2 cars moving from the same point, related rates.












0












$begingroup$


2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?



Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$



Plugging in the values after 2 hours I got



$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$



I am not sure where to go from because I can't find the value of z.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
    $endgroup$
    – David K
    Dec 10 '18 at 13:37
















0












$begingroup$


2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?



Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$



Plugging in the values after 2 hours I got



$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$



I am not sure where to go from because I can't find the value of z.






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
    $endgroup$
    – David K
    Dec 10 '18 at 13:37














0












0








0





$begingroup$


2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?



Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$



Plugging in the values after 2 hours I got



$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$



I am not sure where to go from because I can't find the value of z.






calculus







share|cite|improve this question









$endgroup$




2 cars start from the same point. Car A moves south at 60 $frac{mi}{h}$ and car B travels west at 25 $frac{mi}{h}$. At what rate is the distance between the cars increasing after 2 hours?



Using the Pythagorean theorem I got $2x frac{dx}{dt}+2y frac{dy}{dt}=2z frac{dz}{dt}$



Plugging in the values after 2 hours I got



$$2(120)(60)+2(50)(25)=2z frac{dz}{dt}$$



I am not sure where to go from because I can't find the value of z.






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 0:38









Eric BrownEric Brown

757




757












  • $begingroup$
    In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
    $endgroup$
    – David K
    Dec 10 '18 at 13:37


















  • $begingroup$
    In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
    $endgroup$
    – David K
    Dec 10 '18 at 13:37
















$begingroup$
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
$endgroup$
– David K
Dec 10 '18 at 13:37




$begingroup$
In fact, because both cars move at constant speed along straight lines, the rate of increase of the distance between them is constant. That is, the rate after two hours is the same as the rate after one hour or after one microsecond.
$endgroup$
– David K
Dec 10 '18 at 13:37










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$



We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$



By using Pythagorean theorem we have



$x^2+y^2=z^2$



Now implicitly differentiate with respect to $t$ to get



$$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$



Now plug in the numbers. After $2$ hours



$$x=60(2)=120$$
and $$y=25(2)=50$$



$$z^2=(120)^2+(50)^2$$



Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 0:53










  • $begingroup$
    You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
    $endgroup$
    – N. F. Taussig
    Dec 9 '18 at 16:54



















0












$begingroup$

Hint.



The distance between the two cars is given by



$$
delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
$$



so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$



NOTE



Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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/3.0/"u003ecc by-sa 3.0 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%2f3031862%2f2-cars-moving-from-the-same-point-related-rates%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$



    We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$



    By using Pythagorean theorem we have



    $x^2+y^2=z^2$



    Now implicitly differentiate with respect to $t$ to get



    $$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$



    Now plug in the numbers. After $2$ hours



    $$x=60(2)=120$$
    and $$y=25(2)=50$$



    $$z^2=(120)^2+(50)^2$$



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
      $endgroup$
      – Eric Brown
      Dec 9 '18 at 0:53










    • $begingroup$
      You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
      $endgroup$
      – N. F. Taussig
      Dec 9 '18 at 16:54
















    0












    $begingroup$

    Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$



    We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$



    By using Pythagorean theorem we have



    $x^2+y^2=z^2$



    Now implicitly differentiate with respect to $t$ to get



    $$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$



    Now plug in the numbers. After $2$ hours



    $$x=60(2)=120$$
    and $$y=25(2)=50$$



    $$z^2=(120)^2+(50)^2$$



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
      $endgroup$
      – Eric Brown
      Dec 9 '18 at 0:53










    • $begingroup$
      You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
      $endgroup$
      – N. F. Taussig
      Dec 9 '18 at 16:54














    0












    0








    0





    $begingroup$

    Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$



    We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$



    By using Pythagorean theorem we have



    $x^2+y^2=z^2$



    Now implicitly differentiate with respect to $t$ to get



    $$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$



    Now plug in the numbers. After $2$ hours



    $$x=60(2)=120$$
    and $$y=25(2)=50$$



    $$z^2=(120)^2+(50)^2$$



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$



    Let the rate of change of the distance between the two cars is $dfrac{dz}{dt}$



    We know that $$dfrac{dx}{dt}=60, dfrac{dy}{dt}=25$$



    By using Pythagorean theorem we have



    $x^2+y^2=z^2$



    Now implicitly differentiate with respect to $t$ to get



    $$2xdfrac{dx}{dt}+2ydfrac{dy}{dt}=2z dfrac{dz}{dt}$$



    Now plug in the numbers. After $2$ hours



    $$x=60(2)=120$$
    and $$y=25(2)=50$$



    $$z^2=(120)^2+(50)^2$$



    Can you take it from here?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 19:47

























    answered Dec 9 '18 at 0:45









    Key FlexKey Flex

    8,64361233




    8,64361233












    • $begingroup$
      Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
      $endgroup$
      – Eric Brown
      Dec 9 '18 at 0:53










    • $begingroup$
      You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
      $endgroup$
      – N. F. Taussig
      Dec 9 '18 at 16:54


















    • $begingroup$
      Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
      $endgroup$
      – Eric Brown
      Dec 9 '18 at 0:53










    • $begingroup$
      You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
      $endgroup$
      – N. F. Taussig
      Dec 9 '18 at 16:54
















    $begingroup$
    Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 0:53




    $begingroup$
    Yup, I solve for z then plug it back into the differentiated formula. Thanks for your help!
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 0:53












    $begingroup$
    You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
    $endgroup$
    – N. F. Taussig
    Dec 9 '18 at 16:54




    $begingroup$
    You are using two different variables for the distance between the cars. To be consistent, you should change $frac{ds}{dt}$ to $frac{dz}{dt}$.
    $endgroup$
    – N. F. Taussig
    Dec 9 '18 at 16:54











    0












    $begingroup$

    Hint.



    The distance between the two cars is given by



    $$
    delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
    $$



    so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$



    NOTE



    Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint.



      The distance between the two cars is given by



      $$
      delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
      $$



      so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$



      NOTE



      Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint.



        The distance between the two cars is given by



        $$
        delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
        $$



        so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$



        NOTE



        Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$






        share|cite|improve this answer











        $endgroup$



        Hint.



        The distance between the two cars is given by



        $$
        delta(t) = || vec v_1 t - vec v_2 t|| = ||vec v_1 - vec v_2|| t
        $$



        so the rate for $delta(t)$ is $frac{d}{dt}delta(t) = ||vec v_1 - vec v_2||$



        NOTE



        Given $vec v = (v_x, v_y)$ then $||vec v|| = sqrt{v_x^2+v_y^2}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 12:18

























        answered Dec 9 '18 at 0:45









        CesareoCesareo

        9,3963517




        9,3963517






























            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%2f3031862%2f2-cars-moving-from-the-same-point-related-rates%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more