#$&* course Mth 277 11.6*********************************************
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Given Solution: grad(f) = del f = e^(x + y + z) i + e^(x + y + z) j + e^(x + y + z) k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Find the directional derivative of f(x,y) = x^2 + xy at the point (1, -1) in the direction of the vector v = i - j. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f_x=2x+y f_y=x D_v f(1,-1)=f_x(1,-1)(1)+f_y(1,-1)(-1) =[2(1)+(-1)](1) + [1](1) =1-1=0 Therefore, the directional derivative is 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Find a unit vector which is normal to the surface given by the equation 2 = x^3 + 2xy^2 + 3y - z at the point P = (1,1,1). Also find the equation of the tangent plane at this point using this information. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 = x^3 + 2xy^2 + 3y - z P = (1,1,1) grad(f)=f_x i + f_y j +f_z k f_x=3x^2+2y^2 i f_y=4yx+3 j f_z=-1 k Now I will plug in point P(1, 1, 1) at the deriv in respect of the each var. 3(1)+2(1)=5i 4(1)(1)+3=7j -1=-k Unit vect normal to the surface given by the equation is 5i+7j-1k To find tang plane, we do the dot prod to = 0 5(x-1)+7(y-1)-1(z-1)=0 5x+7y-z=11 confidence 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:OK ********************************************* 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: P = (1,e,-1) f(x,y,z) = z ln (y/x) f_x=(-z/x) f_y=(z/y) f_z=ln (y/x) [-z/x]i + [-z/y]j + [ln (y/x)]k (1,e,-1):[1/1]i + [1/e]j + [ln (e/1)]k i + (1/e)j + k unit vect i/(sqrt(2+(1/e)^2))+j(1/e)/(sqrt(2+(1/e)^2))+k/(sqrt(2+(1/e)^2)) .684i+.252j+.684k magn=.9996
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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 think I did well untill I came to the magnitude of the vector. I probably should not of used the rounded values because my answer was considerably off. However, it seemed to far to be a rounding error so Im not sure exactly what happened. ------------------------------------------------ Self-critique rating:OK ********************************************* 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: r^2 = x^2 + y^2 + z^2 r=sqrt(x^2 + y^2 + z^2) 1/r=sqrt(1/x^2 + 1/y^2 + 1/z^2) d/dx*(1/r) = -x/r^3, d/dy*(1/r) = -y/r^3 d/dz(1/r) = -z/r^3 F = -G(m1(xi + yj + zk))/r^3 V = -G*m1/r This is where I got confused. 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): I see where I got to and how I had to continue to relate things to achieve the final answer. It would take a little bit of rearanging, but it is obvious that these functions are identical. ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!