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







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$endgroup$












  • $begingroup$
    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

737




737












  • $begingroup$
    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











$endgroup$













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    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

    7,77461232




    7,77461232












    • $begingroup$
      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

        8,6243516




        8,6243516






























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