QUIZ ANSWERS

BRIEF QUIZ ANSWERS FOR CALCULUS III STUDENTS

In this document you will find brief answers to quiz questions.

QUIZ

1 (Aug 27)
2 (Sep 3)
3 (Sep 10)
4 (Sep 17)
5 (Sep 24)
6 (Oct 1)
7 (Oct 8)
8 (Oct 15)
9 (Oct 22)
10 (Oct 29)
11 (Nov 5)
12 (Nov 12)
13 (Nov 19)
14 (Nov 26)
15 (Dec 3)

Week 1

Length and direction of (3,4,12) are 13 and (3/13, 4/13, 12/13) respectively.
Unit tangent vector to y = arctan(x) at x=1 is plus or minus (2/sqrt(5), 1/sqrt(5))
If v_1 points in the direction of the minute hand, v_2 points in the direction of the hour hand, it is three o'clock, and <v_1, v_2, v_3> is a right-handed system, then v_3 must point inward, INTO THE CLOCK.

Week 2

The way to express i + 2j - 3k as a product of its length and direction is: sqrt(14)((1/sqrt(14))i + (2/sqrt(14))j - (3/sqrt(14))k).
< k, i, j > is a right-handed system while < k, j, i > is not. (Answer: (a))
T or F

T, u dot v = v dot u. We proved this in class.
T, v dot v is greater than or equal to zero. This is because v dot v = square of length, and the square of any real number is greater than or equal to zero.
F, it is not true that (v dot v)/v = v. We cannot divide by a vector.

Week 3

P=(1,-1,2), Q=(2,0,-1), R=(0,2,1) are points in space.

The area of triangle PQR is (1/2)|vec[PQ] x vec[PR]| = (1/2)|(8,4,4)| = 2sqrt[6].
A unit normal vector to the plan determined by P, Q and R is N = (vec[PQ] x vec[PR])/|vec[PQ] x vec[PR]| = (8,4,4)/(4sqrt[6]) = (2/sqrt[6],1/sqrt[6],1/sqrt[6]).

Now P=(1,-1,2), Q=(2,0,-1), R=(0,2,1) are vectors. To determine the handedness of < P,Q,R > we evaluate a determinant (equal to the triple product (P x Q) dot R) and get the value +12. Since this is positive, < P,Q,R > is right-handed.
If A x B = 0, then A and B are parallel. If A dot B = 0, then A and B are perpendicular. The only way for A and B to be both parallel and perpendicular is for one of them to equal 0.

Week 4

TEST 1

Week 5

v(t) = ((2t)/(t^2+1),1/(t^2+1),t/square_root(t^2+1)), so v(0) = (0,1,0). The cosine of the angle between A=(0,1,0) and B=(1,1,1) is (A dot B)/|A||B| = 1/square_root(3). The angle is cos^{-1}(1/square_root(3)).
The position function is p(t) = (t,square_root(t^2+1)). Thus p'(t) = (1,t/square_root(t^2+1)). The tangent line at time t = 1 is given parametrically by L(t) = p(1) + tp'(1) which is just (1,square_root(2)) + t(1,1/square_root(2)).
Comparing the motions of two particles:

Second particle travels on curve C in the opposite direction of the first particle and meets the first particle at time t=0.
Second particle travels on different curve, -C, obtained by reflecting C through the origin. The motion of the second particle is the reflection of the motion of the first particle.
Second particle travels on curve C in the same direction as the first particle, but twice as fast. They meet at time t=-1.

Week 6

The length of the curve parametrized by p(t) = (t,t,2t^(3/2)) from t = 0 to 1 is the integral of |p'(t)| = square_root(2+9t) from t = 0 to 1. The answer is (2/27)[11^(3/2) - 2^(3/2)].
The radius of curvature of y = x^2 at (0,0) can be calculated using the formula rho = 1/kappa where kappa = |v x a|/|v|^3. To use this formula you need v and a, and so you need to parametrize the curve. You can take p(t) = (t, t^2, 0), and then calculate rho at t = 1. The answer is rho = 1/2.
True or False.

False. The acceleration of a particle moving at constant speed along a curve of constant curvature need not be constant, since direction of acceleration may change. This happens when a particle travels around a circle at a constant speed.
True. A plane curve has constant binormal, so torsion is zero.
True. A particle travelling a curve at constant speed experiences only centripetal acceleration. The magnitude of this is proportional to curvature and to the square of speed. Since we are considering two particles travelling the same curve when they are at the same point on the curve, the curvature is the same in both cases. Thus the acceleration of the particle moving twice as fast will be four times the acceleration of the slow particle.

Week 7

TEST 2

Week 8

f_x = -y/(x^2+y^2) and f_y = x/(x^2 + y^2).
grad(h) = (e^(x+y)cos(z),e^(x+y)cos(z),-e^(x+y)sin(z)), so grad(h(0,0,pi/6)) = (square_root(3)/2,square_root(3)/2,-1/2).
The tangent plane is z = 5 + 8(x-1) + 2(y-1).
There are many possible examples which are correct answers. One is f(x,y) = 2xy/(x^2+y^2).
Integrate f_x with respect to x to get f(x,y) = x^2y^2 + xcos(y) - ysin(x) + x + C(y). Note that C is only constant with respect to x, so it may be a function of y. To determine C(y), differentiate this expression with respect to y and compare it with f_y. You find by doing this that C'(y) = 0, so C(y) does not depend on y after all. Therefore, f(x,y) = x^2y^2 + xcos(y) - ysin(x) + x + C, where the constant C can be chosen arbitrarily.

Week 9

The Chain Rule tells us that the derivative of f(h(t)) at t = pi/4 is grad(f(h(pi/4))) dot h'(pi/4). Since grad(f(x,y)) = (2x,2y) and x=cos(t), y=sin(t), the first vector we need is grad(f(h(pi/4))) = (2cos(pi/4),2sin(pi/4)). The second vector we need is h'(pi/4) = (-sin(pi/4),cos(pi/4)). The dot product of the two vectors is easily seen to be zero.
The direction that f increases most rapidly at (-1,1) is the direction of grad(f(-1,1)) = (-1,1). The derivative in this direction is the length of grad(f(-1,1)), which is square_root(2).
Since we want a vector normal to a level surface, the gradient will work. The answer should be grad(f(1,1,1)). Since grad(f(x,y,z)) = (1,2y,3z^2), the vector (1,2,3) is normal to the surface at (1,1,1).

Week 10

The only critical point in this region is at (-1/4,0), so that is the only potential extreme point.
The Hessian is:
[8,0
0,2].
Therefore the critical point found in Problem 1 is a minimum.
Solving grad f = lambda grad g leads to y = 0 or x = -1/3. If y = 0, then the possible points are (+/-1,0), both of which are local maxima. If x = -1/3, then the possible points are (-1/3,+/- square_root(8/9)), both of which are minima for f restricted to x^2+y^2=1.
Testing the four plausible values (+/-1,0), (-1/3,+/- square_root(8/9)) shows that the absolute maximum occurs at (1,0), and that the maximum value is 6.

Week 11

TEST 3

Week 12

The Jacobian is the following 3x3 matrix:
[cos(t),-rsin(t),0
sin(t),rcos(t),0
0,0,1].
The determinant is r.
The figure is a certain parallelogram in the xy-plane.
The change of variables transforms the integral into the integral of usquare_root(v) over the uv-rectangle [0,2]x[1,2]. The Jacobian determinant is 1. The final answer is: (4/3)(2^(3/2) - 1).

Week 13

To integrate this you should change to polar coordinates and then use the u-substitution u = r^2 + 1. The final answer is (Pi/2)[2ln(2) - 1].
Parametrize the circle by p(t) = (cos(t),sin(t)) with t ranging from 0 to 2Pi. Note that |p'(t)| = 1. The integral reduces to the integral of cos^2(t) + sin(t) from 0 to 2Pi, and the value of this integral is Pi.
Parametrize the curve by p(t) = (t,t^2) with t ranging from -1 to 1. Note that |p'(t)| = square_root(1+4t^2). The integral reduces to 2 times the integral from 0 to 1 of the polynomial t + 4t^3. The final value is 3.

Week 14

The answer is ZERO. Parametrizing the circle by (x(t),y(t)) = (cos(t),sin(t)), the flux integral (integral around C of: M dy - N dx = x dy - (x-y) dx) reduces to the integral of cos^2(t) + cos(t)sin(t) - sin^2(t) from 0 to 2Pi, which is zero.
The answer is 2Pi. Parametrizing the circle by (x(t),y(t)) = (cos(t),sin(t)), the integral reduces to the integral of cos^2(t) + sin^2(t) = 1 from 0 to 2Pi, which is zero.

The direction of F at v = (x,y,z) is toward the origin. The unit vector in that direction is v/|v|. The length is given to be 1/|v|^2. Therefore F at v is v/|v|^3 (length times direction). This is:
(-x/(x^2+y^2+z^2)^(3/2),-y/(x^2+y^2+z^2)^(3/2),-z/(x^2+y^2+z^2)^(3/2)).
To find an f such that F = grad(f) = (f_x,f_y,f_z) we need to integrate the equation f_x = -x/(x^2+y^2+z^2)^(3/2) with respect to x. If we do we get
f = (x^2+y^2+z^2)^(-1/2) + C(y,z).
Here C(y,z) is some function which is constant with respect to $x$. Differentiating this equation with respect to y we get
f_y = -y/(x^2+y^2+z^2)^(3/2) + C_y(y,z).
But since we already know that f_y = -y/(x^2+y^2+z^2)^(3/2)
we get that C_y(y,z) = 0. A similar argument shows that C_z(y,z) = 0. Thus C = C(y,z) is also constant with respect to y and z. The conclusion is that any potential function has the form
f = (x^2+y^2+z^2)^(-1/2) + C.
One can check by differentiation that any function of this form is a potential function.

Week 15

There are three blanks to fill in. I will put brackets around what goes in the blank.

The [line integral over C of F dot dr (circulation integral)] is zero ... etc.
There is an f such that [grad f] = F(x,y).
The differential form M dx + N dy is [exact].

The answer is 4/5. To evaluate the line integral from (1,0) to (3,4), it is enough to calculate f(3,4) - f(1,0) since f is a potential function for F. Here f(3,4) = -1/5 and f(1,0) = -1, so we get 4/5.
The answer is 1. Here the vector field F(x,y) = (xy,xy) is zero along the left and bottom edges of the unit square, and it is invariant under interchanging x and y. Therefore the flux across the lefthand and bottom edges is zero and the flux across the righthand edge will equal the flux across the top edge. Hence the flux integral around the boundary will be twice the flux integral across the righthand edge. We can parametrize that edge by (x(t),y(t)) = (1,t), with t ranging from 0 to 1. Our answer will be twice the integral of t as t ranges from from 0 to 1, which is just 1.

Last modified on August 26, 1997.