299ReviewProblemSolutions.nb 1. Review Problems. Final Exam: Wednesday, 12/16/2009 1:30PM. Mathematica 6.0 Initializations

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1 99ReviewProblemSolutions.nb Review Problems Final Exam: Wednesday, /6/009 :30PM Mathematica 6.0 Initializations R.) Put = t - and y@td = t -. Sketch on the axes below the curve traced out by 8x@tD, y@td< as t advances from - to.

2 99ReviewProblemSolutions.nb y x - - Out[7895]= -3 y x R.) Come up with the specific values of the coordinates 8x, y< of the highest point on the curve parameterized by 8x@tD, y@td< = 9t e 3 t + 5, t - t = with - < t <. At the highest point on the curve, y'[t]=0. So we find y'[t]: y'[t]=-t. y'[t]=0 when t=. y''[t]=-<0, so we have found the max. Thus the max is at {x[],y[]} = {e 3 + 5, }.

3 99ReviewProblemSolutions.nb 3 R.3) Put 8x@r, td, y@r, td, z@r, td< = 8, -3, < + 84 r Cos@tD, 0, r Sin@tD<. Describe what you get when you plot 8x@r, td, y@r, td, z@r, td< for 0 r and 0 t p. Is this a curve or a surface? This is a surface. It is an elliptical disk centered at {,-3,} and parallel to the xz-plane. The disk stretches 4 units parallel to the x-axis and unit parallel to the z-axis. Put 8x@tD, y@td, z@td< = 8, -3, < + 84 Cos@tD, 0, Sin@tD<. Describe what you get when you plot 8x@tD, y@td, z@td< for 0 t p. Is this a curve or a surface? What relation does it have to what you said immediately above? This is a curve. It is an ellipse centered at {,-3,} and parallel to the xzplane. The ellipse stretches 4 units parallel to the x-axis and unit parallel to the z-axis. It is the boundary curve of the above surface. R.4) Write down the parametric formulas that describe the ellipse: I x 5 M + I y- M =. Don't forget to include the range of t. Is your ellipse traced out clockwise or counterclockwise? Parametric formula: {x[t], y[t]} = {5 Cos[t], Sin[t] + }, for 0 t p. For this parameterization, the ellipse is traced out counterclockwise. R.5) Calculate Ÿ 0 e -x x We take the limit as t goes to infinity of the integral: Ÿ0 t e -x x = -e -t + The limit as t goes to infinity of -e -t + is. Thus Ÿ0 e -x x=. R.6) Why would it be a mark of calculus illiteracy to write down Ÿ - x? x because the function has an asymptote at x=0. x

4 99ReviewProblemSolutions.nb 4 because the function x has an asymptote at x=0. R.7) Here is the plot of a certain function g@xd: g@xd x -4 Explain how the plot of g@xd signals that the plot of f@td = Ÿ 0 t g@xd x goes down as t advances from 0 to ; goes up as t advances from to 3; and goes down as t advances from 3 to 4. We know that f[t]=g[t]-g[0], and thus f'[t]=g'[t]-0=g[t]. Thus, since g[t] is negative between 0 and and between 3 and 4, we know that f[t] must decrease on those intervals. Since g[t] is positive between and 3, f[t] must be increasing on that interval. R.8) Here is the plot of a certain function g@xd: g@xd x Here are the plots of three other functions:

5 99ReviewProblemSolutions.nb t One of the plotted curves is the actual plot of Ÿ t g@xd x where g@xd is the function plotted immediately above and the other two are bogus. Pick out the genuine plot of f@td and say how you came to your decision. Since g[x] is always positive, we know that we are looking for a function which is always increasing. this rules out the bottom line. However, we also see that g[x] is close to 0 near x=.5, so we are looking for a function which is nearly flat near.5. So the middle function is the right f[t]. R.9) Measure the area that is under the curve y = x and is over the curve y = x Calculate: Ÿ 0 x - x x = (/)-(/3) = (/6). The limits are from 0 to since x=x when x=0,. R.0) Calculate the following integrals without the help of any machine. a) Ÿ 0 pê e Sin@xD Cos@xD x b) Ÿ t x x c) Ÿ t x x d) Ÿ t -3 Sin@xD Cos@xD+5 x e) Ÿ 0 5 e -xê x a) I'd make the substitution u=sin[x], du=cos[x]dx, then calculate: Ÿ 0 e u u = E-. b) Log[t]-Log[] = Log[t]. c) -(/t)+ d) I'd make the substitution u=cos[x]+5, du=-sin[x]dx, then calculate: Cos@tD+5 3

6 99ReviewProblemSolutions.nb 6 a) I'd make the substitution u=sin[x], du=cos[x]dx, then calculate: Ÿ 0 e u u = E-. b) Log[t]-Log[] = Log[t]. c) -(/t)+ d) I'd make the substitution u=cos[x]+5, du=-sin[x]dx, then calculate: Cos@tD+5 3 Ÿ Cos@D+5 u u = 3Log[Cos[t]+5]-3Log[Cos[]+5]. e) - e -5ê +. R.) All the linear dimensions of a certain solid are doubled. What is the ratio of the new volume to the original volume? What is the ratio of the new surface area to the original surface area? The new volume is 8 times the old volume. The new surface area is 4 times as large as the original surface area. R.) The base of a certain solid is the circle x + y = 4 in the xy-plane. Each cross section cut by a plane perpendicular to the x-axis is a square. Measure the volume of the solid. I'll take slices perpendicular to the x-axis. Since the base is a circle, at a particular value of x, the length of the baseline is Sqrt[4-x ]. So a square cross-section has area 4(4-x ). Thus we integrate: Ÿ - Area@xD x = Ÿ - 4 I4 - x M x = 8 3. R.3) Measure the volume of the solid obtained by rotating the curve y=x around the x-axis for x between 0 and 4. Measure the volume of the solid obtained by rotating the curve y=x around the y-axis for x between 0 and 4. In the first case, we are integrating the area of circular cross-sections with radius = y. So we get: 4 Ÿ 0 p IIx M^M x = (p/5)(4^5) = 04 p. 5

7 99ReviewProblemSolutions.nb 7 R.4) Explain how to transform the integal 5 Ÿ 0 CosAx E x x into Ÿ 5 0 Cos@uD u Use the second integral to calculate the first. I'd make the substitution u=x, du=xdx. So xdx = du. Also, the limits of integration change from 0 to 0^ =0, and from 5 to 5^=5. Let's calculate the simpler integral. Ÿ 5 0 Cos@uD u = Sin[5]. R.5) Here are three integrals: Ÿ 0 e -x 3 3 x x, Ÿ 0 e -x x x, and Ÿ 0 e -x x Use transformations to explain why all these integrals calculate out to the same value. In the first integral, make the substitution u=x 3, du=3 x x. Thus the integral becomes: Ÿ 0 e -u u. In the second integral, make the substitution u=x, du= x x. Thus the integral becomes: Ÿ 0 e -u u. Thus all the integrals are equivalent with a substitution and I am not surprised to hear that they all calculate out to the same value. R.6) Here's a closed curve in true scale:

8 99ReviewProblemSolutions.nb This curve is traced out by the parametric formula 8x@tD, y@td< = 86 t H3 - tl, t H3 - tl Ht + 4L< = 98 t - 6 t, t - t - t 3 = as t advances from 0 to 3. No part of the curve is traced out more than once. Use an integral to measure the area enclosed by this curve. Determine whether the curve is traced out in the clockwise or the counterclockwise fashion as t advances from 0 to 3. at t= we are at (, 0) and at t= we are at (, ), so it looks like the curve is traced out conterclockwise. So we do the integration: Ÿ 0 3 -y@td x '@td t = Ÿ 0 3 -I t - t - t 3 M H8 - tl t = 4.3 So the area in the encosed curve is 4.3 units squared. R.7) Calculate the following by integration by parts: a) Ÿ e Log@xD x b) Ÿ 0 x Sin@xD x c) Ÿ x Log@xD x (a) The integration by parts formula is b Ÿ a u@xd v'@xd x = u@xd v@xd b b a -Ÿ a v@xd u'@xd x. Make the assignments: u@xd = Log@xD and v'@xd = dx. This gives u'@xd = HêxL dx and v@xd = x. With these assignments,

9 (a) The integration by parts formula is 99ReviewProblemSolutions.nb b Ÿ a u@xd v'@xd x = u@xd v@xd b b a -Ÿ a v@xd u'@xd x. 9 Make the assignments: u@xd = Log@xD and v'@xd = dx. This gives u'@xd = HêxL dx and v@xd = x. With these assignments, Ÿ E Log@xD x = Ÿ E u@xd v'@xd x = u@xd v@xd E -Ÿ E v@xd u'@xd x = x Log@xD E -Ÿ E x HêxL x = x Log@xD E -x E = HE - 0L - HE - L =. (b) The integration by parts formula is Ÿ a b u@xd v'@xd x = u@xd v@xd a b -Ÿ a b v@xd u'@xd x. Make the assignments: u@xd = x and v'@xd = Sin@ xd dx. This gives u'@xd = dx and v@xd = -Cos@ xd. With these assignments, Ÿ 0 x Sin@ xd x = Ÿ 0 u@xd v'@xd x = u@xd v@xd 0 -Ÿ 0 v@xd u'@xd x = -x Cos@ xd 0 +Ÿ 0 Cos@ xd x = -x Cos@ xd 0 +Sin@ xd 0 = (-Cos[]-0)+(Sin[]-0) = Sin[]-Cos[]. (c) The integration by parts formula is b Ÿ a u@xd v'@xd x = u@xd v@xd b b a -Ÿ a v@xd u'@xd x. Make the assignments: u@xd = Log@xD and v'@xd = xdx. This gives u'@xd = HêxL dx and v@xd = HêL x^.

10 = Sin[]-Cos[]. 99ReviewProblemSolutions.nb (c) The integration by parts formula is 0 Ÿ a b u@xd v'@xd x = u@xd v@xd a b -Ÿ a b v@xd u'@xd x. Make the assignments: u@xd = Log@xD and v'@xd = xdx. This gives u'@xd = HêxL dx and v@xd = HêL x^. With these assignments, Ÿ x Log@ xd x = Ÿ 0 u@xd v'@xd x = u@xd v@xd 0 -Ÿ 0 v@xd u'@xd x = 0.5 Hx^L Log@ xd -Ÿ 0.5 x x = (Log[]-0)-(-0.5) = Log[] R.8) Here is the surface z = f@x, yd = Sin@x + yd +.5 plotted over the rectangle R in the xy-plane consisting of the points 8x, y, 0< with - x and -3 y 3: Measure by hand calculation the volume of the solid whose top skin is the plotted surface and whose bottom skin is the rectangle R. We calculate the following double integral: 3 f@x, yd y x Ÿ -Ÿ-3 =Ÿ -Ÿ-3 3 Sin@x + yd +.5 y x =Ÿ - H- Cos@x + 3D + Cos@x - 3D + 5L x =-Sin[5]+Sin[-]+30-(-Sin[]+Sin[-5]-30) = 60.

11 We calculate the following double integral: 3 99ReviewProblemSolutions.nb Ÿ -Ÿ-3f@x, yd y x 3 Sin@x + yd +.5 y x =Ÿ -Ÿ-3 =Ÿ - H- Cos@x + 3D + Cos@x - 3D + 5L x =-Sin[5]+Sin[-]+30-(-Sin[]+Sin[-5]-30) = 60. R.9) Here is the surface z = f@x, yd = x y + 3 plotted above the region R inside the ellipse I x M + I y 3 M = in the xy-plane. Come up with a counterclockwise parameterization 8x@tD, y@td< of the ellipse. Use your parameterization and the Gauss-Green formula to measure by hand calculation the volume of the solid whose top skin is the plotted surface and whose bottom skin is the rectangle R. You can use that the antiderivative of Cos@tD is t + Sin@ td+c, if you 4 like. Let {x[t],y[t]}={cos[t], 3Sin[t]}, which is a counterclockwise parameterization. x x Let n[x,y]=ÿ 0 f@s, yd s = y + 3 x. Then, by Gauss-Green, we calculate: Ÿ 0 p n@x@td, y@tdd y'@td t =Ÿ 0 p I8 Cos@tD 3 Sin@tD + 8 Cos@tD M t

12 Let {x[t],y[t]}={cos[t], 3Sin[t]}, which is a counterclockwise 99ReviewProblemSolutions.nb parameterization. x x Let n[x,y]=ÿ 0 f@s, yd s = y + 3 x. Then, by Gauss-Green, we calculate: Ÿ 0 p n@x@td, y@tdd y'@td t =Ÿ 0 p I8 Cos@tD 3 Sin@tD + 8 Cos@tD M t = 8p. (I used a substitution of u=cos[x] on the first part of the integral, and the hint formula on the second part.) R.0) Here is a D region R whose left boundary is the curve left@yd = y Hy - 6L + and whose right boundary is the curve right@yd = -y + : Out[630]= 5 4 y R Calculate Ÿ Ÿ R H3 + yl x y by hand. We slice the region horizontally, thus we integrate with respect to x first. 5 -y Ÿ 0 Ÿy Hy-6L+ H3 + yl x y =Ÿ 0 5 H3 x + yxl y =Ÿ 0 5 I-y 3 + y + 5 ym y = 375 x

13 99ReviewProblemSolutions.nb 3 R.) Here is a D region R with bottom boundary curve low@xd = x Hx - L and with top boundary curve high@xd = x H - xl:

14 99ReviewProblemSolutions.nb y R y = x H - xl x y = x H x - L Calculate Ÿ Ÿ R x y by hand and say what Ÿ Ÿ R x y measures. We slice the region vertically, thus we integrate with respect to y first. x H-xL y x Ÿ 0 Ÿx Hx-L =Ÿ 0 y x =Ÿ 0 I3 x - 3 x M y = 0.5 This is the area of the region R. R.) When you plot a certain 3D vector X with its tail at 80, 0, 0<, it turns out that the tip of X is sitting on 89, -6, 4<. When you plot the same vector with its tail at 8-, 7, 5<, where will the tip of X be? The tip should be at {9,-6,4}+{-,7,5}={7,,9}. R.3) Give the number t that makes X = 8, -4, < and Y = 8-,, t< perpendicular. We need t such that X.Y=0. Well X.Y=--4+t=-6+t. Thus X.Y=0 when t=3.

15 99ReviewProblemSolutions.nb 5 R.4) Here are parametric formulas for two 3D lines: = 85,, -< + t 8-,, <, and = 85,, -< + t 8, 3, -<. Are they perpendicular? How can you tell? We know they go through the same point, so we just take the dot product of the direction vectors. {-,,}.{,3,-}=-+3-=0. So yes, they are perpendicular. R.5) Given X={,3} and Y={0.5, 0.5}. Calculate the push of X in the direction of Y. X.Y=, Y.Y=0.5, so (X.Y)/(Y.Y)=4 Thus the push is 4Y={,}. R.6) If X.Y > 0, then is the push of X in the direction of Y with Y or against Y? WITH Y If X.Y < 0, then is the push of X in the direction of Y with Y or against Y? AGAINST Y R.7) Write a parametric formula for the line that passes through the points 83, < and 85, <. Give a vector parallel to this line. Give a vector perpendicular to this line. A parametric formula: line[t]={3,}+t{,0}. A vector parallel to the line is {,0}. A vector perpendicular to the line is {-0,}.

16 99ReviewProblemSolutions.nb 6 R.8) Are the lines with parametric formulas = 8, 3< + t 8-, 4<, and = 8, 3< + t 8-4, 8< the same line or different lines? Are the lines with parametric formulas = 8, 3< + t 8-3, 5<, and = 8-4, 3< + t 8-3, 5< the same line or different lines? The first set is the same since they go through the same point and their direction vectors are parallel. The second set is also the same. The two lines are clearly parallel, so we check that they go through the same points. Both go through the point (-,8) with t= for the first line and t=- for the second line. R.9) Give a parametric formula for the 3D-line through the points 8, 5, 0< and 84,, 3<. line[t]={,5,0}+t{,-4,3}, for example. R.30) Here are xyz-equations for two planes: x + 3 y - z = 9, and - x + y + z = 3. Say how you can tell that these planes cut each other at right angles. We can see that their normals are perpendicular, so the planes must also be perpendicular. {,3,-}.{-,,}=0. R.3) Given X = 8,, < and Y = 8, 0, 0<, calculate X.Y and X äy. X.Y=. XäY={0,,-} R.3) Come up with a unit vector that points in the same direction as the vector X = 8, 3, 0<. X.X=4+9=3, so a unit vector is: {, 3, 0}. 3 3

17 99ReviewProblemSolutions.nb 7 X.X=4+9=3, so a unit vector is: { 3, 3 3, 0}. R.33) Using X = 8,, < and Y = 8, 0, 0<, as in problem R.3, and the point (,,3). Find an xyz-equation for the plane described by this point and vectors X and Y. Find a parametric equation for the same plane. By R.3, we know that the normal to the plane is given by: XäY={0,,-} So we write: 0(x-)+(y-)-(z-3)=0. R.34) A plane has xyz-equation x - 4 y + z = 6. Give a vector perpendicular to the plane, and give a point on the plane. The vector {,-4,} is perpendicular to the plane. The point {,0,4} is on the plane, for example. R.35) Given that X and Y are perpendicular unit vectors in 3D, describe the curve traced out by = Cos@tD X + 4 Sin@tD Y as t runs from 0 to p. This is an ellipse. The ellipse is centered at the origin, and is stretched units in the direction of vector X and 4 units in the direction of vector Y. It lies in the plane defined by the origin, X and Y. R.36) Here is a curve parameterized by a certain formula = 8x@tD, y@td< for 0 t p:

18 99ReviewProblemSolutions.nb 8 y x As t advances from 0 to p, you are given that moves along this curve in the counterclockwise direction. Pencil in some of the unit normals + with their tails at 8x@tD, y@td<. The normals will point OUT of the curve (since we have a counterclockwise parameterization), perpendicular to the curve, at each point on the curve. R.37) Calculate the gradient, f@x, yd = gradf@x, yd, of f@x, yd = e -3 x y. Calculate the gradient, f@x, y, zd = gradf@x, y, zd, of f@x, y, zd = x Sin@x y zd. gradf@x, yd = 9-3 ye -3 x y, -3 xe -3 x y = gradf@x, y, zd = 9Sin@xyzD + xyzcos@xyzd, x zcos@xyzd, x ycos@xyzd= R.38) You are at the point 8, 0<. In the direction of what vector should you step off 8, 0< in order to get the greatest initial increase in the function f@x, yd = x + 7 x e y? In the direction of what vector should you step off 8, 0< in order to get the greatest initial decrease in the function f@x, yd = x + 7 x e y? gradf[x, y] = {x+7 e y, 7 x e y } gradf[, 0] = {9,7}, thus we should leave {,0} in the direction {9,7} to get the greatest possible initial increase of f[x, y]. We should leave {, 0} in Printed the by direction Mathematica for {-9.-7} Students to get the greatest possible initial decrease of f[x, y].

19 99ReviewProblemSolutions.nb gradf[x, y] = {x+7 e y, 7 x e y 9 } gradf[, 0] = {9,7}, thus we should leave {,0} in the direction {9,7} to get the greatest possible initial increase of f[x, y]. We should leave {, 0} in the direction {-9.-7} to get the greatest possible initial decrease of f[x, y]. R.39) Go with f@x, yd = x +.4 y, and look at a plot of gradf@x, yd with tails at 8x, y< for some selected points 8x, y< in the vicinity of 80, 0<:.0 y x Why are these gradient vectors so attracted to the point 80, 0<? We can see that {0,0} minimizes the denominator of f[x,y], so it maximizes the function. The gradient vectors point in the direction of greatest increase for the function, so they are attracted to the max. R.40) The point 8, < is on the level curve x y 3 e x + y = : Here's a plot of part of this level curve with axes going through the point 8, <:

20 99ReviewProblemSolutions.nb 0 Pencil in a vector with its tail at 8, < that points in the same direction as gradf@, D. gradf[x, y] = {( x y 3 + x y 3 ) e x + y, (3 x y + x y 3 ) e x + y } Thus, gradf[, ] has both entries positive (although you were probably able to tell this without calculating gradf out). The gradient vector will be perpendicular to the curve (as they are always perpendicular to level curves) and will point up and to the right. Any vector meeting this description is fine here. R.4) Here is the level surface f@x, y, zd = x + I y M + z = together with a selection of gradient vectors of f@x, y, zd with tails at the points at which they are calculated:

21 99ReviewProblemSolutions.nb Explain: Ø Why you are not surprised by the fact that the plotted gradient vectors are perpendicular to the level surface. Ø Why these gradient vectors point out away from the surface and not in toward the interior of the surface. I am not surprised because gradient vectors are always perpendicular to level curves and surfaces. They point in the direction of greatest initial increase, so they point outwards here (f[x,y,z] is larger for larger x, y, and z ). R.4) Does f@x, yd = x 6-3 x y + y 4-6 x + 5 y have a maximizer or minimizer? How do you know? When x and y are large then the dominant terms x 6 + y 4 make f@x, yd really huge, so the global scale plot of f@x, yd looks like a cup. This means f@x, yd has no tallest crest but does have a deepest dip. As a result, f@x, yd has no maximum value but does have a minimum value. Does f@x, yd = e -x ë - y ë Ix y 3 + 4M have a maximizer or a minimizer? How do you know? When x and y are large then the exponential function takes over and makes f@x, yd really close to 0, so the global scale plot of f@x, yd looks flat to the xy-plane, except for a few interesting bits close to the origin. Since f[x,0] is positive, and f[0,y] is negative for some negative y's, we know that f[x,y] has both a maximizer and a minimizer.

22 99ReviewProblemSolutions.nb When x and y are large then the exponential function takes over and makes f@x, yd really close to 0, so the global scale plot of f@x, yd looks flat to the xy-plane, except for a few interesting bits close to the origin. Since f[x,0] is positive, and f[0,y] is negative for some negative y's, we know that f[x,y] has both a maximizer and a minimizer. Does f@x, yd = x5 + 3 y + x 4 + y 6 have a maximizer or a minimizer? How do you know? We know f@x, 0D > 0 for x > 0 and f@x, 0D < 0 for x < 0. Since for large positive x's f[x,0] runs off to infinity, and for large negative x's, f[x,0] runs off to minus infinity, there is neither a maximizer nor a minimizer. R.43) Here's a plot of a certain vector field shown with the plots of two curves:

23 99ReviewProblemSolutions.nb 3 One of these curves is a genuine trajectory in the vector field. The other is bogus. Identify the trajectory and discuss the information that led to your choice. The one on the right is a genuine trajectory since it is clearly tangent to the vectors in the vector field. R.44) You are given that the point 8, < is on a certain trajectory in the vector field Field@x, yd = 8x + y, x - y<. Write down a vector that is tangent to this trajectory at the point 8, <. {3,} is tangent to the trajectory at {,}. R.45) Here is a plot of the gradient field of a certain function near its maximizer, which is plotted as a heavy dot:

24 99ReviewProblemSolutions.nb 4.0 y x Pencil in a couple of trajectories. Where are the trajectories in this gradient field trying to go? Why are they trying to go there? The trajectories are trying to go to the maximum. Gradient vectors point in the direction of greatest initial increase, so we expect them to point towards the maximum. You should draw some curves that start at any point and are tangent to the gradient field vectors. They should move generally towards the max. R.46) Suppose f@x 0, y 0 D > f@x, yd for all other 8x, y< near 8x 0, y 0 <. If you center a small circle C at 8x 0, y 0 <, why do you expect that the net flow of gradf@x, yd across C is from outside to inside? Suppose f@x 0, y 0 D < f@x, yd for all other 8x, y< near 8x 0, y 0 <. If you center a small circle C at 8x 0, y 0 <, why do you expect that the net flow of gradf@x, yd across C is from inside to outside? In the first case, I expect the net flow to be from outside to inside since gradient vectors point in the direction of greatest initial increase - which means they will point generally towards the max. In the second case I expect the net flow to be from inside to outside since gradient vectors will point away from the minimum.