首页
网站开发
桌面应用
管理软件
微信开发
App开发
嵌入式软件
工具软件
数据采集与分析
其他
首页
>
> 详细
辅导COMP2421语言程序、Python设计程序讲解、data编程语言调试 讲解SPSS|辅导Database
项目预算:
开发周期:
发布时间:
要求地区:
School of Computing, University of Leeds
COMP2421: Numerical Computation
Coursework Seven: Derivatives and Differential Equations (Summative)
These exercises are intended to give you more practice at sketching derivatives and solving
ordinary differential equations: they are intended to reinforce the material that was covered
in lectures. Your solutions should be submitted on Gradescope by 10:00AM on Friday
11th December 2020. This coursework is summative and is worth 20% of the final grade
in this module.
Gradescope instructions:
1) Where numerical answers are requested, short answers will be set up on Gradescope and
the questions below list the answers you will need to provide. For all such answers, please
a) Use decimal notation only.
b) Use either exact answers or round to 4 significant figures.
c) Use numerical answers only – Do not use LaTeX. Do not use letters or symbols, e.g.
to denote recurring decimals or powers.
2) All other answers will require you to upload your answer in pdf format. There is no need
to upload code.
1. [16 marks]
For each of the four graphs on the attached pages plot sketches of the function which
is their derivative at every point. (Please sketch on top of the original graph and hand
these in as your solution to this question.)
2. [9 marks]
The decay of a radio-active material can be modelled by the following differential
equation,
q
′
(t) = −k × q(t) ,
where q(t) is the quantity of the material (in grams, say) that is present at time t.
Assuming that k = 2.0 and that at t = t0 = 0 we know that q(t0) = 10.0, take four
steps of Euler’s method (with dt = 0.25) to estimate q(1) (i.e. q(t) when t = 1.0).
Answer: q(t1) = , t1 =
q(t2) = , t2 =
q(t3) = , t3 =
q(t4) = , t4 =
3. [7 marks]
Let y(t) satisfy the following differential equation and initial condition:
y
′
(t) = y
3 + t
2
; y(1) = −1 .
Take three steps of Euler’s method (with dt =
1
3
) to estimate y(2) (i.e. the solution
when t = 2.0).
1
Answer: y(t1) = , t1 =
y(t2) = , t2 =
y(t3) = , t3 =
4. [8 marks]
Use two steps of the midpoint rule, with dt = 0.5, to estimate the solution of the
problem in the previous question (i.e. the same equation and initial condition) at
t = 2.0.
Answer: k = , temp = in step 1
y(t1) = , t1 =
k = , temp = in step 2
y(t2) = , t2 =
5. [10 marks] As shown in lectures, the Euler algorithm applies even in the case of vector
equations (as does the midpoint rule but this is not considered in this question): we can apply Euler’s method with:
f(t, y) = Ay .
This gives the following implementation of Euler’s method at each step:
Writing this in component form, and using Python notation, this gives the following
at each step:
y1[k + 1] = y1[k] + dt × (−2 × y1[k] + y2[k])
y2[k + 1] = y2[k] + dt × (y1[k] − 4 × y2[k])
t[k + 1] = t[k] + dt .
Hence take two steps of Euler’s method, with dt = 0.5, to approximate y(1) (i.e. the
solution at t = 1.0).
Answer: y1(t1) = , and
y2(t1) = , t1 =
y1(t2) = , and
y2(t2) = , t2 =
2
6. [20 marks]
For this question, you may wish to submit figures with your graphical analysis in
addition to your written answers to the questions.
Let y(t) = 4e
t−1 − t
2 − 2t − 2 (where e, the base of the natural logarithm, is a constant
equal to 2.7182818284. . .). The function y(t) satisfies the differential equation:
It is easy to see that for t = 1, y(1) = −1. We will use this as our initial condition.
(a) Write a Python script, using the Euler method function provided to you for numerical
integration, to numerically integrate this differential equation over 10 seconds,
to a final time of t = 11. Extend your code to include an error analysis. Note that you
will need to make some judgements in writing the code (in particular, choosing a time
step and which errors might be expedient to compute.)
Hints:
(i) Use the formula given for y(t) to calculate errors.
(ii) Write your code to conveniently compare different choices of dt.
(iii) You might wish to include graphical analysis of the numerical errors, plotting the
relevant errors as a function of time (note for Python, you would need to import matplotlib
to generate the figures. Alternatively, you might prefer to export the errors and
plot them with your favourite plotting tool). This is not required, but will help you to
answer the following questions.
(b) Based on your analysis, describe the behaviour of the errors after a long integration
time (t ≫ dt and in this case t ≫ 2). How do the errors obtained with different choices
of dt compare?
(c) Specifying dt to one significant figure only, determine the most efficient choice of
dt that should provide an error of at most 1% at t = 11.
The following optional questions are for extension only, and carry no marks.
(d) Optional: How does the behaviour of the absolute and relative errors change for
t < 2, t = 2 and t > 2 and why?
(e) Optional: Discuss the implicatons of the choice of dt you made in item (c).
Consider the following points: (i) The criterion used: this was an accuracy requirement
on the solution at t = 11; (ii) The computational cost. How would you expect this
cost to change if you halved the time step? What if you doubled the time step? How
does the error scale at times t < 2, t = 11?
(f) Optional: Based on your experience above, provide advice for choosing an appropriate
value of dt (for a given differential equation and integration scheme).
(g) Optional: Repeat your analysis with a different integration scheme (midpoint
scheme or the Runge Kutta, RK4 scheme) and compare the results. How do the
choices of dt change? How does the error behave after a long time with this new choice
of integration scheme?
软件开发、广告设计客服
QQ:99515681
邮箱:99515681@qq.com
工作时间:8:00-23:00
微信:codinghelp
热点项目
更多
代写math 1151, autumn 2024 w...
2024-11-14
代做comp4336/9336 mobile dat...
2024-11-14
代做eesa01 lab 2: weather an...
2024-11-14
代写comp1521 - 24t3 assignme...
2024-11-14
代写nbs8020 - dissertation s...
2024-11-14
代做fin b377f technical anal...
2024-11-14
代做ceic6714 mini design pro...
2024-11-14
代做introduction to computer...
2024-11-14
代做cs 353, fall 2024 introd...
2024-11-14
代做phy254 problem set #3 fa...
2024-11-14
代写n1569 financial risk man...
2024-11-14
代写csci-ua.0202 lab 3: enco...
2024-11-14
代写econ2226: chinese econom...
2024-11-14
热点标签
mktg2509
csci 2600
38170
lng302
csse3010
phas3226
77938
arch1162
engn4536/engn6536
acx5903
comp151101
phl245
cse12
comp9312
stat3016/6016
phas0038
comp2140
6qqmb312
xjco3011
rest0005
ematm0051
5qqmn219
lubs5062m
eee8155
cege0100
eap033
artd1109
mat246
etc3430
ecmm462
mis102
inft6800
ddes9903
comp6521
comp9517
comp3331/9331
comp4337
comp6008
comp9414
bu.231.790.81
man00150m
csb352h
math1041
eengm4100
isys1002
08
6057cem
mktg3504
mthm036
mtrx1701
mth3241
eeee3086
cmp-7038b
cmp-7000a
ints4010
econ2151
infs5710
fins5516
fin3309
fins5510
gsoe9340
math2007
math2036
soee5010
mark3088
infs3605
elec9714
comp2271
ma214
comp2211
infs3604
600426
sit254
acct3091
bbt405
msin0116
com107/com113
mark5826
sit120
comp9021
eco2101
eeen40700
cs253
ece3114
ecmm447
chns3000
math377
itd102
comp9444
comp(2041|9044)
econ0060
econ7230
mgt001371
ecs-323
cs6250
mgdi60012
mdia2012
comm221001
comm5000
ma1008
engl642
econ241
com333
math367
mis201
nbs-7041x
meek16104
econ2003
comm1190
mbas902
comp-1027
dpst1091
comp7315
eppd1033
m06
ee3025
msci231
bb113/bbs1063
fc709
comp3425
comp9417
econ42915
cb9101
math1102e
chme0017
fc307
mkt60104
5522usst
litr1-uc6201.200
ee1102
cosc2803
math39512
omp9727
int2067/int5051
bsb151
mgt253
fc021
babs2202
mis2002s
phya21
18-213
cege0012
mdia1002
math38032
mech5125
07
cisc102
mgx3110
cs240
11175
fin3020s
eco3420
ictten622
comp9727
cpt111
de114102d
mgm320h5s
bafi1019
math21112
efim20036
mn-3503
fins5568
110.807
bcpm000028
info6030
bma0092
bcpm0054
math20212
ce335
cs365
cenv6141
ftec5580
math2010
ec3450
comm1170
ecmt1010
csci-ua.0480-003
econ12-200
ib3960
ectb60h3f
cs247—assignment
tk3163
ics3u
ib3j80
comp20008
comp9334
eppd1063
acct2343
cct109
isys1055/3412
math350-real
math2014
eec180
stat141b
econ2101
msinm014/msing014/msing014b
fit2004
comp643
bu1002
cm2030
联系我们
- QQ: 9951568
© 2021
www.rj363.com
软件定制开发网!