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

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)

 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

 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

 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

 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

 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)*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]| 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|(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: |xy^2-100|=|xy^2-4*5^2|=|xy^2-4y^2+4y^2-4*5^2| 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: |xy^2-4y^2+4y^2-4*5^2|<=|xy^2-4y^2|+|4y^2-4*5^2| 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: |4y^2-4*5^2|=4|y^2-5^2|=4|y-5|*|y+5|. 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

 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) and 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) so 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