Here are the quizzes and their associated problem sets (Quiz 1, Quiz 2, Quiz 3, Quiz 4, Quiz 5, Quiz 6, Quiz 7, Quiz 8, Quiz 9):

- Vectors and curves:
- Date taken: January 12 Thursday.
- Problems from the Chapter 11 Practice Exercises (pages 638&639): 17, 19, 25, 29, 31, 35, 37, 43, 50.
- Extra-credit essay question: Explain your background in mathematics and what you are going to use this course for.
- Problems from §15.1 (page 826): 1, 3, 5, 7.
- Problems from §12.1 (pages 648–650): 1, 4, 6, 7, 11, 14, 15, 17, 19, 20, 23.
- Problems from §12.2 (pages 654–656): 1, 4, 6, 12, 15, 17, 21, 22.
- Problems from §12.3 (page 660): 1, 5, 8, 9, 11, 15, 18.

- Functions of several variables:
- Date taken: January 23 Monday.
- Problems from §13.1 (pages 682–684): 3, 4, 7, 9, 10, 15, 17, 18, 19, 23, 24, 30, 31–36, 39, 40, 42, 52, 54, 59, 62.
- Problems from §13.2 (pages 690–693): 2, 6, 11, 18, 23, 28, 31, 32, 36, 39, 43, 46, 55.
- Additional extra-credit problem:
Prove that the two definitions of continuity
spanning pages 3 and 4 of
my handout on functions of several variables
(one in terms of the continuity of composite functions
and the other in terms of
*ε*and*δ*) are equivalent. (In other words: Given a function*f*of several variables and a point*P*_{0}, show that if the conditions in the first definition of the continuity of*f*at*P*_{0}are met, then the conditions in the second definition must also be met,*and*the other way around. To provide the link between these definitions, you will need to refer to the*ε*-*δ*definition of continuity for an ordinary function of one variable; see the very top of page 6 in this handout from my Calculus 1 class last term if you don't know precisely how that goes.) - Problems from §15.2 (pages 838–840): 5, 6, 39, 41, 43.
- Additional problems
(you
*must*attempt these to get full credit for the problem set):- Given
*α*= 3*x*d*x*+ 4*x*^{2}*y*d*y*, evaluate*α*at (*x*,*y*) = (2, 6) along ⟨d*x*, d*y*⟩ = ⟨0.003, 0.005⟩. (Answer.) - Given
*α*= 2*x**y*d*x*+ 2*y**z*d*y*+ 2*x**z*d*z*, evaluate*α*at (*x*,*y*,*z*) = (−1, 3, 2) along ⟨d*x*, d*y*, d*z*⟩ = ⟨0.01, 0.02, −0.01⟩. - Given
*β*=*x*^{2}d*x*+*x**y*d*y*+*x**z*d*z*, evaluate*β*at (*x*,*y*,*z*) = (4, 3, −2). (Answer.) - Given
*β*= 5*x*^{2}d*x*− 3*x**y*d*y*, evaluate*β*at (*x*,*y*) = (1, 2).

- Given

- Differentiation:
- Date taken: January 30 Monday.
- Problems from §13.3 (pages 702–704): 3, 4, 10, 12, 24, 26, 30, 39, 43, 46, 55, 57, 75, 82, 91.
- Problems from §13.4 (pages 711&712):
- Use any method (including differentials or gradients): 2, 4, 7, 10;
- 19, 20, 27, 28, 33, 41.

- Additional extra-credit problem:
If you don't know about matrices (at least how to multiply them),
read
`http://www.mathsisfun.com/algebra/matrix-multiplying.html`

first. Now, if you have*m*functions of*n*variables each, then you can put their partial derivatives into an*m*-by-*n*matrix; for example, if you have 2 functions of 3 variables each, say*u*=*f*(*x*,*y*,*z*) and*v*=*g*(*x*,*y*,*z*), then the partial derivatives fit into a 2-by-3 matrix:

We may call this matrix d(⎡ ∂ *u*/∂*x*∂ *u*/∂*y*∂ *u*/∂*z*⎤ ⎣ ∂ *v*/∂*x*∂ *v*/∂*y*∂ *v*/∂*z*⎦ *u*,*v*)/d(*x*,*y*,*z*).- If you have an ordinary function
*y*=*f*(*x*), think of this as a group of only 1 function of only 1 variable each, so that d(*y*)/d(*x*) in the notation above is a 1-by-1 matrix, consisting of a single entry. Check that the usual derivative d*y*/d*x*is this entry. (You don't have to even write anything down for this if you don't want to; all that you'd write down anyway is d(*y*)/d(*x*) = [d*y*/d*x*]. But make sure that it makes sense before you go on.) - If you have a parametrized curve in 3 dimensions,
say
*P*= (*x*,*y*,*z*) = (*f*(*t*),*g*(*t*),*h*(*t*)), think of this as a group of 3 functions of only 1 variable each, so that d(*x*,*y*,*z*)/d(*t*) is a 3-by-1 matrix, consisting of a single column with 3 entries. Check that the components of the velocity vector d*P*/d*t*are the same as the entries of this matrix. (For this reason, ordinary vectors that represent change of position are sometimes called*column vectors*.) - If you have a function of 3 variables,
say
*u*=*F*(*x*,*y*,*z*), think of this as a group of only 1 function of 3 variables each, so that d(*u*)/d(*x*,*y*,*z*) is a 1-by-3 matrix, consisting of a single row with 3 entries. Check that the components of the gradient vector ∇*F*(*x*,*y*,*z*) are the same as the entries of this matrix. (For this reason, vectors such as gradients that represent change with respect to position are sometimes called*row vectors*.) - If you have both
(
*x*,*y*,*z*) = (*f*(*t*),*g*(*t*),*h*(*t*)) and*u*=*F*(*x*,*y*,*z*), then composition makes*u*an ordinary function of*t*. Show that d(*u*)/d(*t*) = d(*u*)/d(*x*,*y*,*z*) d(*x*,*y*,*z*)/d(*t*), using matrix multiplication, and check that this matches the defining property of the gradient from page 6 of my handout on functions of several variables. - If you have both
**V**= ⟨*f*(*u*),*g*(*u*),*h*(*u*)⟩ and*u*=*F*(*x*,*y*,*z*), then composition in the other order makes a vector field; that is,**V**is a vector-valued function of*x*,*y*, and*z*. Use matrix multiplication to get a 3-by-3 matrix; this matrix is called the*total derivative*of the vector field. (By the way, even if a vector field is not obtained as a composite in this way, it still has a matrix-valued function like this as its total derivative.)

- If you have an ordinary function
- Problems from §13.5 (pages 720&721): 2, 3, 7, 8, 14, 15, 16, 20, 23, 28.
- Problems from §15.2 (pages 838): 1, 4.
- Problems from §13.6 (pages 727–730): 3, 6, 10, 13, 14.

- Applications of differentiation:
- Date taken: February 6 Monday.
- Problems from §13.6 (pages 727–730): 19, 21, 29, 30, 33, 35, 39, 50, 54.
- Additional extra-credit problem:
Let
*f*be the function of two variables given by*f*(*x*,*y*) = 3 sin(*x*+*y*) + 4 cos(*x*−*y*). Evaluate*f*, both of its partial derivatives, and all four of its second partial derivatives at (0, 0). Then use these results to approximate*f*near (0, 0) with a quadratic polynomial (that is one whose degree is at most 2). - Problems from §13.7 (pages 737–739): 2, 7, 9, 15, 27, 32, 34, 37, 43, 52, 57.
- Problems from §13.8 (pages 746–748): 1, 5, 10, 11, 16, 23, 29.

- Integration on curves:
- Date taken: February 13 Monday.
- Problems from §15.2 (pages 838–840): 10, 11, 14, 16, 17.A&B, 19, 22, 23, 24, 29.
- Problems from §15.1 (pages 826–828): 10, 13, 16, 22, 30, 35.
- Problems from §15.3 (pages 849–851): 1, 3, 6, 7, 8, 11, 14, 17, 21, 25.
- Additional extra-credit problem:
Suppose that
**F**is a conservative vector field defined on all of 3-dimensional space; then there exists a scalar field*f*such that**F**= ∇*f*. Let*U*= −*f*. In physics, if**F**is a force field, then we call*U*a*potential energy field*for**F**. Recall that, if an object travels along a curve*C*in the force field**F**, then the work done on that object by that force field, or in other words the energy transferred to that object by that force field, is the integral ∫_{P∈C }**F**(*P*) ⋅ d*P*(or ∫_{C}**F**⋅ d**r**for short). If the curve*C*begins at the point*P*_{1}and ends at the point*P*_{2}, then use that**F**= −∇*U*to express the value of this work using values of the scalar field*U*at those points. If you imagine that*U*(*P*) is the amount of ‘potential’ energy held by an object at*P*by virtue of its position within this force field, then check that the amount of energy transferred to the object by the field (the work) is the opposite of the change in the object's potential energy. (In other words, we have*conservation of energy*: the total change in energy is zero. This conservation is why conservative vectors fields are called ‘conservative’.)

- Multiple integrals:
- Date taken: February 20 Monday.
- Problems from §14.1 (pages 759&760): 3, 7, 10, 17, 22, 27.
- Problems from §14.2 (pages 767–769):
- 1, 2, 7, 9, 12, 14, 17, 19, 23, 35, 41, 47, 51, 57, 61;
- Extra credit: 80.

- Problems from §14.5 (pages 785–788): 3, 6, 9, 15, 21, 25, 29, 34, 37.
- Problems from §14.3 (page 772): 1, 4, 7, 12, 13, 14, 17, 20, 21.

- Applications of multiple integrals:
- Date taken: February 27 Monday.
- Problems from §14.6 (pages 793–795): 3, 14, 19, 25, 29.
- Extra-credit problem from §14.8 (page 815):
To show work, show at least
the integral in
*u*and*v*that you evaluate: 16. - Problems from §14.4 (pages 777–786): 1, 3, 5, 7, 9, 17, 20, 23, 24, 28, 29, 34, 37.
- Problems from §14.7 (pages 803–806): 1, 2, 8, 12, 14, 23, 37, 43, 46, 57, 77.

- Integration on surfaces:
- Date taken: March 6 Monday.
- Problems from §15.5 (pages 872–874): 2, 3, 6, 9, 13, 20, 23.
- Additional extra-credit problem:
Consider the surface given by
*r*=*f*(*z*) in cylindrical coordinates, where*f*is a differentiable function defined on the interval [*a*,*b*]. Use the methods of §15.5 (or the handout) to show that the area of this surface is 2π ∫_{a}^{b}*f*(*z*) √(*f*′(*z*)^{2}+ 1) d*z*. - Problems from §15.6 (pages 883–884): 1, 5, 8, 11, 16, 17, 19, 23, 25, 34, 35, 37, 41, 45.

- The Stokes theorems:
- Date taken: March 13 Monday.
- Problems from §15.4 (pages 861–863): 1, 4, 7, 9, 12, 15, 21, 24, 26, 33.
- Problems from §15.7 (pages 895–897): 1, 3, 5, 6, 9, 14, 17, 21, 28.
- Problems from §15.8 (pages 906–908): 1, 2, 6, 7, 8, 13, 17.
- Extra credit problem based on Exercise 15.8.31 from the 2nd Edition:
To describe a scalar quantity
that not only takes values throughout space but also changes with time,
we need a function of
*four*variables, one (say*t*) to represent time and three (say*x*,*y*, and*z*) to represent space. The overall 4-dimensional space whose coordinates are*t*,*x*,*y*, and*z*is called*spacetime*. Let*δ*=*p*(*t*,*x*,*y*,*z*) be the mass density, at a given time and place, of some fluid substance, and let**v**=**F**(*t*,*x*,*y*,*z*) be the velocity, at a given time and place, of the fluid. The components of*δ***v**tell you the speed at which mass is flowing in particular directions, and we can put all of this information together into a single exterior differential pseudoform*Φ*of rank 3 in 4 variables:*Φ*=*δ*d*V*−*δ***v**⋅ d*t*∧ d**S**. (More explicitly, if**v**= ⟨*v*_{1},*v*_{2},*v*_{3}⟩, then*Φ*=*δ*d*x*∧ d*y*∧ d*z*−*δ**v*_{1}d*t*∧ d*y*∧ d*z*−*δ**v*_{2}d*t*∧ d*z*∧ d*x*−*δ**v*_{3}d*t*∧ d*x*∧ d*y*using the right-hand rule.) If the fluid is just flowing and not undergoing any physical, chemical, or nuclear reactions, then its mass should be conserved. This means not only that its total mass should be the same at any two times, but also that if its mass in any region of space changes between two times, then that change should be accounted for by flow through the region's boundary. That is, if*D*is any region of (*x*,*y*,*z*)-space,*S*is its boundary (pseudooriented outwards as usual), and*t*_{1}and*t*_{2}are two times (two values of*t*), then the integral of*δ*on*D*at*t*_{2}minus the integral of*δ*on*D*at*t*_{1}should equal the integral over the time from*t*_{1}to*t*_{2}of the integral of*δ***v**across*S*. In symbols, ∫_{(x,y,z)∈D}*p*(*t*_{2},*x*,*y*,*z*) d*V*− ∫_{(x,y,z)∈D}*p*(*t*_{1},*x*,*y*,*z*) d*V*= ∫_{t=t1}^{t2}(∫_{(x,y,z)∈S}*p*(*t*,*x*,*y*,*z*)**F**(*t*,*x*,*y*,*z*) ⋅ d**S**) d*t*. We can combine the region*D*at time*t*_{1}, the region*D*at time*t*_{2}, and the boundary surface*S*at the times in between into a single closed 3-dimensional hypersurface*H*in the 4-dimensional spacetime, and then this equation states that the integral of*Φ*out of*H*is 0. By the generalized Stokes Theorem, this integral is equal to the integral of the exterior differential d ∧*Φ*on the 4-dimensional region of spacetime bounded by*H*; by making*D*arbitrarily small around a given point in space and making*t*_{1}and*t*_{2}arbitrarily close on either side of a given time, this means that d ∧*Φ*itself must be 0. Now here is your assignment: Work out d ∧*Φ*and write the equation d ∧*Φ*= 0 in terms of the partial derivatives with respect to time and the gradient, curl, and/or divergence with respect to space of*δ*=*p*(*t*,*x*,*y*,*z*) and**v**=**F**(*t*,*x*,*y*,*z*). (The equation that you should end up with is called the*continuity equation*of conserved fluid flow.)

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