + x j I wasn't sure how to do the rest, so I looked at the given solution. grad(1,-1) = (2 x + y) i + x j = 2(1) -1 i + j = i + j magnitude of the gradient is sqrt(2) unit vector in the direction of the vector v = i - j is 'u = 1/sqrt(2) * <1,-1> = <1/sqrt(2), -1/sqrt(2)> directional derviative in the direction of v <1,1> dot <1/sqrt(2), -1/sqrt(2)> = 1/sqrt(2) - 1/sqrt(2) = 0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: grad(f) = del f = (2 x + y) i + x j. At (1, -1) the gradient is therefore i + j. The unit vector in the direction of v is sqrt(2) / 2 * (i - j). The directional derivative in the direction of v is the dot product of the gradient and the unit vector. In this case the directional derivative is zero. At the point (1, -1) the direction of the vector v is the one in which f(x, y) has zero rate of change. v would be tangent to a level curve at (1, -1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still trying to wrap my head around this concept. I understand the math, the visuals are hard to see. But I understand the a directional derivative is the slope at a given point, and since the point is tangent to the curve, the slope would equal zero. ------------------------------------------------ Self-critique rating:3
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Given Solution: The gradient of f(x, y, z), evaluated at a point, is normal to the surface.at that point. The gradient is easily found to be grad f = (3 x^2 + 2 y^2) i + (4 x + 3) j - k. At (1, 1, 1) the gradient is therefore (5 i + 7 j - k). The tangent plane passes through (1, 1, 1) and is normal to 5 i + 7 j - k. A point (x, y, z) lies on the tangent plane if the vector (x - 1) i + (y - 1) j + (z - 1) k is perpendicular to the normal vector. So (x, y, z) lies on the plane if the dot product of this vector and the normal vector is zero. The dot product is 5 (x - 1) + 7 ( y - 1) - (z - 1), so the equation of the tangent plane is 5 (x - 1) + 7 ( y - 1) - (z - 1) = 0, which simplifies to 5 x + 7 y - z - 11 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Find the direction from the point P = (1,e,-1) in which the function f(x,y,z) = z ln (y/x) increases the most rapidly and compute the magnitude of the greatest rate of increase. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z ln (y/x) = z [lny - lnx] f_x = -z/x f_y = z/y f_z = ln(y/x) grad(f) = -z/x i + z/y j + ln(y/x) k grad(1,e,-1) = i + -1/e j + k ||grad|| = sqrt(2+1/e^2) 1/sqrt(2+1/e^2) <1,-1/e,1> =
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Given Solution: The gradient is -z/x i + (z/y) j + ln(y/x) k. At (1, e, -1) we get i + 1/e j + k. A unit vector in this direction is u = i / sqrt(2 + 1/e^2) + j ./ (e sqrt(2 + 1/e^2)) + k / sqrt(2 + 1/e^2), approximately .68 i + .25 j + .68 k. The magnitude of this vector is sqrt(2 + 1/e^2). The magnitude of the gradient is the greatest rate of increase. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the unit vector wrong. I understand what I did wrong. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. A particle P1 with mass m1 is located at the origin, and a particle P2 with mass 1 unit is located at the point (x,y,z). According to Newton's law of universal gravitation, the force P1 exerts on P2 is modeled by F = -G(m1(xi + yj + zk))/r^3 where r is the distance between P1 and P2 and G is the gravitational constant. Starting from the fact that r^2 = x^2 + y^2 + z^2, show that d/dx*(1/r) = -x/r^3, d/dy*(1/r) = -y/r^3 and d/dz(1/r) = -z/r^3. (Here d/dx denotes partial with respect to x) The function V = -G*m1/r is called the potential energy function for the system. Show that F = -grad(V). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = -G(m1(xi + yj + zk))/r^3 r^2 = x^2 + y^2 + z^2 r_x = 2x r_y = 2y r_z = 2z grad(r) = 2xi + 2yj + 2zk sqrt(4x^2 + 4y^2 + 4z^2) is magnitude of gradient u' = 1/sqrt(4x^2 + 4y^2 + 4z^2) <2x, 2y, 2z> =
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Given Solution: 1/r = 1 / sqrt(x^2 + y^2 + z^2) = (x^2 + y^2 = z^2)^(-1/2), so d/dx (1/r) = 2 x * (-1/2) (x^2 + y^2 + z^2)^(-3/2) = -x / (sqrt(x^2 + y^2 + z^2)) ^3 = -x / r^3. The results for the y and z derivatives are acquired by a completely analogous series of steps. It follows that the gradient of V = - G m1 / r is V_x i + V_y j + V_z k = -G m1 * ( -x / r^3 i - y / r^3 j - z / r^3 k ). Simple rearrangement shows that this is identical to the force function F. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): YES SIR! :) ------------------------------------------------ Self-critique rating:3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!