MAT 127: Calculus C, Fall 2010 Solutions to Midterm I

Transcription

1 MAT 7: Calculus C, Fall 00 Solutions to Midterm I Problem (0pts) Consider the four differential equations for = (): (a) = ( + ) (b) = ( + ) (c) = e + (d) = e. Each of the four diagrams below shows a solution curve for one of these equations: I II III IV Match each of the diagrams to the corresponding differential equation (the match is one-to-one): diagram I II III IV equation b c a d Answer Onl: no eplanation is required. Eplanation is on the net page.

2 Of the four diagrams, the most distinctive is I; it shows the graph of the constant function ()=0. Plugging in this function into the four equations, we get (a) 0? = ( + 0 ) (b) 0? = 0( + ) (c) 0? = e +0 (d) 0? = e 0. Of these four potential equalities, onl (b) is satisfied for all ; so diagram I must correspond to (b). Of the three remaining graphs, the most distinctive is III; the slope of the graph at (0, 0) there is 0. This is the case for the slope of equation (a) at (0,0) (because 0(+0 )=0), but the slopes for the other two remaining equations are alwas positive (because e is alwas positive). So diagram III must correspond to (a). The two remaining graphs look rather similar. One distinguishing feature is that the graph in II becomes steeper as, increase, while the graph in IV less steep as, increase. Considering the two remaining equations, e + becomes larger as, increase (thus making the slope of the graph steeper), while e becomes smaller as, increase (thus making the slope of the graph less steep). So, diagram II must correspond to (c), while diagram IV must correspond to (d). Alternativel, diagram II depicts the graph of a function with > 0, while diagram IV depicts the graph of a function with <0. If =() is an function satisfing equation (c), then b the Chain Rule = ( ) = ( e +) = e + ( + ) = e + ( + ) = e + ( + e + ) > 0, because e + > 0. On the other hand, if = () is an function satisfing equation (d), then b the Chain Rule = ( ) = ( e ) = e ( ) = e ( + ) = e ( + e ) < 0, because e >0. So, II must correspond to (c), while IV must correspond to (d). Grading: correct repeats 0-4 points

3 Problem (5pts) (a; 7pts) Show that the function () = e is a solution to the initial-value problem = 0, = (), (0) = 0, (0) =. Show our work and/or eplain our reasoning. Compute () and () to check that the differential equation is satisfied: () = e = () = e + e ( ) = e e = () = e ( ) ( e + e ( ) ) = 4e + 4e = = ( 4e + 4e ) + 4 ( e e ) + 4e = ( 4 + 4)e + ( )e = 0. It remains to check that the initial condition are satisfied: (0) = 0 e 0 = 0, (0) = e 0 0e 0 =. Grading: () pt; () from () pts; plug in into DE and simplif pt each; check initial conditions pt each; no carr-over penalties (b; 8pts) Find the general solution of the differential equation = 0, = (). Show our work and/or eplain our reasoning. Since () = e is a solution of this equation (b above), () = e is also a solution (b the structure theorem for solutions of such equations) and the general solution of the differential equation is () = C e + C e Alternativel, the associated polnomial equation for this differential equation is r + 4r + 4 = 0 (r + ) = 0. So there is a double root of r=, which again gives the above general solution. Grading: in the nd approach, pts for the roots, pt each for () containing e and e, with remainder for the correct formula; if the roots are obtained incorrectl, pt for setting up the polnomial and up to 4 additional points for converting the roots to the corresponding equation for (); in the st approach, a little bit of eplanation is required (anthing in parentheses above is not required); if no eplanation is given, 4pts for correct answer, pts if () contains onl one constant or missing e or e ; pt for () = Ce or () = Ce. Solution to this problem continues on the net page.

4 Alternativel, one could first use the second approach in (b) to find that the general solution to the differential equation is () = C e + C e and then find C and C so that the two initial conditions in (a) hold. In order to do this, compute (), (0), and (0) and set them equal to the given initial condition: () = C e ( ) + C ( e + e ( ) ) = (C C )e C e, (0) = C e 0 + C 0 e 0 = C (0) = (C C )e 0 C 0 e 0 = (C C ), { (0) = C = 0 = = C = 0, C =. (0) = C C = So the solution to the initial-value problem in (a) is which is the function given in (a). () = 0 e + e = e, Grading: finding the general solution as in the second approach to part (b) on the previous page; after that, finding () pts, setting up the sstem pts; finding C and C pts, and concluding the argument pt. Remark: using the last approach indicates solid understanding of how to solve initial-value problems, but also misunderstanding of the basic concept of how to check that a given function solves a given initial-value problem. 4

5 Problem (5pts) A sample of tritium- (a radioactive substance) decaed to 90% of its original amount in ears. (a; 0pts) Let =(t) be the ratio of the amount of tritium- remaining after t ears to the original amount. Find a formula for (t). Show our work and/or eplain our reasoning. Since the deca rate is proportional to, (t) satisfies the eponential deca equation: (t) = (0)e rt = e rt, where r is the relative deca rate. B the last assumption, Thus, (t) = e (ln.9)t/ =.9 t/ () = e r =.9 r = ln.9 r = (ln.9)/. Grading: (t) = e rt pts; e r =.9 pts; finding r pts; final answer pts (either form acceptable, as well %, but there should be no units otherwise); finding r not necessar for full credit (something like e rt = (e r ) t/ would do instead); correct answer with no eplanation 4pts. (b; 5pts) If the initial weight of the sample was 00 mg (at time t = 0), what is the weight of the sample after 6 ears (from t=0)? Show our work and/or eplain our reasoning. Since ever ears the amount drops b a factor of.9, in 6 = ears this happens times; so the amount drops b a factor of =.79. Since the original amount was 00 mg, the amount after 6 ears is then = 7.9 mg Alternativel, plug in t = 6 in (t) in part (a) and multipl b 00 mg: 00 (t) = / = 7.9 mg Grading: correct answer in eactl the same form as above (or 79/0 mg) with no eplanation is pts; pts off if the units are missing or incorrect; plugging in t = 6 and multipling b 00 in the second approach is point each; if the final answer involves e (ln.9), then 0pts for part (b); if (t) is found incorrectl in (a) and is used in the second approach to (b), then the same grading principles appl to the function (t) found in (a). 5

6 Problem 4 (0pts) The direction field for a differential equation is shown below. (0)= (0)= (0)=0 (0)=.5 (a; 6pts) On the direction field, sketch and clearl label the graphs of the four solutions with the initial conditions (0) =.5, (0) = 0, (0) =, and (0) = (each of these four conditions determines a solution to the differential equation). No eplanation is required. The four curves must pass through the points (0,.5), (0, 0), (0, ), and (0, ), respectivel. These curves must be tangent to the slope lines at those points and roughl approimate the slopes everwhere else. The should never intersect. The solution curves with (0) =.5 and (0) = 0 ascend toward the line = as and descend toward the line = as. The solution curve with (0) = is the horizontal line =. The solution curve (0) = descends toward the line = as and ascends rapidl as decreases. Grading: 4pts for each curve (passing through specified point and roughl tangent there pt each; general shape ok pts; the rd curve must be the correct line to receive the last pts); curves clearl intersect (as opposed to being asmptotic) 4pts off; one curve not labeled pt off, -4 curves not labeled pts off. Solution to this problem continues on the net page. 6

7 (b; 4pts) The direction field above is for one of the following differential equations for = (): (i) =, (ii) =, (iii) = 4. Which of these three equations does the direction field correspond to and wh? The slopes for (, ) are positive in the diagram, while the slopes for < and for > are negative. Each of these three properties holds onl for the first of the three equations. Grading: wrong answer 0pts regardless of eplanation; correct answer pts; justification clearl ruling out the other answers up to pts (one of several possible reasons would suffice). 7

8 Problem 5 (0pts) (a; 5pts) Let = () be the solution to the initial-value problem =, = (), () = 0. Use Euler s method with n = steps to estimate the value of (). Show our steps clearl and use simple fractions (so 5/4 or 5 4, not.5). The step size is h = ( )/ = /, so we need to obtain estimates,, for the -value at = 4/, = 5/, and = starting with the value 0 =0 of at 0 =: s 0 = 0 0 = 0 = = = 0 + s 0 h = 0 + = s = = = 5 s = = = 67 7 = = + s h = = 8 9 = = + s h = = = Alternativel, this can be done using a table: i i i s i = i i i+ = i + s i = 0 + = = = = = = 9 8 The first column consists of the numbers i running from 0 to n, where n is the number of steps ( in this case). The second column starts with the initial value of ( in this case) with subsequent entries in the column obtained b adding the step size h ( in this case); it ends just before the final value of would have been entered ( = 5 + in this case). Thus, the first two columns can be filled in at the start. The first entr in the third column is the initial -value (0 in the case). After this, one computes the first entries in the remaining two columns and copies the first entr in the last column to the third column in the net line. The process then repeats across the second and third rows. The estimate for the final value of is the last entr in the table. Grading: h, 0,, pt each; correct recursive setup 5pts (pt each for number of steps, slope and change equations, left end points, end of the procedure); each of steps in each of computations pt each Solution to this problem continues on the net page. 8

9 (b; 5pts) Sketch the path in the -plane that represents the approimation carried out in part (a) and indicate its (path s) primar relation to the graph of the actual solution = () of the initialvalue problem in (a). (, ) (, ) 4/ 5/ This path consists of the three connected line segments from ( i, i ) to ( i+, i+ ) with i = 0,,. The first of these is tangent to the solution curve at (, 0). Grading: line segments pt, correct end points pts, solution curve through (,0) pt, tangent to the st line segment pt; aes not labeled pt off; comments not required (c; bonus 5pts, all or nothing) Is our estimate for () in part (a) an over-estimate (larger than ()) or an under-estimate? Justif our answer; no credit for correct answer onl. If =() is a solution of the differential equation in (a), then = ( ) = ( ) = = ( ) = (6 ) + 9. Thus, > 0 if (0, 6) and > 0. Furthermore, if ( 0 ) = 0 for some 0 0, then ( 0 ) = 0 0 > 0; so if ( 0 ) 0 for some 0 > 0, then () 0 for all 0. It follows that if ( 0 ) 0 for some 0 >0, then ()>0 for all ( 0, 6). B the previous paragraph, the first line segment, which is tangent to the solution curve at ( 0, 0 ), lies below this solution curve. Similarl, the second line segment, which is tangent to the solution curve at (, ), lies below this new solution curve. Finall, the third line segment, which is tangent to the solution curve at (, ), lies below this last solution curve. Since solution curves do not intersect, it follows that the entire -segment approimation path lies below the solution curve through ( 0, 0 ) and so <( ), i.e. our estimate for () is an under-estimate Grading: no partial credit; eplanation must be complete on the substance in order to receive the 5 bonus points. 9

10 Problem 6 (0pts) (a; 5pts) What are the constant solutions of the differential equation Show our work and/or eplain our reasoning. = ( ), = ()? The constant solutions correspond to the values of so that ( ) = 0 (for all in general, but there is no in this case to worr about). Thus, the constant solutions are () = 0 and () = Grading: correct answer pts (can be written as = 0,, etc.); minimal eplanation, such as setting RHS=0, pts; listing = 0 or = onl pt total, regardless of eplanation. (b; 5pts) Find the general solution to the differential equation = ( ), = (). Simplif our answer as much as possible. Show our work and/or eplain our reasoning. This differential equation is separable, so write as d/d and move all terms involving to LHS and all terms involving to RHS, and integrate: d d = ( ) d ( ) = d d ( ) = d = + C. In order to do the -integral, we need to use partial fractions: ( ( ) = ) = ( ) ( ) ( 0 ) 0 Thus, = ( ). d ( ) = ( ) d = ( ) ln ln + C = ln + C. Combining this with the first line gives ln = 4 + C = e4+c = e 4 e C = ±e 4C e 4 = Ae 4, where A is an nonzero constant. This equation defines implicitl as a function of, but it can be simplified further: = Ae4 Ae 4 = = Ae 4. We now have to remember the constant solutions found in (a): = 0 and =. The latter corresponds to A = 0 in the above formula. So the general solution of the differential equation is: () = 0, () = Solution to this problem continues on the net page. Ae 4 0

11 Grading: integrating the equation to something like = (/) or = ( ) 0pts for the entire question; separation of variables pt; RHS integral pt; LHS integral 5pts with penalties for computational errors; missing absolute value -pt; RHS is integrated to something like ln ( ) loss of entire 5pts; simplification from anti-derivatives to the final form 5pts; simplifing e 4+C to e 4 +e C, etc. loss of entire 5pts; incorporating the constant solutions into the final answer pts (if the are listed twice or = is not absorbed into the formula -pts); if separation of variables is done incorrectl in a significant wa (as opposed to minor coping errors, such as factors of ) 5pts ma for the entire question (with reductions for missing constant solutions, etc.) Alternativel, this differential equation can be re-written as a logistic growth equation, ( = r ), = (), K the general solution of which is In this case, () = K, () = 0. Ae r ( = ( ) = ( ) = 4 ), and so r = 4 and K =. Putting these into the general solution again gives () =, () = 0 Ae4 While in an actual logistic growth equation r, K > 0, this does not matter in terms of finding the general solution of the equation (K 0 is needed, since there is division b K). Grading: restating DE as logistic equation with clear indication that this is the purpose 5pts; formula with correct r and K 7pts (5pts off if the sign in eponent is wrong); pts for the zero constant solution (pts off if = is also listed separatel)