### Students' answers to review problems for the first exam in Math 291, spring 2003

 #1 BERGKNOFF #2 BERKOWITZ #3 BROWN #4 #5 ELKHOLY #6 #7 #8 HEDDY #9 KLEYZIT #10 MCGOWAN #11 #12 PASHKOVA #13 TOZOUR #14 #15 WILSON #16

 1.Suppose that F(s,t) is a differentiable function of two variables and that F(5,3)=A, dF/ds(5,3)=B, dF/dt(5,3)=C, d^2 F/ds^2(5,3)=D, d^2 F/ds dt(5,3)=E, and d^2 F/dt^2(5,3)=F. If G(x,y)=F(x^2+y^2,x^2-y^2), compute d^2 G/dx^2(2,1) in terms of the information given. There wasn't much chicanery involved in my solution, it was just a straightforward application of the chain rule twice. I differentiated G with respect to x first, getting dg/dx = (dF/ds)(ds/dx)+(dF/dt)(dt/dx). Then I used the product rule and chain rule to differentiate with respect to x again. d^2 G/dx^2=2(dF/ds)+(2x)((d^2 F/ds^2)(ds/dx)+(d^2 F/dsdt)(dt/dx))+2(dF/dt)+(2x)((d^2 F/dt^2)(dt/dx)+(d^2 F/dtds)(ds/dx)) because the derivatives exist (and assumedly are continuous), d^2 F/dtds = d^2 F/dsdt. d^2 G/dx^2 evaluated at (x=2,y=1) is 2B + 2C + 16D + 32E + 16F (assuming that the last line on my problem is a typo and that should be F rather than a second E). Sent by Jonathan Bergknoff on Sun, 9 Mar 2003 20:40:33 Message from the management The error has been corrected (at least in the version posted on the web). It should indeed have been an F rather than an E. I thank Mr. Bergknoff for his alertness. Of course (sigh!), the F as a constant then conflicts typographically and logically with the F for the function.

 2. The tangent plane of the graph of a function f(x,y) at point P(xo,yo,zo) is: equation 1: z-zo = f_x(x0,y0)(x-xo)+f_y(xo,yo)(y-yo) where P(x_0,y_0,z_0)= (3,0,9) (we find z_0 by plugging in x_0 and y_0 in f). Here f_x(3,0)=6 and f_y(3,0)=9 which makes equation 1 become: equation 2: (z-9) = 6(x-3) + (9y) since the equation for a plane is a(x-xo)+b(y-yo)+c(z-zo)=0 where is the normal vector, we rewrite equation 2 to look like this: 6(x-3)+9(y)+(-1)(z-9)=0 The normal vector, , is <6,9,-1>. Because we know that this is the normal vector at the point P(3,0,9), we can find parametric equations for the normal line using the equations: x=x_0-at y=y_0-bt z=z_0-ct plugging the known numbers into these 3 equations gives us the paremetric equations: x=3-6t y=9t z=9-t Sent by Ron Berkowitz on Sun, 9 Mar 2003 18:37:33

 3.The surfaces x^3+5y^2z-4z^2=3 and y^4-7z-2x^2=5 intersect in a curve, C. The point p=(3,2,-1) is on C. Find parametric equations for a line tangent to C at p. Let's call this surface A: x^3+5zy^2-4z^2=3 and this one B: y^4-7z-2x^2 =5. It is very difficult to find out the exact equation for where A and B intersect, but it is not necessary. Your original intuition could follow that this line tangent to C (lets call it T) is a line that is tangent to both A and B. Therefore, this line is going to reside in the tangent planes to A and B, and go through the point (3,2,-1). Unless both tangent planes are equal, there will be 1 distinct line through tangent planes to A and B at (3,2,-1). We start by finding each tangent plane. The book says that a tangent plane to a curve S through (X_0,Y_0,Z_0) is Z-Z_0=(dz/dx)(X-X_0)+(dz/dy)(Y-Y_0). A: (we must do implicit differentiation of x^3+5zy^2-4z^2=3) dz/dx: 3x^2+5y^2(dz/dx)-8z(dz/dx)=0 so dz/dx=3x^2)/(8z-5y^2) and at (3,2,-1), dz/dx=-27/28. dz/dy: 10yz + 5y^2(dz/dy)-8z(dz/dy)=0 so dz/dy=(10yz)/(8z-5y^2) and at (3,2,-1), dz/dy=20/28. B: (regular partial differentiation from y^4-7z-2x^2=5) So dz/dx=(-4/7)x and at (3,2,-1), dz/dx=-12/7. And dz/dy=(2/7)y and at (3,2,-1), dz/dy=4/7. The normal vector for the tangent plane to A at (3,2,-1) is <-(27/28),(20/28),-1>. The normal vector for the tangent plane to B at (3,2,-1) is <-(12/7),(4/7),-1>. Well all lines in the tangent plane are perpendicular to the normal vector, so our tangent line T must be perpendicular to both normal vectors of the planes A and B. The cross product of these normal vectors will be a vector in the direction of T. The cross product is <9/7,3/4,33/49>. We can use <3/7,1/4,11/49> (just divided by common factor of 3 which does not affect the direction). Now we have for the tangent line x=X_0+at, y=Y_0+bt,z=Z_0+ct (as described in the book). The answer(I hope I didn't make any mistakes): x=3+(3/7)t y=2+(1/4)t z=-1+(11/49)t Sent by Matt Brown on Fri, 7 Mar 2003 04:03:57 Message from the management I might have just taken the gradient of the left-hand sides of each of the equations x^3+5y^2z-4z^2=3 and y^4-7z-2x^2=5, remembering that gradient vectors are perpendicular to level "sets" (curves in two dimensions, surfaces in three dimensions, etc.). Then I would "plug in" (3,2,-1). That way I have to think and remember less. You can try these Maple commands:```with(plots): F:=(x,y,z)->x^3+5*y^2*z-4*z^2; G:=(x,y,z)->y^4-7*z-2*x^2; RED:=implicitplot3d(F(x,y,z)=3,x=1..5,y=0..4,z=-3..1,axes=normal,color=red,grid=[20,20,20]): BLUE:=implicitplot3d(G(x,y,z)=5,x=1..5,y=0..4,z=-3..1,axes=normal,color=blue,grid=[20,20,20]): display3d({RED,BLUE});``` A view of what is produced is included here, so you perhaps can "see" the two surfaces and the intersection curve. It is remarkable to me that the tangent line can be described so easily. The curve itself is difficult to describe algebraically.

 5.Consider the function f(x,y,z)=x-2y+5z if (x,y,z) is not (0,0,0) and 10 if (x,y,z)=(0,0,0). a) Use the definition of continuity to verify that f is continuous if (x,y,z)=(2,6,-7). b) Use the definition of continuity to verify that f is not continuous if (x,y,z)=(0,0,0). The point that we are asked to find whether the function is continuous or not lies on the part of the function where f(x,y,z)=x-2y+5z. To verify that the function was continuous at (2,6,-7), we need to check the following statement: Given K>0, there is H>0 so that if sqrt((x-2)2+(y-6)2 +(z+7)2) This satisfies the condition, and therefore when (x,y,z) doesn't equal (0,0,0), the function is continuous. Now we must prove that the function is discontinuous when (x,y,z)=(0,0,0). Here P2=(a,b,c)=(0,0,0) and f(P2)=10. For (x,y,z) near (0,0,0) but not equal to (0,0,0), f(x,y,z)=x-2y+5z. This is very small. So let's take K=4. I claim that no H>0 will work. For example, if H=1/(10,000), then (x,y,z)=(1/(20,000),0,0) is close enough to (0,0,0), and f at that (x,y,z) is 1/(20,000). |1/(20,000)-10|>4. If we try any H>0, then take (x,y,z)=(*,0,0) where * is the minimum of 1/(20,000) and H/2. (H could be smaller than 1/(20,000): we need to consider this possibility). Then |f at this (x,y,z)-f(0,0,0)|>4, so f is not continuous at (0,0,0). (The management helped with this case a bit.) Sent by Emir Elkholy on Tue, 11 Mar 2003 11:51:11

 9. Suppose S is the surface defined by the equation z cos(x^3+y^2)+xz^2+2=0. The point (-1,1,2) is on S. Assume that the equation defines x implicitly as a differentiable function of y and z near (-1,1,2). If y is changed from 1 to 1.03 and z is changed from 2 to 1.96, use linear approximation to find an approximate x so that (x,1.03,1.96) is on S. Note You are not asked to solve a nonlinear equation. You are asked to use a certain approximation strategy to get an answer. In this problem x is written implicitly as a function x=f(y,z) in the form F(x,y,z)=0 with z*cos(x^3+y^2)+x*z^2+2=0. By Taylor's Theorem: x+w=f(y+h,z+k)=f(y,z)+f_y(y,z)*h+f_z(y,z)*k+Error, where f(y,z)+f_y(y,z)*h+f_z(y,z)*k is called linear approximation. By implicit differentiation: x_y=-F_y/F_x so x_y=-(z*sin(x^3+y^2)*2y)/(z*sin(x^3+y^2)*3x^2+z^2) and x_y(-1,1,2)=0 x_z=-F_z/F_x so x_z=(cos(x^3+y^2)+2xz)/(z*sin(x^3+y^2)*3x^2+z^2) and x_z(-1,1,2) = -3/4. Returning to the linear approximation: f(1,2)+f_y(1,2)*h+f_z(1,2)*k, where h=-.03 and k=.04. f(1,2)=-1 f_y(1,2)=x_y(-1,1,2)=0 f_z(1,2)*k=x_z(-1,1,2)*(-.04)=-(3/4)*.04=-.03 Hence x+w=-1-.03=-1.03 where (-1.03,1.03,1.96) is the approximate point on S. Sent by Max Kleyzit on Sun, 09 Mar 2003 12:15:41 Message from the management The Maple command fsolve allows approximation of roots to equations. The reported root is -1.030533773: the ``higher order'' effects are not felt until the 4th digit!

 10. Find the area of the triangle whose vertices are the points A=(-1,2,1) and B=(2,-1,0) and C=(0,0,4) in R^3. Also find the coordinates of a point D which together with A and B and C will form the vertices of a parallelogram. The area is one half of the square root of two hundred thirty The points that would make the triangle a parallelogram are (1,1,-3),(3,-3,3)and(-3,3,5). The area of the triangle is half of the length of the cross product of two vectors that define the triangle. To find the vectors, subtract the x, y, and z coordinates of a two of the points. Make sure one is a reference point, EXAMPLE use A-B, and A-C to get vectors, calculate their cross product, use the Pythagorean Theorem to find the length of the vector and divide by 2. We divide by 2 because the length of the cross product gives the area of a parallelogram, and that parallelogram has twice the area of the triangle. To find another point that would make the triangle a parallelogram, we use the parallelogram method of adding vectors. EXAMPLE Take the point A as a reference point. Subtract the coordinates of point B from A to find a vector that connects A to B. Do the same for A and C. Add the components of these vectors to get a resultant. Add the resultant to the point A, and you will have a new point that will cause the triangle to be a parallelogram. Sent by William McGowan on Thu, 6 Mar 2003 10:45:24 Message from the management For those who may not believe that there can be more than one answer to the second part of the problem, here is a possible picture of the situation, drawn in the plane containing the points A and B and C. Only one possible solution (one of the D's) is requested in the problem.

 12. Find the point (x,y) in the plane for which the sum of the squares of the distances from the three points (a_i,b_i), i=1,2,3, is a minimum. Give a physical interpretation of your answer.    Source An MIT calculus exam. The distance between each point (a_i, b_i) and (x,y) is given by: sqrt( (x-a_i)^2 + (y-b_i)^2) ). So the sum of all three distances squared is: f(x,y)=(x-a_1)^2+(y-b_1)^2+(x-a_2)^2+(y-b_2)^2+(x-a_3)^2+(y-b_3)^2. Differentiatng, with respect to x we get: 2(x-a_1)+2(x-a_2)+2(x-a_3)=0 and with respect to y: 2(y-b_1)+2(y-b_2)+2(y-b_3)=0 We can divide the x equation by 2 and then combine the x's, so we end up with: 3x-a_1-a_2-a_3=0 and x=(a_1+a_2+a_3)/3. Similarly, y=(b_1+b_2+b_3)/3 What we found to be the point (x,y) is the centroid of a triangle with vertices (a_1,b_1), (a_2,b_2), (a_3,b_3). The centroid can be found by constructing 3 medians (connecting a midpoint of a side to the opposite vertex). The centroid is known to be the center of gravity of a triangle i.e. you can balance the triangle on that point. Sent by Anastasia Pashkova on Tue, 11 Mar 2003 13:00:15

 13. Find and classify all the critical points of these functions. a) K(x,y)=3x^2+6xy+12y^3+12x-24y.    Source An MIT calculus exam. b) J(x,y)=8xy-x^4-y^4. K(x,y)=3x^2+6xy+12y^3+12x-24y so dK/dx=6x+6y+12 and dK/dy=6x+36y^2-24. We want to find x and y when both partial derivatives equal zero. First subtract one equation from the other to eliminate the x term. Then solve for y. 6y-36y^2+36=0 and then y=-.920 or 1.087. Plugging these values into one of the original partial equations, x = -1.08 and -.913, respectively. Using the second derivative test, H=(d^2K/dx^2)(d^2K/dy^2)-(d^2K/dxdy)^2, we can determine the nature of these critical points. d^2K/dx^2=6 and d^2K/dy^2=72y and d^2K/dxdy=6 so H=6*72y-36. When y=1.087, H>0, and d^2K/dx^2 is always greater than zero, so the function is at a local minimum. When y=-.920, H<0, so the function is neither a maximum or a minimum at this point: it has a saddle point. We can use the same method in this part as we did in Part a to find the critical points. dJ/dx=8y-4x^3 and dJ/dy=8x-4y^3. d^2J/dx^2=-12x^2 and d^2J/dxdy=8 and d^2J/dy^2=-12y^2. y=(x^3)/2 and then 8x=4((x^3)/2)^3=x^9/2 so x=0,+/-sqrt(2) and y=0,+/-sqrt(2) respectively. H=[(-12x^2)*(-12y^2)]-8 At (0,0), the critical point is a saddle point since H<0. At (+/-sqrt(2),+/-sqrt(2)), the critical points are local maxima since H>0 and d^2J/dx^2=-24<0. Sent by Elizabeth Tozour on Thu, 13 Mar 2003 00:50:40

 15. Sketch z=x^2+y in R^3. The normal line to the surface at a point is the line perpendicular to the tangent plane to the surface at that point. Find all normal lines to z=x^2+y which pass through (0,0,3). I will rewrite the equation as an implicit function of three variables, F(x,y,z)=x^2+y-z=0. The normal vector N to a tangent plane at P=(a,b,c) equals grad F(P)=. D_1 F(P)=2a, D_2 F(P)=1, D_3 F(P)=-1. The normal line is in the direction of N and contains the point P, so its parametric equations could be x=a+2a*t=a*(1+2t)=0 y=b+1*t=b+t=0 z=c+-1*t=c-t=3 Since this line must pass through the point (0,0,3), we can set x=0, y=0, and z=3. We solve the first equation assuming that a is not 0; 1+2t=0, t=-1/2. Therefore, b=-t=1/2, and c=3+t=5/2. P is on the surface of F, so we can solve the function for a. a^2+b-c=0, a^2=c-b=2, a =+/-sqrt(2). Plugging these values of a, b, and c back into the equations for the normal line, we get x=+/-sqrt(2)(1+2*t) y=1/2+t z=5/2-t We must also solve if a=0, in which case x=0 always, and -b=c-3 (solve y and z for t and set them equal). We then plug a, b, and c back into the function equation again and find 0^2+(3-c)-c=0 or 3=2c, c=3/2. Therefore, b=3-c=3/2. Plug these values into the parametric representation of the normal line. x=0 y=3/2+t z=3/2-t Each of the above mentioned lines must be normal to a tangent plane of z=x^2+y (which is represented (grad F)·r=0, where r is the position vector), and passes through (0,0,3). Sent by Neil Wilson on Sun, 9 Mar 2003 22:55:56 Message from the management I think this is a difficult problem. To the right is an approximate sketch of the situation. To the left is the output of a Maple command Mr. Wilson suggested: plot3d(x^2+y,x=-5..5,y=-5..5,axes=normal);