If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
11.6
Question: `q001. Find grad(f) when f(x,y,z) = e^(x+y+z).
Your solution:
Confidence rating:
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):
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.
Your solution:
Confidence rating:
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):
Self-critique rating:
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.
Your solution:
Confidence rating:
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):
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.
Your solution:
Confidence rating:
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):
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.
Your solution:
Confidence rating:
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):
Self-critique rating: