Why do I get two different answers when solving for arclength?What am I missing when solving this integral with trigonometric substition?Arclength of parametric curveWhy are these two answers different?Indefinite integral vs definite integral: Why the different answers?I am getting two different answers for a basic integration problemDifferent answers for integral of $sin^3x$Equality of tw0 arclengthsA definite integral with two different answersLoophole? I'm getting 2 different answers when solving a differential equation in 2 different methodsWhy are these two ways of measuring the length of the groove in a phonograph record different?

Does obfuscation give any measurable security benefit?

I need an automatic way of making a lot of numbered folders

What is the gold linker?

Is is possible to externally power my DSLR with the original battery that is connected to the DSLR by means of wires?

What fantasy book has twins (except one's blue) and a cloaked ice bear on the cover?

How can I check the implementation of a builtin function?

How will the crew exit Starship when it lands on Mars?

How (and if) to include name change for transgender person in genealogy?

Word for 'most late'

How to figure out key from key signature?

Is oxygen above the critical point always supercritical fluid? Would it still appear to roughly follow the ideal gas law?

Car as a good investment

Why do Computer Science degrees contain a high proportion of mathematics?

Why is Mars cold?

What are the differences between prismatic, lensatic, and optical sighting compasses?

How should I understand FPGA architecture?

Does any politician honestly want a No Deal Brexit?

Did it take 3 minutes to reload a musket when the second amendment to the US constitution was ratified?

Why are second inversion triads considered less consonant than first inversion triads?

Do more Americans want the Bidens investigated than Trump impeached?

Can you use a virtual credit card to withdraw money from an ATM in the UK?

Does immunity to fear prevent a mummy's Dreadful Glare from paralyzing a character?

Interaction between casting spells and the Attack action

Digit Date Range



Why do I get two different answers when solving for arclength?


What am I missing when solving this integral with trigonometric substition?Arclength of parametric curveWhy are these two answers different?Indefinite integral vs definite integral: Why the different answers?I am getting two different answers for a basic integration problemDifferent answers for integral of $sin^3x$Equality of tw0 arclengthsA definite integral with two different answersLoophole? I'm getting 2 different answers when solving a differential equation in 2 different methodsWhy are these two ways of measuring the length of the groove in a phonograph record different?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








8














$begingroup$


I am given that $fracdxdt=8tcos(t)$ and $fracdydt=8tsin(t)$. I tried solving for the arclength from $t=0$ to $t=1.$



Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$



Method 2:



$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ However, when I solve using method 2, I get $1.22619,$ when the answer should be $4.$ What is causing this difference?










share|cite|improve this question












$endgroup$











  • 1




    $begingroup$
    About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
    $endgroup$
    – John Hughes
    May 13 at 1:39











  • $begingroup$
    Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
    $endgroup$
    – Tojrah
    May 13 at 1:39






  • 15




    $begingroup$
    The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
    $endgroup$
    – MathIsFun7225
    May 13 at 1:40






  • 1




    $begingroup$
    A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
    $endgroup$
    – David K
    May 13 at 10:47











  • $begingroup$
    Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
    $endgroup$
    – infinitezero
    May 13 at 14:40

















8














$begingroup$


I am given that $fracdxdt=8tcos(t)$ and $fracdydt=8tsin(t)$. I tried solving for the arclength from $t=0$ to $t=1.$



Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$



Method 2:



$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ However, when I solve using method 2, I get $1.22619,$ when the answer should be $4.$ What is causing this difference?










share|cite|improve this question












$endgroup$











  • 1




    $begingroup$
    About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
    $endgroup$
    – John Hughes
    May 13 at 1:39











  • $begingroup$
    Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
    $endgroup$
    – Tojrah
    May 13 at 1:39






  • 15




    $begingroup$
    The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
    $endgroup$
    – MathIsFun7225
    May 13 at 1:40






  • 1




    $begingroup$
    A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
    $endgroup$
    – David K
    May 13 at 10:47











  • $begingroup$
    Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
    $endgroup$
    – infinitezero
    May 13 at 14:40













8












8








8





$begingroup$


I am given that $fracdxdt=8tcos(t)$ and $fracdydt=8tsin(t)$. I tried solving for the arclength from $t=0$ to $t=1.$



Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$



Method 2:



$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ However, when I solve using method 2, I get $1.22619,$ when the answer should be $4.$ What is causing this difference?










share|cite|improve this question












$endgroup$




I am given that $fracdxdt=8tcos(t)$ and $fracdydt=8tsin(t)$. I tried solving for the arclength from $t=0$ to $t=1.$



Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$



Method 2:



$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ However, when I solve using method 2, I get $1.22619,$ when the answer should be $4.$ What is causing this difference?







integration arc-length






share|cite|improve this question
















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 13 at 10:43









David K

60.8k4 gold badges48 silver badges137 bronze badges




60.8k4 gold badges48 silver badges137 bronze badges










asked May 13 at 1:33









JayJay

1057 bronze badges




1057 bronze badges










  • 1




    $begingroup$
    About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
    $endgroup$
    – John Hughes
    May 13 at 1:39











  • $begingroup$
    Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
    $endgroup$
    – Tojrah
    May 13 at 1:39






  • 15




    $begingroup$
    The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
    $endgroup$
    – MathIsFun7225
    May 13 at 1:40






  • 1




    $begingroup$
    A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
    $endgroup$
    – David K
    May 13 at 10:47











  • $begingroup$
    Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
    $endgroup$
    – infinitezero
    May 13 at 14:40












  • 1




    $begingroup$
    About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
    $endgroup$
    – John Hughes
    May 13 at 1:39











  • $begingroup$
    Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
    $endgroup$
    – Tojrah
    May 13 at 1:39






  • 15




    $begingroup$
    The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
    $endgroup$
    – MathIsFun7225
    May 13 at 1:40






  • 1




    $begingroup$
    A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
    $endgroup$
    – David K
    May 13 at 10:47











  • $begingroup$
    Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
    $endgroup$
    – infinitezero
    May 13 at 14:40







1




1




$begingroup$
About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
$endgroup$
– John Hughes
May 13 at 1:39





$begingroup$
About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
$endgroup$
– John Hughes
May 13 at 1:39













$begingroup$
Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
$endgroup$
– Tojrah
May 13 at 1:39




$begingroup$
Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
$endgroup$
– Tojrah
May 13 at 1:39




15




15




$begingroup$
The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
$endgroup$
– MathIsFun7225
May 13 at 1:40




$begingroup$
The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
$endgroup$
– MathIsFun7225
May 13 at 1:40




1




1




$begingroup$
A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
$endgroup$
– David K
May 13 at 10:47





$begingroup$
A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
$endgroup$
– David K
May 13 at 10:47













$begingroup$
Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
$endgroup$
– infinitezero
May 13 at 14:40




$begingroup$
Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
$endgroup$
– infinitezero
May 13 at 14:40










3 Answers
3






active

oldest

votes


















17
















$begingroup$

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.



Playing a bit loose with differentials, we have
$$
fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
$$

Then
$$
sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
=frac1cos t,8t,cos t,dt=8t,dt.
$$

So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
$$
int_0^18t,dt = 4.
$$






share|cite|improve this answer










$endgroup$






















    3
















    $begingroup$

    Your first method requires a change. (It is $dt$ not $dx$)



    $$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$



    Now, for the 2nd method.



    It is actually an equivalence of the first one. It can be deduced like this.



    $$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$



    So, the second method also yields 4.






    share|cite|improve this answer










    $endgroup$






















      3
















      $begingroup$

      The second method should give you the correct answer as well.



      Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$



      so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$






      share|cite|improve this answer










      $endgroup$
















        Your Answer








        StackExchange.ready(function()
        var channelOptions =
        tags: "".split(" "),
        id: "69"
        ;
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function()
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled)
        StackExchange.using("snippets", function()
        createEditor();
        );

        else
        createEditor();

        );

        function createEditor()
        StackExchange.prepareEditor(
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader:
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        ,
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        );



        );














        draft saved

        draft discarded
















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3223929%2fwhy-do-i-get-two-different-answers-when-solving-for-arclength%23new-answer', 'question_page');

        );

        Post as a guest















        Required, but never shown


























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        17
















        $begingroup$

        Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.



        Playing a bit loose with differentials, we have
        $$
        fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
        $$

        Then
        $$
        sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
        =frac1cos t,8t,cos t,dt=8t,dt.
        $$

        So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
        $$
        int_0^18t,dt = 4.
        $$






        share|cite|improve this answer










        $endgroup$



















          17
















          $begingroup$

          Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.



          Playing a bit loose with differentials, we have
          $$
          fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
          $$

          Then
          $$
          sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
          =frac1cos t,8t,cos t,dt=8t,dt.
          $$

          So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
          $$
          int_0^18t,dt = 4.
          $$






          share|cite|improve this answer










          $endgroup$

















            17














            17










            17







            $begingroup$

            Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.



            Playing a bit loose with differentials, we have
            $$
            fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
            $$

            Then
            $$
            sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
            =frac1cos t,8t,cos t,dt=8t,dt.
            $$

            So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
            $$
            int_0^18t,dt = 4.
            $$






            share|cite|improve this answer










            $endgroup$



            Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.



            Playing a bit loose with differentials, we have
            $$
            fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
            $$

            Then
            $$
            sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
            =frac1cos t,8t,cos t,dt=8t,dt.
            $$

            So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
            $$
            int_0^18t,dt = 4.
            $$







            share|cite|improve this answer













            share|cite|improve this answer




            share|cite|improve this answer










            answered May 13 at 1:49









            Martin ArgeramiMartin Argerami

            134k12 gold badges86 silver badges191 bronze badges




            134k12 gold badges86 silver badges191 bronze badges


























                3
















                $begingroup$

                Your first method requires a change. (It is $dt$ not $dx$)



                $$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$



                Now, for the 2nd method.



                It is actually an equivalence of the first one. It can be deduced like this.



                $$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$



                So, the second method also yields 4.






                share|cite|improve this answer










                $endgroup$



















                  3
















                  $begingroup$

                  Your first method requires a change. (It is $dt$ not $dx$)



                  $$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$



                  Now, for the 2nd method.



                  It is actually an equivalence of the first one. It can be deduced like this.



                  $$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$



                  So, the second method also yields 4.






                  share|cite|improve this answer










                  $endgroup$

















                    3














                    3










                    3







                    $begingroup$

                    Your first method requires a change. (It is $dt$ not $dx$)



                    $$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$



                    Now, for the 2nd method.



                    It is actually an equivalence of the first one. It can be deduced like this.



                    $$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$



                    So, the second method also yields 4.






                    share|cite|improve this answer










                    $endgroup$



                    Your first method requires a change. (It is $dt$ not $dx$)



                    $$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$



                    Now, for the 2nd method.



                    It is actually an equivalence of the first one. It can be deduced like this.



                    $$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$



                    So, the second method also yields 4.







                    share|cite|improve this answer













                    share|cite|improve this answer




                    share|cite|improve this answer










                    answered May 13 at 1:52









                    Ak19Ak19

                    5,4467 silver badges28 bronze badges




                    5,4467 silver badges28 bronze badges
























                        3
















                        $begingroup$

                        The second method should give you the correct answer as well.



                        Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$



                        so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$






                        share|cite|improve this answer










                        $endgroup$



















                          3
















                          $begingroup$

                          The second method should give you the correct answer as well.



                          Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$



                          so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$






                          share|cite|improve this answer










                          $endgroup$

















                            3














                            3










                            3







                            $begingroup$

                            The second method should give you the correct answer as well.



                            Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$



                            so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$






                            share|cite|improve this answer










                            $endgroup$



                            The second method should give you the correct answer as well.



                            Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$



                            so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$







                            share|cite|improve this answer













                            share|cite|improve this answer




                            share|cite|improve this answer










                            answered May 13 at 1:59









                            Mohammad Riazi-KermaniMohammad Riazi-Kermani

                            55.1k4 gold badges27 silver badges76 bronze badges




                            55.1k4 gold badges27 silver badges76 bronze badges































                                draft saved

                                draft discarded















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid


                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.

                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3223929%2fwhy-do-i-get-two-different-answers-when-solving-for-arclength%23new-answer', 'question_page');

                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown









                                Popular posts from this blog

                                Tamil (spriik) Luke uk diar | Nawigatjuun

                                Align equal signs while including text over equalitiesAMS align: left aligned text/math plus multicolumn alignmentMultiple alignmentsAligning equations in multiple placesNumbering and aligning an equation with multiple columnsHow to align one equation with another multline equationUsing \ in environments inside the begintabularxNumber equations and preserving alignment of equal signsHow can I align equations to the left and to the right?Double equation alignment problem within align enviromentAligned within align: Why are they right-aligned?

                                Training a classifier when some of the features are unknownWhy does Gradient Boosting regression predict negative values when there are no negative y-values in my training set?How to improve an existing (trained) classifier?What is effect when I set up some self defined predisctor variables?Why Matlab neural network classification returns decimal values on prediction dataset?Fitting and transforming text data in training, testing, and validation setsHow to quantify the performance of the classifier (multi-class SVM) using the test data?How do I control for some patients providing multiple samples in my training data?Training and Test setTraining a convolutional neural network for image denoising in MatlabShouldn't an autoencoder with #(neurons in hidden layer) = #(neurons in input layer) be “perfect”?