#$&* course Mth 277 11/19 9 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: grad(f) = del f = e^(x + y + z) i + e^(x + y + z) j + e^(x + y + z) k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Find the directional derivative of f(x,y) = x^2 + xy at the point (1, -1) in the direction of the vector v = i - j. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I'm getting del f to be (2x+y)i + xj at the point 1,-1, which is i + j if it is the direction of the vector v, this implies to dot product the two, which i'm just getting i^2 -j^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ?? unclear about the directional derivative. Shouldn't the dot product be i^2 - j^2 ? how do you know is has 0 rate of change? ------------------------------------------------ Self-critique rating:
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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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gradient is -z/x i + (z/y) j + ln(y/x) k. At (1, e, -1) we get i + 1/e j + k. A unit vector in this direction is u = i / sqrt(2 + 1/e^2) + j ./ (e sqrt(2 + 1/e^2)) + k / sqrt(2 + 1/e^2), approximately .68 i + .25 j + .68 k. The magnitude of this vector is sqrt(2 + 1/e^2). The magnitude of the gradient is the greatest rate of increase. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm getting gradient to be zln(y) i + zln(1/x)j + ln(y/x)k which when point P is plugged in I get -i + k how are you getting that answer? my magnitude is messed up because of this. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. A particle P1 with mass m1 is located at the origin, and a particle P2 with mass 1 unit is located at the point (x,y,z). According to Newton's law of universal gravitation, the force P1 exerts on P2 is modeled by F = -G(m1(xi + yj + zk))/r^3 where r is the distance between P1 and P2 and G is the gravitational constant. Starting from the fact that r^2 = x^2 + y^2 + z^2, show that d/dx*(1/r) = -x/r^3, d/dy*(1/r) = -y/r^3 and d/dz(1/r) = -z/r^3. (Here d/dx denotes partial with respect to x) The function V = -G*m1/r is called the potential energy function for the system. Show that F = -grad(V). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: i can solve for 1/r as the starting point, and get all the partial derivatives easily. but when i multiply, i'm getting ... d/dx(F) = (-x/r^3)( 1 / sqrt(x^2 + y^2 + z^2) ) = -x / (r^3 * sqrt ( x^2 + y^2 + z^2)). I cant see how this = -x / r^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Still having trouble with this. ------------------------------------------------ Self-critique rating:"