Students' answers to review problems for the first exam
in Math 291, fall 2002

#1 Ellway #2 Graziadio #3 Grubin #4 Gurkovich #5 Hort
#6#7 #8 Kravtsov #9 Lunemann #10 Odoi
#11 Pergament #12 #13 Ryslik #14 Sullivan #15
#16 Tokayer #17a Vander Valk #17b #18 Walsh #19 Zhu

List of textbook problems students have asked me to review on Monday, October 14: 14.5:51-53, 14.3:85, 14.4:41,42.

A draft of formulas requested by students on the first exam [Postscript | PDF]
(first version posted 10/14/2002; revised 10/15/2002)

1. Sketch two level curves of f(x,y)= (x-y)/(x+y). Explain why lim_{(x,y)->(0,0)}(x-y)/(x+y) does or does not exist.

The graph of the two level curves is found by selecting values of k to set the equation (x-y)/(x+y) equal to. For example when k=2: 2x+2y=x-y so x=-3y and y= -(1/3)x
The sketch of this curve is a straight line that passes through the origin with a slope of -1/3.
A second level curve can be found by selecting k=0: x-y=0 so y=x.
The sketch of this curve is a straight line that passes through the origin with a slope of 1.

lim_{(x,y)->(0,0)}(x-y)/(x+y) does not exist because as you approach the origin along where k=0 the value of the function is equal to 0. As you approach the origin along the curve k=2, the function is equal to 2. Therefore since 0 does not equal 2, the limit cannot exist.

Sent by Jordana Ellway on Sun, 13 Oct 2002 19:38:11

2. Let f(x,y)=xy^2+2x^3.
a) If one stands on the graph of f at (1,2,6), what is the slope of the graph at that point in its steepest direction?
b) Find D_u f(1,2) if u is the unit vector in the direction from (1,2) to (2,4).

2)a) f(x,y)=xy^2 + 2x^3
grad f = (y^2 + 6x)i + (2xy)j
grad f(1,2,6) = 10i + 4j
Slope = SQRT(10^2 + 4^2) = SQRT(116)

b) u = (1/SQRT(5))i + (2/SQRT(5))j
D_u   f = grad f dot u
D_u   f(1,2) = (10)(1/SQRT(5)) + (4)(2/SQRT(5))
D_u   f(1,2) = 18/SQRT(5)

Sent by Chris Graziadio on Mon, 14 Oct 2002 15:19:42

3. Find an equation of the plane containing the points (1,2,0) and (0,2,1) and parallel to the line x=1+t, y=-1+t, z=2t.

The parametric equations x=1+t, y=-1+t, z=2t can be converted into a vector v. Taking the coefficients of t in each parametric equation yields v =i+j+2k. The plane should be parallel to this vector. Using the point (1,2,0), define the vector r as i+2j. Then, define the vector s as r + tv. This vector represents a line parallel to vector v that lies on the plane. s equals (1+t)i+(2+t)j+(2t)k. Next, plug in t=1 (this value is arbitrary) to s to obtain the point (2,3,2). This point is on vector s which is on the plane, so this point must be on the plane. Now, with (1,2,0) and (0,2,1) given and (2,3,2) obtained, it is easy to find the equation of the plane. Find the vector from (1,2,0) to (0,2,1), which equals -i+k. Also find the vector from (1,2,0) to (2,3,2), which equals i+k+2k. Take the cross product of these two vectors to get the vector normal to the plane. This vector is equal to -i+3j-k. Using this vector and the point (1,2,0), the equation of the plane is found to be -(x-1)+3(y-2)-z=0. Check this equation by plugging in the three points above. Also, dot the normal vector with vector v. The result equals zero, so the two are indeed perpendicular.

Sent by Jeremy Grubin on Mon, 14 Oct 2002 01:53:17

4. If z=f(x,y), x=3uv, and y=u-v^2, express z_v in terms of z_x, z_y, u, and v.

dz/dv = (dz/dx) (dx/dv) + (dz/dy) (dy/dv)
dx/dv = 3u
dy/dv = -2v
dz/dv = (dz/dx) 3u + (dz/dy) -2v

Sent by Corey Gurkovich on Mon, 14 Oct 2002 15:38:14

5. For the surface defined by x^5+y^5+z^3+xyz=0, find an equation of the tangent plane at (1,1,-1), and the value of z_x at (1,1,-1).

Suppose f(x,y,z)=x^5+y^5+z^3+xyz=0. Equation of the tangent line: use the grad of the function: grad f = (5x^4+yz)i + (5y^4+xz)j + (3z^2+xy)k = <5x^4+yz, 5y^4+xz, 3z^2+xy> Evaluated at (1,1,-1)= 4i + 4j + 4k = <4,4,4>
Thus, the equation of the plane is: 4(x-1)+4(y-1)+4(z+1)=0.
Value of z_x at (1,1,-1): z_x= -(f_x)/(f_z) so z_x= -(5x^4+yz)/(3z^2+xy). Evaluated at (1,1,-1), this =-(5+(-1))/(3+1) = -1. Value of z_x at (1,1,-1)= -1.

Sent by Greg Hort on Sat, 12 Oct 2002 16:26:50

8. Use linear approximation at the point (64,9) for an appropriate function of two variables to approximate 63^(1/3)*sqrt(10).

Let f be the function of two variables f(x,y)=x^(1/3)*y^(1/2). Using the linear approximation we will know that f(x,y) is approximately equal to f(a,b)+f_x(a,b)*(x-a)+f_y(a,b)*(y-b) at some point (a,b) which is (64,9) in our case. Therefore, f(63,10) is approximatley f(64,9)+f_x(64,9)*(x-64)+f_y(64,9)*(y-9), where f_x(x,y)=y^(1/2)/(3*x^(2/3)) and f_y(x,y)=x^(1/3)/(2*y^(1/2)).
Here f_x(64,9)=3/(3*64^(2/3))=1/16 and f_y(64,9)=64^(1/3)/(2*9^(1/2))=4/6=2/3 and f(64,9)=64^(1/3)*9^(1/2)=4*3=12.
So the approximation is 12-(1/16)+(2/3) which is (about!) 12.60416667. Maple reports the "true value" (also, of course, an approximation because this is an irrational number!) to be 12.58288372 .

Sent by Andrey Kravtsov on Sun, 13 Oct 2002 12:01:51

9. Find an equation of the tangent plane to the surface z=x^3*y at (1,2,2).

The tangent plane can be found by using the equation:
z- z_0 = (f_x(x_0,y_0))(x-x_0) + (f_y(x_0,y_0))(y-y_0) which is the equation of the tangent plane to the surface z = f(x,y) at the point P(x_0,y_0,z_0).
In this case, f_x = 3yx^2 and f_y = x^3. At P is (1, 2, 2) so f_x(1, 2) = 6 f_y(1, 2) = 1.
Therefore, the equation of the tangent plane to the surface z=x^3*y at (1,2,2) is z-2 = 6(x-1) + (y-2) OR (for those who wish to simplify) z = 6x + y - 6.

Sent by Matthew Lunemann on Sun, 13 Oct 2002 22:42:06

10. Suppose that w=f(x,y), where f is a function satisfying f(1,2)=3, f_x(1,2)=1, f_y(1,2)=-2, f_{xx}(1,2)=3, f_{xy}(1,2)=2, and f_{yy}(1,2)=0. Suppose further that x=u+v-1 and y=3uv-1. Find w_u|{u=1,v=1} and w_vu|{u=1,v=1}.

By applying the chain rule to w_u, we get w_u=w_x(x_u)+w_y(y_u) =w_x*1+w_y*3v =f_x(x,y)*1+f_y(x,y)*3v. Call this equation (#).
Notice that x=1 and y=2 when u=1 and v=1.
Therefore, w_u|{u=1,v=1}=f_x(1,2)*1+f_y(1,2)*(3*1) =1*1+(-2)*3 =-5

Using w_u in (#) from the previous part and applying the product rule to the part of w_y*3v, we rewrite: w_vu=(w_u)_v =(w_x*1+w_y*3v)_v =(w_x)_v+(w_y)_v*3v+w_y*3.
Computing (w_x)_v by the chain rule, we get (w_x)_v=(w_x)_x*x_v+(w_x)_y*y_v =w_xx(x_v)+w_yx(y_v) =f_xx(x,y)*1+f_xy(x,y)*3u.
Similarly, for (w_y)_v, we get, (w_y)_v=(w_y)_x*x_v+(w_y)_y*y_v =w_xy(x_v)+w_yy(y_v) =f_xy(x,y)* 1+f_yy(x,y)*3u.
Therefore, w_vu|{u=1,v=1}=f_xx(1,2)*1+f_xy(1,2)*3*1+{f_xy(1,2)*1+f_yy(1,2)*3*1}*3*1 +f_y(1,2)*3 =3*1+2*3+(2+0)*3+(-2)*3 =3+6+6-6 =9

Sent by Atsuko Odoi on Sun, 13 Oct 2002 05:40:14

11. A point is moving along the curve below in the direction indicated. Its motion is parameterized by arc length, s, so that it is moving at unit speed. Arc length is measured from the point P (both backward and forward). The curve is intended to continue indefinitely both forward and backward in s, with its forward motion curling more and more tightly about the indicated circle, and, backward, closer to the indicated line.
Sketch a graph of the curvature, kappa, as a function of the arc length, s. What are lim_{s-->+infinity}kappa(s) and lim_{s-->-infinity}kappa(s)$? Use complete English sentences to briefly explain the numbers you give.

At the point P, curvature is somwhere between 0 and 1, but closer to 0. As s approaches +infinity, there is a horizontal asymptote at kappa=1. As s approaches negative infinity, there is a horizontal asymptote at kappa=0. The limit as s approaches infinity is 1, and the limit as s approache as negative infinity is 0. We know this because the curvature of a straight line is 0. Going in the negative s direction, the graph of the function gets closer and closer to a straight line, and so the curvature approaches 0. As you travel along the curve in the positive s direction, the curve starts spinning around a circle. We know that the curvature of a circle is 1/r, and since the radius of the circle is 1, the curvature approaches 1.

Sent by Alexander Pergament on Mon, 14 Oct 2002 16:18:35 Special added bonus!

The management is happy to supply this picture:

13. Find & classify as well as you can all critical points of f(x,y) = x^2-2yx^2+2y^2.

Eq. 1 f_x = 2x-4yx = 2x(1-4y) and Eq. 2 f_y = -2x^2 +4y -2(x^2-2y).
Now to find the critical points I will set these equal to 0.
So 2x(1-4y) = 0. Therefore either 2x or 1-4y = 0.
First I will deal with the situation where 2x = 0.
Hence, 2x=0 --> x = 0. Now if I plug 0 into x in Eq. 2 the equation becomes 2*0^2 + 4y = 0 so 4y =0 then y = 0. Therefore one critical point is (0,0).
The other situation is that 1-4y = 0. Hence 1 = 4y --> 1/4 = y. now if I plug this y into Eq. 2 the equation becomes -2(x^2 - 2(1/4)) = 0. Then x^2 - 1/2 = 0 and x^2 = 1/2. Therefore x = sqrt(1/2) and another critical point is (sqrt(1/2), 1/4).
Now because F_x and F_y are polynomials, they are defined everywhere. Thus there are no critical points at undefined locations. Hence, there are only 2 critical points.
Now to calculate whether these points are maximums, minimums or saddle points, I will calculate D which is defined as the determinant of

| f_xx    f_xy |
|              |
| f_yx    f_yy |
at the critical point.
Assuming that the second partial derivatives of f(x,y) are continuous (which they are) and evaluating at the critical points (a,b) found above,
If D>0 and f_xx (a,b) (the second partial derivative with respect to x and to x again at the point (a,b) ) >0 then f(a,b) is a local minimum.
If D>0 and f_xx (a,b) < 0 then f(a,b) is a local minimum.
If D<0, then f(a,b) neither a maximum or a minimum and is therefore a saddle point.
Eq. 3 f_xx = 2-4y
Eq. 4 f_yy = 4
Eq. 5 f_xy = -4x
Eq. 5 f_yx = -4x
Therefore the determinant (since f_xy is equal to f_yx) is f_xx*f_yy-[f_xy]^2. Therefore D = D(x,y) = 8-16y-16x^2.
At the critical point (0,0) D = 8 and f_xx = 2. Thus since D>0 and f_xx(0,0)>0, the point (0,0) is a local minimum.
At the critical point (sqrt(1/2),1/4), D = -4. Since D<0, the point is automatically a saddle point.

Sent by Greg Ryslik on Thu, 10 Oct 2002 20:38:55

14. Find and classify as well as you can the critical points of f(x,y)=(x^2+y^2-1)^10.

grad f==<20x(x^2+y^2-1)^9,20y(x^2+y^2-1)^9>.
The gradient is the zero vector for (0,0), and for any point on the circle x^2+y^2=1.
Now, I have to find whether these points are max, min, or saddle point. To test (0,0,1), I can use the second derivative test.
f_xy & f_yx=360xy(x^2+y^2-1)^8
To find whether (0,0,1) is a local max/min, or saddle, find H, and plug in (0,0).

          |f_xx  f_xy|
H(x,y)=det|f_yx  f_yy|=400(x^2+y^2-1)^18+7200(x^2+y^2)(x^2+y^2-1)^17
H(0,0)=400 > 1, so (0,0,1) is either a local min or max. Since f_xx(0,0)=-20 < 0, (0,0,1) is a local maximum.
The other critical points, (x,y) satisfying the equation x^2+y^2=1, cannot be classified using the second derivative test (because H=0 for them!), I used algebraic reasoning.
f(x,y)=(x^2+y^2-1)^10 is raised to an even power, which means that f(x,y) is greater than or equal to zero for all x,y. Since the minimum value of f(x,y)=0, set (x^2+y^2-1)^10=0. By setting f(x,y) equal to its minimum value and solving, I get x^2+y^2=1, verifying that the points on that circle are the only points on f(x,y) that touch zero, and that they are absolute minimums.

Sent by Jason Sullivan on Sat, 12 Oct 2002 15:00:31

16. Use the epsilon-delta definition to verify that g(x,y)=x^2y^2 is continuous at (-1,2).

Definition of continuity: A function f is continuous at x0 if, given any eps>0, there is a delta>0 so that if |x-x0| Applying the definition to this problem: vector_x=, vector_x0=<-1,2>.
We need to verify:
IF: sqrt((x+1)^2+(y-2)^2)<delta, then |(x^2*y^2)-(-1^2*2^2)|<epsilon

sqrt((x+1)^2+(y-2)^2)<delta: Since both sqrt((x+1)^2) and sqrt((y-2)^2) are greater than or equal to zero, if the restriction is satisfied, then each of the parts satisfies it. So then |x+1|<delta and |y-2|<delta are true also (because (sqrt((x+1)^2)=|x+1| and sqrt((y-2)^2)=|y-2|). We can look for delta restrictions on |x+1| and |y-2| separately.
|(x^2*y^2)-(-1^2*2^2)|<epsilon =>*1
In order to control the changes of x&y individually, we can rewrite this equation as: |((x^2*y^2)-(x^2*2^2))[part1]+((x^2*2^2)-(-1^2*2^2))[part2]|<epsilon.
We are allowed to do this because: -(x^2*2^2)+( x^2*2^2)=0. Therefore, we are not actually adding anything to the equation). For part 1 of the equation, only the y-value is changing, while in part 2 only the x-value is changing.
According to the triangle inequality: |(x^2*y^2)-(x^2*2^2)+(x^2*2^2)-(-1^2*2^2)|<=|(x^2*y^2)-(x^2*2^2)|+|(x^2*2^2)-(-1^2*2^2)|
Since I have just split the equation into 2 pieces, I know that each piece makes up some fraction of the original formula *1. Therefore, if I can verify that |(x^2*y^2)-(x^2*2^2)|<epsilon/2 and if I can verify that |(x^2*2^2)-(-1^2*2^2)|<epsilon/2, then I will be done. (Adding these 2 equations will again give *1.)
|(x^2*y^2)-(x^2*2^2)|<epsilon/2 Factoring out x^2 gives: |x^2|*|y*2-4|<epsilon/2 or |x^2|*|y+2|*|y-2|<epsilon/2
I want this inequality to be true. Suppose I know that |x|<=2 and |y|<=3. If these conditions are true, then I know that the maximum value of |x^2|=4 and the maximum value of |y+2|=5. Therefore, the equation, solved for |y-2|, now becomes: |y-2|<epsilon/40.
Factoring out a |2^2| gives: 4|(x^2)-1|<epsilon/2, so |(x+1)*(x-1)|<epsilon/8. Using the same estimates that I suggested before, I know that the maximum value of |x-1| is 1. Plugging this in to the equation I now have |x+1|<epsilon/8.
Now I collect the conditions that must be satisfied for this inequality to apply to the more general condition. I need:
|x|<=2 and |y|<=3, |x+1|<epsilon/8, |y-2|<epsilon/40.
What I am able to specify is sqrt(|x+1|^2+|y-2|^2)<delta.
I now choose delta to be the minimum of eps/8 and eps/40 and 2 and 3. I just really need the minimum of epsilon/40 and 2. Collecting all the restrictions on the variables shows that:
If |(x,y)-(-1,2)|<delta, then |f(x,y)-f(-1,2)|<eps, where delta is the minimum of epsilon/40 and 2.

Sent by Jason Tokayer on Sun, 13 Oct 2002 10:29:19

Special added bonus!

Here is the substance of several e-mail messages I wrote to a student explaining another example. Maybe what's written here will be helpful to some people.

let's work through a 1 variable example. the algebra will be irritating, but, what the heck. i believe that the limit of x^3 as x--> 2 is 8. so i should be able to make |x^3-8| small by making |x-2| small. but how could i verify that? (yes, i could graph things, etc., but this is really preparation for a more than 1 variable example, so i would prefer to stick with algebraic techniques.)
the connection between x^3-8 and x-2 is "simple". i hope i have it right: (x^3-8)=(x^3-2^3)=(x-2)(x^2+2x+4) (you aren't watching me, thank goodness ... i am painfully checking what i just typed ... yes, it IS correct.)
suppose i wanted |x^3-8| to be less than, say, 1/(10,000). (i am picking a specific number just to have something specific to work with). well, if i look at the equation above, all i seem to need to do is make |x-2| really really small. but i do need to deal with the other factor. is IS possible that x-2 could be small while the other stuff (x^2+2x+4) could be really big, so the product wouldn't be small. well, i need some sort of control. so right now, i will insist that |x-2|<1. (again, the "1" is pulled out of the air, but i do need something.) if |x-2|<1, then x must be between 1 and 3, i think. and then i will look at x^2+2x+4: the pieces are all positive, so if i want to estimate the largest possible size i just need to find the largest size of each piece. and the pieces are increasing. so, in this case (which is "easy") i can just plug in 3 for x. and i have the following logical implication:
IF |x-2|<1, THEN |x^2+2x+4|<19 (i put x=3 in the quadratic, i hope correctly.)
well then, if |x-2|<1, i know that |x^3-8|= |x-2|*|x^2+2x+4|< 19|x-2|. if i want this less than 1/(10,000), i should choose |x-2|<1/(190,000).
o.k. now put it together. if i want |x^3-8| to be less than some epsilon (the "output error") then i'd better make |x-2| less than ... let's see: i need epsilon/19 AND i also need it to be less than 1 (otherwise i can't control |x^2+2x+4| enough to get the estimate |x^2+2x+4|<19 which i need.
so there seem to be two possible restrictions: |x-2|<epsilon/19 AND |x-2|<1. the traditional name for the restriction on |x-2| is delta. so if i say that delta is the MINIMUM of 1 and epsilon/19, this logical statement is correct:
IF |x-2|<delta, THEN |x^3-8|<epsilon.
i think there's very little "higher math" in this, but certainly some intricate logic.
well, what we did here was a rather straightforward computation. in more complicated examples, especially in more than 1 variable, there may be more restrictions needed.
let me try a 2 variable example and see if it helps. look at xy^2 as (x,y)-->(4,5). i am confident that xy^2 will get close to 100. (i think that is 4*5^2.) but i would like to control the closeness. in fact, i would like to know restrictions to put on x and y so that i am GUARANTEED that |xy^2-100|<1/(10,000). i will use the techniques i showed in class yesterday. again, this willbe tedious, but what i'm doing has these advantages: 1. it is essentially elementary (no fancy stuff needed!). and 2. it works -- it gives an effective answer.
so now we start:
here in this step i have replaced "0" by "-4y^2+4y^2" and the aim is to make a 2 variable change into a succession of 1 variable changes. now i will use the triangle inequality in order to make things easier to handle:
i use <= for "less than or equal to" because the ascii character set doesn't have the appropriate thing.
there are two pieces. i need to keep my eye on the final goal, which is to get things less than 1/(10,000). since there are two pieces, i know that if i can guarantee each of the pieces is less than 1/(20,000), i will have won. (yes, i could split things up differently, but i just want to "win" here.)
so i want |4y^2-4*5^2|<1/(20,000) and i will do this by "controlling" |y-5|. here is the algebra:
now i need to control the size of |y+5|. i will make an assumption about |y-5|. just to be a bit different, let me try |y-5|<.5 (yeah, i could have used 1, but let me do this in an effort to convince you that other choices work.) if |y-5|<.5, i know that -.5<y-5<.5 so 4.5<y<5.5 and so 9.5<y+5<10.5 and so |y+5|<10.5. so:
IF |y-5|<.5, THEN |4y^2-4*5^2|=4|y-5|*|y+5|<4*|y-5|*10.5=42|y-5|.
but i want this to be less than 1/(20,000). i can achieve this if i know |y-5|<1/(840,000). the 840,00 is the product of 42 and 20,000. (hey, now you may begin to see why people invented algebra in the first place: the numbers are getting weirder and weirder.)
i am NOT done yet. i still must make the other term "small". that is, i still want |xy^2-4y^2|<1/(20,000). so let's see:
now the multiplier is y^2. but, thank goodness, i already have some control over y because of the previous work. i know that 4.5<y<5.5, so 20.25<y^2<30.25 (in another window i just used a calculator. i was NOT too intelligent when i chose that darn .5 instead of 1. but now you may see why people make simple choices like 1 instead of things like .5).
i now know:
if |y-5|<.5, then |xy^2-4y^2|=y^2|x-4|<30.25|x-4|.
this is less than 1/(20,000) if i require |x-4|<1/(30.25*20,000)=1/(605,000).
sigh. the numbers are getting more and more ludicrous.
o.k. let me sum up the EASY EXAMPLE (which i think i have managed to make complicated enough!).
IF |x-4|<1/(605,000) and IF |y-5|<.5 and IF |y-5|<1/(840,000) THEN |xy^2-100|<1/(10,000).
this is too complicated. why not look carefully at the numbers. the "tightest" restriction is with the smallest number, in this case 1/(840,000). so a simpler statement is the following:
IF the distance from (x,y) to (4,5) is less than 1/(840,000), THEN |xy^2-100|<1/(10,000) is guaranteed.
notice please that this may not be "best possible" -- i did NOT ask for that. what i mean is that maybe a looser restriction may serve just as well to control the output. i wanted a simple way to get a simple restriction on the input error to guarantee the output error.
in this example, if instead of 1/(10,000) i had written epsilon, what would i need to get the output restriction? let's see: if i stick with the darn |y-5|<.5, i get that i need |y-5|<epsilon/84 and i also need |x-4|<epsilon/60.5. there are 3 restrictions. now if i make delta equal to the minimum of .5 AND epsilon/84 AND epsilon/60.5 (stupid choice, that original .5, darn it!) then i will know:
IF the distance from (x,y) to (4,5) is less than this delta described above, THEN |xy^2-100|<epsilon is guaranteed.
We need to control |y-5| and |x-4|. Specifically, we need several assumptions to get the sequences of inequalities to work.
So: |x-4|<epsilon/60.5, |y-5|<epsilon/84 and |y-5|<.5 are all needed.
It would be nice to "package" this better, so that anyone who actually wants to use it could easily verify that all three (or four or five or sixty-eight, however many in the application being considered) inequalities are correct.
CLAIM: if sqrt((x-4)^2+(y-5)^2) < K then |x-4|<K and |y-5|<K.
Why is this correct? Well, take sqrt((x-4)^2+(y-5)^2) < K and square it. Then (x-4)^2+(y-5)^2 < K^2 of course. Both of the terms on the left-hand side of the inequality are non-negative. Certainly each of them is < K^2, otherwise the stated inequality would be wrong. SO we know (x-4)^2<K^2. Take the square root. Then sqrt((x-4)^2)<K. But sqrt((x-4)^2) is actually |x-4|. So |x-4|<K. The same is true for the other term, so |y-5|<K.
If we use the CLAIM, then
|x-4|<epsilon/60.5, |y-5|<epsilon/84 and |y-5|<.5
will be verified if sqrt((x-4)^2+(y-5)^2) < K where K is selected so that K is the smallest of the numbers epsilon/60.5, epsilon/84, and .5 and that suggests that we just take K to be the minimum of those three numbers.
The standard name for K is delta. Another reason that people like to use sqrt((x-4)^2+(y-5)^2) to measure "error" is that it corresponds to distance between (x,y) and (4,5), so that there is a picture (?) of what it supposedly measures.

17a. Find the following limits, if they exist. lim_{(x,y) \to (0,0)} (10x^4y)/(x^4+y^4).

If we approach (0,0) along the line y=x, we get lim_{(x,y)->(0,0)}(10x^5)/(2x^4) = lim {(x,y)->(0,0)}(10x)/2 = 0. So if the limit exists, its value must be 0. Now we must show the limit does exist.
Let eps>0. We want to find delta>0 such that: |(10x^4*y)/(x^4+y^4) - 0| < eps whenever 0 < (x^4+y^4)^(1/4) < delta.
That is, (10x^4* |y| )/(x^4+y^4) < eps whenever 0 < (x^4+y^4)^(1/4) < delta.
We know that x^4 <= x^4+y^4 since y^4>=0, so (x^4) / (x^4+y^4) <= 1 and therefore (10x^4 * |y|) / (x^4+y^4) <=10*|y| = 10*(y^4)^(1/4) <= 10(x^4 + y^4)^(1/4).
If we choose delta = eps/10 and let 0 < (x^4+y^4)^(1/4) < delta, then |(10x^4*y)/(x^4+y^4) - 0| <= 10(x^4 + y^4)^(1/4) < 10*delta = (10*eps)/10 = eps.
By the definition of a limit: lim_{(x,y)->(0,0)}(10x^4*y)/(x^4+y^4) = 0

Sent by Nicholas Vander Valk on Mon, 14 Oct 2002 20:03:53

Special added bonus!

The work above is certainly excellent. Another way to analyze this limit is to use polar coordinates (this will work because of the coincidences of various degrees of the top and bottom). So if x=r  cos(theta) and y=r  sin(theta), then 10x^4y=10r^5 (cos(theta))^4 sin(theta) and x^4+y^4=r^4[(cos(theta)^4+sin(theta)^4)]. So the fraction has the form (after cancelling r's): r (trig stuff), where the "trig stuff" looks like [(cos(theta))^4 sin(theta)]/[(cos(theta)^4+sin(theta)^4)]. theta varies between 0 and 2pi. But notice that the bottom is never 0 (because cosine and sine can't be 0 together since the sum of their squares is 1). So the trig stuff is a continuous function which is defined on [0,2pi] and is certainly bounded by some constant, M, which I don't know but whose specific value is not necessary. As (x,y)-->(0,0), r-->0. The absolute value of the original fraction is squeezed between 0 and Mr, and the upper bound --> as r-->0. So the limit exists and is 0.
This certainly isn't "better" than Mr. Vander Valk's approach, just different.

18. Suppose G(u,v) is a differentiable function over two variables and g(x,y) = G(x/y,y/x). Prove xg_x(x,y) + yg_y(x,y) = 0.

For easy understandability, I will use u = x/y, and v = y/x. Thus, g(x,y) = G(u,v).
By applying the chain rule,

g_x(x,y) = D_1 G(u,v)*(du/dx) + D_2 G(u,v)*(dv/dx)
               = D_1 G(u,v)*(1/y) + D_2 G(u,v)*(-y/x^2)
g_y(x,y) = D_1 G(u,v)*(du/dy) + D_2 G(u,v)*(dv/dy)
         = D_1 G(u,v)*(-x/y^2) + D_2 G (u,v)*(1/x)
xg_x(x,y) + yg_y(x,y) =  x*[D_1 G(u,v)*(1/y) + D_2 G(u,v)*(-y/x^2)]
                       + y*[D_1 G(u,v)*(-x/y^2) + D_2 G(u,v)*(1/x)]
                      =  (x/y)D_1 G(u,v) - (y/x)D_2 G(u, v)
                       - (x/y)D_1 G(u,v) + (y/x)D_2 G(u,v)
                      =  0.
Sent by Joseph Walsh on Sat, 12 Oct 2002 15:04:34

19. a) Find parametric equations for the line through the points (3,2,7) and (-1,1,2).
b) At what point does this line intersect the plane x+y+z=22?

a. Let vectors A = (3, 2, 7) and B = (-1, 1, 2). Then the position vector (x, y, z) for a point on the line AB is given by A + t(A - B). So x = 3 - 4t, y = 2 - t, z = 7 - 5t.
b. Solve simultaneously the equations from a and x + y + z = 22. Substituting for x, y, z gives t = -1. Then (x, y, z) = (7, 3, 12).

Sent by Siwei Zhu on Sun, 13 Oct 2002 19:17:03