Math 251 Final Review Pack

# Now what is the work done by the force in moving a

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Unformatted text preview: be a smooth curve given by the vector function and let the function, , be differentiable whose gradient vector, , is continuous over the smooth curve, , then Proof QED Theorem: is conservative, then 53 Proof Then By Clairaut’s Theorem: QED Assume: - have continuous first-order partial derivatives The domain on which connected Theorem: Let - is defined is open and simply (the curve doesn’t cross itself) be a vector field defined on domain, , then the following are equivalent is conservative is independent of path for every closed path in Assume: - is continuous is open and connected 54 16.4 Green’s Theorem Theorem (Green’s Theorem): Let D be the area bounded by a positively oriented (the arrows are directed in the anti-clockwise direction), smooth, simple curve c. If P and Q have continuous partial derivatives over the open region that contains D then 55 Review Problems w/ Solutions Section 14.3 p. 914 The ellipsoid intersects the plane equations for the tangent line of the ellipse at the point in an ellipse. Find the parametric Solution First we need to find the tangent plane to the ellipsoid. It is much easier to move all the terms of the equation to one side, giving us In order to do this we need the normal vector to this tangent plane, we know that the gradient vector is the normal vector to the tangent plane so And we need to know the normal vector of the tangent plane at the given point This vector is the normal vector to the tangent plane at the ellipsoid at We can now reapply this method to find the tangent plane to the plane at the given point, since it is a plane the gradient vector at the point is simply We have now found the normal to the second surface at the given point. Now we can take the cross product of these two points to find the vector that lies in both of the tangent plants and is the tangent line to the ellipse formed due to the intersection We know the direction vector of the line that is tangent to the ellipse at the given point. We know that the parametric equations are of the form Where the direction vector, and and we have both these quantities 56 So, the parametric equations to the ellipse at the given point are: Section Chapter 15 Review p. 1050 Evaluate Solution As we can see the integrand with respect to x is difficult to solve, therefore we must use a change of order of integration to simplify the process. We graph the curves defined by our bounds of integration for x, in this case the curves are: The area of integration is shaded in the graph above. As we can see in the above graph, x ranges from 0 to 1 and y ranges from 0 to . We can see that this is a type 1 region and thus we can now change the order of integration to make it simpler to evaluate our integral. 57 The integrand with respect to y simply integrates into Section 15 Review p. 1051 Evaluate Where H is the hemisphere which lies above the xy-plane and centered on the origin with radius 1. Solution We identify the equation of the whole circle as: We can see that this is more easily evaluated when we convert to spherical coordinates. So knowing our conversion of coordinate systems, we can first convert the z coordinate: We can see that the other part of the integrand just simply converts to ρ, or the radius of the circle in question. We can see that ρ ranges from 0 to 1 as the radius is 1. Secondly we see that Θ ranges from 0 to 2π as the circumference of the circle is complete. Finally we see that φ varies from 0 to π/2 as the angle sweeps from the z axis to the xy-plane, subtending a right angle. So we have H defined in spherical coordinates as: So using our change of variable for the spherical coordinates and substituting the elements in the integrand to their spherical counterparts, our integral now looks like: We can expand out the terms and group like terms 58 Then we can integrate with respect to ρ We must use substitution to solve the remaining integrand taking the following substitutions We plug the boundaries into the substitutions and get the new boundaries of 1 to 0. Our integrand now looks like 59...
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## This note was uploaded on 02/03/2014 for the course CMPT 150 taught by Professor Dr.anthonydixon during the Spring '08 term at Simon Fraser.

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