Math 280 exam notes

Exam #3

1. 1. It is fine to leave the linearization in the form \(L(x,y,z)=27+3.2(x-1)-1.3(y-4)+2.3(z-2)\) rather than expressing it as \(L(x,y,z)=3.2x-1.3y+2.3z+24.4\). The first form has the advantage of letting you read off the base point whereas the base point is obscured in the second form.
2. 1. Start with \( d(A)=d(WH+\frac{1}{8}\pi W^2) \) which expresses the statement "a small change in \(A\) is equal to a small change in \(WH+\frac{1}{8}W^2\).
      
      
   2. Computing the percentage change is not necessary here.
      
      
   3. Since \(dA=100dH+(20+25π)dW≈100dH+98.54dW\) for \(W=100\) and \(H=20\), we can conclude that a change in \(H\) will contribute more to \(dA\) than an equal-size change in \(W\).
      
      
3. 1. For each piece of the boundary, you must also check for critical points of the function restricted to that boundary. These boundary critical points are distinct from the interior critical points given in the problem statement.
      
      
   2. You must also compare with the outputs at the endpoints of each boundary piece. Here, this means comparing with the outputs at the four corners of the rectangle in this case.
      
      
   3. Since we are looking for the global extremes, we do not need to classify each local critical point. For each critical point, we only need to compute the corresponding output to include in the list of outputs to be compared. So, the second-derivative test is not needed here.
7. 1. Your uses of \((x,y)\) are not consistent with each other. If you use a density proportional to \(x^2+y^2\), then your origin for \((x,y)\) is at the center of the rectangle. If you write \(0\leq x\leq L\) and \(0\leq y\leq W\), then your origin for \((x,y)\) is at one corner of the rectangle. These are not consistent.
      
      

Exam #4

1. 1. Fundamentally, integration is adding up infinitely many infinitesimal contributions to a total.
      
      
   2. The essential difference is that a double integral involves adding up over a region in the plane while a triple integral involves adding up over a solid region in space. A double integral is not necessarily adding just areas. It could be adding height times area or density times area. Similarly, a triple integral could be adding density times volume.
2. 1. How did you determine this is exactly a circle?
      
      
   2. The exact geometric description is that the curve is the circle of radius 3 centered at (x,y)=(0,3). You can deduce this by converting to cartestion coordinates.
      
      
3. 1. How have you set up your coordinate system? In particular, where is the origin in relation to the sphere? Where is the z-axis in relation to the sphere? Where is the x-axis in relation to the slice you are describing?
      
      
   2. The problem asks only to describe the region, not to compute a volume.
      
      
7. 1. If you use constant bounds on z, you will need to split the solid region into two pieces. The upper piece is bounded laterally by the paraboloid. The lower piece is bounded laterally by the sphere. The cut-off value of z is determined by the intersection between the paraboloid and the sphere.
      
      
   2. The projection into the z=0 plane is a circle. The radius of the circle is determined by the intersection of the paraboloid and the sphere. The radius is not √3 and it is not 9.
      
      
8. 1. The bounds on θ should be 0 to π rather than 0 to 2π since y is restricted to y ≥ 0. The range of φ is 0 to π.
      
      
9. 1. For (b), you can determine the total mass in the uniform density case by a simple multiplication. Integration is not needed.