Solutions to Final Exam review problems

• 1.
  • a) In matrix form we have AX = b where A = [[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]] is a 3x3 matrix, and X = (x,y,z) and b = (d1,d2,d3) are column vectors.
  • b) First find A^(-1), the inverse matrix of A. Then X = (A^(-1))b.
  • c) The matrix A is invertible if the determinant of A is not zero.
  • d) If det(A) = 0, then the system either has no solutions, or infinitely many solutions.

• 2.
  • a) Set c1<-1,0,2,1> + c2<3,1,-2,0> + c3<0,1,4,3> = <0,0,0,0> and solve for c1,c2 and c3. One nonzero solution is c1 = 3, c2 = 1, c3 = -1, so the vectors are linearly dependent.
  • b) Set c1<1,2,3> + c2<-1,0,1> + c3<0,0,1> = <0,0,0> and solve for c1,c2 and c3. The only solution is c1 = c2 = c3 = 0, so the vectors are linearly independent is this case, too.
  • c) Set c1v1 + c2v2 + c3v3 = 0. We want to show that the only solution is c1 = c2 = c3 = 0. Let d = 0 and add d*v4 to both sides. This gives the equation c1v1 + c2v2 + c3v3 + d*v4 = 0 + d*v4 = 0 (since d = 0). But {v1,v2,v3,v4} is a linearly independent set of vectors, so we know the only possible solution to the equation c1v1 + c2v2 + c3v3 + d*v4 = 0 is c1 = c2 = c3 = d = 0. So c1 = c2 = c3 = 0. So we have shown that c1v2 + c2v2 + c2v3 = 0 implies that c1 = c2 = c3 = 0 and therefore {v1,v2,v3} is a linearly independent set.

• 3. Let S be the set of all vectors (a,b,c) such that a + b + c = 1. Suppose u = (a1,b1,c1) and v = (a2,b2,c2) are in S. Then a1 + b1 + c1 = 1 and a2 + b2 + c2 = 1. But u + v = (a1 + a2, b1 + b2, c1 + c2) is not in S since (a1 + a2) + (b1 + b2) + (c1 + c2) = 2 not 1. So S is not a subspace of R3. Alternately, the zero vector (0,0,0) is not in S since 0 + 0 + 0 = 0 not 1, so S is not a subspace of R3 (all subspaces must contain the zero vector).

• 4. The matrix that represents T is [[1,1],[1,0],[2,-1]].

• 5.
  • a) The Jacobian matrix of f at (-1,1,2) is J = [[0,1,8],[2,-2,-1]] and the total derivative is J*(x,y,z) = [[y+8z],[2x-2y-z]].
  • b) f(-0.9,1.1,1.9) is approximately equal to f(-1,1,2) + J*(0.1,0.1,-0.1) = (8.3,-1.9).

• 6.
  • a) fx(0,0) = Lim[(f(0+h,0)-f(0,0))/h, h->0]
  • b) fx(0,0) = 0

• 7.
  • a) The gradient (normal) vector to the ellipsoid at the point (1,2,1) is <2,1,2>, while the gradient to the hyperboloid at that same point is <2,4,-2>. Since these two vectors are not multiples of each other, the surfaces are not tangent.
  • b) The tangent plane to the ellipsoid is 2x + y + 2z = 6 and the tangent plane to the hyperboloid is 2x + 4y -2z = 8. The line in which these two planes intersect is given parametrically by r(t) = <1,2,1> + t<-10,8,6>.

• 8.
  • a) Think of the surface as a hill. If you are on the side of the hill there is always a direction (uphill) in which the directional derivative will be positive. The only way this wouldn't happen is if you were at a flat place, such as the top, but then the directional derivative would be zero in every direction.
  • b) The directional derivative is equal to (grad(f))dot(u), where u is a unit vector, and is maximized when u is in the direction of the gradient. This is when u = grad(f)/||grad(f)||, so the maximum value of the directional derivative is (grad(f))dot(grad(f))/||grad(f)|| = (||grad(f)||^2)/||grad(f)|| = ||grad(f)||. In other words, the maximum value of the directional derivative is the magnitude of the gradient vector! And the magnitude of a vector can never be negative.
  • c) If grad(f) is the zero vector, then the directional derivative is zero in all directions. This happens when the surface is horizontal at that point. It could be a maximum or minimum, or some sort of saddle point.

• 9. Make the change of variables u = y - x and v = y + x. Then the bounds for v are 1 and 2 and the bounds for u are -v and v. The correction factor is 1/|d(u,v)/d(x,y)| = 1/2. The integral becomes Integrate[(1/2)cos(u/v),{v,1,2},{u,-v,v}] (Mathematica notation) which is equal to (3/2)Sin[1].

• 10. The two surfaces intersect in the circle x^2 + y^2 = 2, z = 4. Using polar coordinates we have Integrate[(10 - 3r^2 - 4)r,{theta,0,2Pi},{r,0,Sqrt[2]}] = 6Pi.

• 11. Reversing the order of integration gives Intgrate[(x^3)Sin[y^3],{y,0,1},{x,0,Sqrt[y]}] = (1/12)(1 - Cos[1]).

• 12. This is a repeated root problem with lambda = 1. The general solution is Y(t) = k1(e^t)<1,0> + k2(e^t)(t<1,0> + <0,1>).

• 13. We have eigen values lambda1 = -2, lambda2 = 1 and eigenvectors v1 = <1,2>, v2 = <1,-1>. The general solution is Y(t) = k1(e^(-2t))<1,2> + k2(e^t)<1,-1>.

• 14.
  • a) A has two distinct real eigenvalues if the discriminant (-(a+d))^2 - 4(ad-bc) > 0.
  • b) A has a repeated eigenvalue if the discriminant = 0.
  • c) A has complex eigenvalues if the discriminant < 0.
  • d) For part a) we have two straight line solutions, for part b) we have one striaght line solutions and for part c) we get a spiral source or sink (or a center if the eigenvalues are purely imaginary).

• 15.
  • a) m is the mass, b is the damping coefficient and k is the spring constant.
  • b) The system is underdamped (complex eigenvalues) if b^2 < 4mk.
  • c) The system is critically damped (repeated eigenvalue) if b^2 = 4mk.
  • d) The system is overdamped (real eigenvalues) if b^2 > 4mk.
  • e) For the underdamped case the solutions oscillate around the equilibrium. A possible graph of y(t) would be a cosine curve whose amplitude decreases with time. The v(t) graph will be the derivative of y(t).

• 16. The characteristic polynomial has roots lambda1 = -2 and lambda2 = -5 so the general solution of the homogeneous equation is y(t) = k1e^(-2t) + k2e^(-5t). Since the forcing function g(t) = e^(-2t) is a solution of the homogeneous equation we guess a particular solution of the form yp(t) = kte^(-2t). Solving for the constant gives k = 1/3, so the general solution of the nonhomogeneous equation is y(t) = k1e^(-2t) + k2e^(-5t) + (1/3)te^(-2t).

Student questions

Q: I have a question regarding problem number 9. How were you able to come up with the bounds of -v to v? If there is any way that you could go more into detail with this problem I would greatly appreciate it.

A: This is a tough problem. If you look at the trapezoid, the two parallel sides are taken care of (that's how u goes from 1 to 2). It's the other two sides that cause trouble. One side is along y = 0 (the x-axis) and the other is along x = 0 (the y-axis). The thing to do here is to solve for x and y in terms of u and v. We get x = (u-v)/2 and y = (u+v)/2. So when y = 0 we have (u+v)/2 = 0, or v = -u. And when x = 0 we have (u-v)/2 = 0 to v = u. That's how we get that v goes from -u to u.

Q: Can you explain how you figured out part b) of number 7. We've got the equations for the planes but can't figure out the line where they intersect.

A: An important property of a normal vector to a plane is that it is perpendicular to every vector in the plane. The key to this problem is that since the line of intersection lies in both planes, it most be perpendicular to both normal vectors (i.e. perpendicular to the normal vectors of both planes). So the cross product of the normal vectors give us a vector in the direction of the line. Then we can use this vector and the point (1,2,1) to write the equation of the line.

Q: I was looking at #23 in 3.7. I wasn't sure how to tell which way it was increasing or decreasing most rapidly. If you could give me a few pointers that would be great.

A: Most of the problems of this kind that we have done have been two-dimensional (i.e. concerning a function of two variables). The idea is the same though - if we have a function of several variables whose OUTPUT is a single number, then the direction of greatest increase in that output at a particular point is given by the gradient vector of the function at that point. In this problem, the gradient will have three components (dT/dx, dT/dy and dT/dz), as it should, since we are describing the temperature in a three-dimensional piece of metal.

The direction in which the temperature is decreasing most rapidly is just the direction opposite to the gradient vector.