Note that this was done under the assumption ab > 0; if ab < 0, omit the negative sign in the numerator.
Concepts Review
Circular cylinder
Plane
Parabolic cylinder
Two planes
Circular cone
Cylinder
Plane
Hyperbolic paraboloid
Plane
Hemisphere
One sheet of a hyperboloid of two sheets
The equation of the elliptical cross section is
Cylindrical to Spherical
Chapter Review Concepts Test
True: See Section 11.2
True: The vectors are both parallel and perpendicular, so one or both must
True
True: T depends only upon the shape of the curve, hence N and B also
Circular cylinder
Sphere
Parabolic cylinder
Circular paraboloid
Plane
Plane
Hyperboloid of one sheet
Any of the following (as well as others) would be acceptable: sin cos sin sin cos. Note that when we divided by ρ in part c and d, we did not lose the pole since it is also a solution of the resulting equations.). The acute angle between two planes is the same as the acute angle θ between the normal vectors.
Concepts Review
Since the level curve runs for 70 southwest to northeast, you can drive southwest or northeast and stay at about the same temperature. The lowest barometric pressure, 1000 millibars and less, occurred in the Great Lakes region, specifically near Wisconsin. The highest barometric pressure, 1025 millibars and more, occurred on the east coast, from Massachusetts to South Carolina.
Driving northwest would take you to lower barometric pressure, and heading southeast would take you to higher barometric pressure. Louis, the level curves run from southwest to northeast, you can drive southwest or northeast and stay at the same barometric pressure. The set of all points inside (the part containing the z-axis) and on the hyperboloid of a sheet;.
The set of all points inside (the part containing the z-axis) and on the hyperboloid of one sheet;
Since the argument of the natural logarithm function must be positive, we must have xy>0. This happens when the ordered pair (x y, ) is in the first quadrant or the third quadrant of the xy plane. The domain thus consists of all points (x y z, , ) such that x and y are both positive or both negative.
AC is the least steep path and BC is the steepest path between A and C, since the level curves are farthest along AC and closest together along BC.
Completing the squares on x and y yields the equivalent equation
- Concepts Review
The largest rectangle that can be included in the circle is a square with a diameter of 20.
The largest rectangle that can be contained in the circle is a square of diameter length 20. The edge
Moving parallel to the y-axis from the point (1, 1) to the nearest level curve and (1, 1) to the nearest level curve and. Moving parallel to the x-axis from the point (–4, 2) to the nearest level curve and zooming in with,. Moving parallel to the x-axis from the point (–5, –2) to the nearest level curve and approximately z,.
Moving parallel to the y-axis from the point (0, –2) to the nearest level curve and zooming in with,.
The limit does not exist because of Theorem A
Changing to polar coordinates,
The only points at which f might be discontinuous occur when xy = 0
The boundary consists of the points that form the outer edge of the rectangle. The set is closed
The boundary consists of the points of the circle shown. The set is open
The boundary consists of the circle and the origin. The set is neither open (since, for
The boundary consists of the graph of sin 1
The boundary is the set itself along with the origin. The set is neither open (since none of its
- Concepts Review 1. gradient
This means that in the first case one travels from P to the origin and then to Q; in the second case one travels directly from P to Q without passing through the origin, so f is discontinuous on the set. A function of three variables is continuous on an open set S if it is continuous at every point in the interior of the set.
The equation of a plane containing this point and parallel to the x-axis is given by. The intersection of the two planes is the tangent to the surface, passing through the point (2,1,9), whose projection in the xy-plane is parallel to the x-axis. This line of intersection is parallel to the cross product of the normal vectors for the planes.
Using the equation for the tangent plane from the previous part, we now want the vertical plane to be parallel to the y-axis, but still pass through the projected point (2,1,0). Using the equation for the tangent plane from the first part, we now want the vertical plane to be parallel to the line y=x, but still pass through the projected point (2,1, 0). The equation of a plane containing this point and parallel to the x-axis is given by y=2.
The line of intersection between the two planes is the tangent line to the surface passing through the point whose projection in the xy-plane is parallel to the x-axis. Using the equation for the tangent plane from the first part, we now want the vertical plane to be parallel to the line x= −y, but still pass through the projected point (3, 2.72).
Let a be any point of S and let b be any other point of S. Then for some c on the line segment
- Concepts Review
The rate of change of f(x, y) in that direction at that point is the magnitude of the gradient.
The level curves are
He should move in the direction of
The unit vector from (2, 4) toward (5, 0) is
Graph of domain of f
Leave (–2, –5)
The stream carries the boat along at 2 ft/s with respect to the boy
- Concepts Review 1. perpendicular
- Concepts Review 1. closed bounded
A vector in the direction of the line, . dT gdL Ldg gdL Ldg. u is the unit vector in the direction of flight, and . Using the parametric form of the flight line to substitute into the equation for the plane yields t = 3 as the time of intersection with the plane. Therefore, the gradient is defined but never zero in its domain, and the boundary of the domain is outside the domain, so there are no critical points.
Let S denote the area of the sides and bottom of the tank with base l for w and depth h. In the latter case, ax + by + (c – M) is positive in one of the regions and negative in the other.
Let x and y be defined as shown in Figure 4 from Section 12.8. The total cost is given by
The volume of the box can be expressed as
This minimum cost occurs when the length and width are approximately 3.2 feet and the height is approximately 5.8 feet. Geometrically: we are looking for the smallest and largest of all spheres centered at the origin and some point in common with the given triangular area. Without loss of generality, we will assume that α β γ≤ ≤. We will consider it intuitively clear that for the triangle with the largest area, the center of the circle will be inside or on the boundary of the triangle; i.e. , ,α β and γ are in the interval [0, ]π.
The area of an isosceles triangle with congruent sides of length r and included angle θ is 1 2 sin. There are no critical points within the domain of A. The critical points are indicated on the graph of the domain of A.).
- Concepts Review 1. free; constrained
Minimize the square of the distance to the plane,
Minimize the square of the distance,
There is also a maximum value (see Problem 40, Section 12.8), but our Lagrangian approach does not capture this. Finding critical points on the boundary: Solve the system of equations 1− = ⋅y λ 2 ;x 1− = ⋅x λ 2 ;y x2+y2=9 By substitution, we can find that the critical points on the boundary are . There is no minimum or maximum value inside, since there are infinitely many critical points.
Clearly, the maximum will occur for the triangle containing the center of the circle.
It is clear that the maximum will occur for a triangle which contains the center of the circle
- Chapter Review Concepts Test
True: The set (call it S, a line segment) contains all its boundary points, because for every point P not in S (ie, not on the line segment), there is an open neighborhood to P (ie, a circle with P as center) that does not contain any point of S.
Everywhere in the plane except on the parabola
The nature of the problem suggests that the critical point yields a maximum value rather than a minimum value.
