One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph...












0












$begingroup$


One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town. At time $t=0$, the first train is $70$ miles south and the second train is $90$ miles west. Find the rate at which the distance between the trains is changing at time $t=30$ minutes



I have thought to do the following:



I made a drawing that illustrated the situation and gave me a right triangle. Suppose that the distance of the train one to the origin is $x$ and that the distance of the train two to the origin is $y$, then in the time $t=0$ we have that $x=-70$ and $y=-90$ and the distance $d$ between the two is $d=114,02$. I have thought to use the pythagoras theorem and implicit derivation to get to that $x^2+y^2=d^2$ and $2xfrac{dx}{dt}+2yfrac{dy}{dt}=2dfrac{dd}{dt}$ but I do not know what else to do, I know that $frac{dx}{dt}=140$ and $frac{dy}{dt}=150$ but here I am stuck, I would appreciate any collaboration, thank you very much.






calculus algebra-precalculus derivatives applications







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  • 1




    $begingroup$
    I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
    $endgroup$
    – Makina
    Dec 1 '18 at 23:06


















0












$begingroup$


One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town. At time $t=0$, the first train is $70$ miles south and the second train is $90$ miles west. Find the rate at which the distance between the trains is changing at time $t=30$ minutes



I have thought to do the following:



I made a drawing that illustrated the situation and gave me a right triangle. Suppose that the distance of the train one to the origin is $x$ and that the distance of the train two to the origin is $y$, then in the time $t=0$ we have that $x=-70$ and $y=-90$ and the distance $d$ between the two is $d=114,02$. I have thought to use the pythagoras theorem and implicit derivation to get to that $x^2+y^2=d^2$ and $2xfrac{dx}{dt}+2yfrac{dy}{dt}=2dfrac{dd}{dt}$ but I do not know what else to do, I know that $frac{dx}{dt}=140$ and $frac{dy}{dt}=150$ but here I am stuck, I would appreciate any collaboration, thank you very much.






calculus algebra-precalculus derivatives applications







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
    $endgroup$
    – Makina
    Dec 1 '18 at 23:06
















0












0








0


1



$begingroup$


One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town. At time $t=0$, the first train is $70$ miles south and the second train is $90$ miles west. Find the rate at which the distance between the trains is changing at time $t=30$ minutes



I have thought to do the following:



I made a drawing that illustrated the situation and gave me a right triangle. Suppose that the distance of the train one to the origin is $x$ and that the distance of the train two to the origin is $y$, then in the time $t=0$ we have that $x=-70$ and $y=-90$ and the distance $d$ between the two is $d=114,02$. I have thought to use the pythagoras theorem and implicit derivation to get to that $x^2+y^2=d^2$ and $2xfrac{dx}{dt}+2yfrac{dy}{dt}=2dfrac{dd}{dt}$ but I do not know what else to do, I know that $frac{dx}{dt}=140$ and $frac{dy}{dt}=150$ but here I am stuck, I would appreciate any collaboration, thank you very much.






calculus algebra-precalculus derivatives applications







share|cite|improve this question









$endgroup$




One train travels north at $140$ mph towards Traveler's Town, while a second train travels west at $150$ mph away from Traveler's Town. At time $t=0$, the first train is $70$ miles south and the second train is $90$ miles west. Find the rate at which the distance between the trains is changing at time $t=30$ minutes



I have thought to do the following:



I made a drawing that illustrated the situation and gave me a right triangle. Suppose that the distance of the train one to the origin is $x$ and that the distance of the train two to the origin is $y$, then in the time $t=0$ we have that $x=-70$ and $y=-90$ and the distance $d$ between the two is $d=114,02$. I have thought to use the pythagoras theorem and implicit derivation to get to that $x^2+y^2=d^2$ and $2xfrac{dx}{dt}+2yfrac{dy}{dt}=2dfrac{dd}{dt}$ but I do not know what else to do, I know that $frac{dx}{dt}=140$ and $frac{dy}{dt}=150$ but here I am stuck, I would appreciate any collaboration, thank you very much.






calculus algebra-precalculus derivatives applications




calculus algebra-precalculus derivatives applications






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asked Dec 1 '18 at 22:59









user482152user482152

987




987








  • 1




    $begingroup$
    I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
    $endgroup$
    – Makina
    Dec 1 '18 at 23:06
















  • 1




    $begingroup$
    I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
    $endgroup$
    – Makina
    Dec 1 '18 at 23:06










1




1




$begingroup$
I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
$endgroup$
– Makina
Dec 1 '18 at 23:06






$begingroup$
I'm not a physics expert but I believe that since $d$ is the distance, then derivative with respect to $t$ will be the "speed" of, I guess, the rate of change. At $30$ mins your $y$ will be $0$ and $x$ will be $165$ (I directed my x axis to the west). Plug in and that should be it
$endgroup$
– Makina
Dec 1 '18 at 23:06












2 Answers
2






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$begingroup$

Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by:
begin{align}
vec{r_1}(t) &= langle 0, -70+140trangle\
vec{r_2}(t) &= langle -90-150t,0rangle .
end{align}



Hence the distance between the trains is:
begin{align}
d(t) = sqrt{(vec{r_2}(t)-vec{r_1}(t))^2}= 10 sqrt{421t^2+74t+130}.
end{align}
Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.






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    0












    $begingroup$

    The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.



    The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.



    The distance between the two trains is (by Pythagoras):
    $$d=sqrt{(70-140t)^2+(90+150t)^2}.$$
    The distance is changing at time $t=0.5$ hours at the rate:
    $$d'(0.5)=frac{-2(70-140cdot 0.5)cdot (-140)+2(90+150cdot 0.5)cdot 150}{2sqrt{(70-140cdot 0.5)^2+(90+150cdot 0.5)^2}}=150 text{mph}.$$






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by:
      begin{align}
      vec{r_1}(t) &= langle 0, -70+140trangle\
      vec{r_2}(t) &= langle -90-150t,0rangle .
      end{align}



      Hence the distance between the trains is:
      begin{align}
      d(t) = sqrt{(vec{r_2}(t)-vec{r_1}(t))^2}= 10 sqrt{421t^2+74t+130}.
      end{align}
      Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by:
        begin{align}
        vec{r_1}(t) &= langle 0, -70+140trangle\
        vec{r_2}(t) &= langle -90-150t,0rangle .
        end{align}



        Hence the distance between the trains is:
        begin{align}
        d(t) = sqrt{(vec{r_2}(t)-vec{r_1}(t))^2}= 10 sqrt{421t^2+74t+130}.
        end{align}
        Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by:
          begin{align}
          vec{r_1}(t) &= langle 0, -70+140trangle\
          vec{r_2}(t) &= langle -90-150t,0rangle .
          end{align}



          Hence the distance between the trains is:
          begin{align}
          d(t) = sqrt{(vec{r_2}(t)-vec{r_1}(t))^2}= 10 sqrt{421t^2+74t+130}.
          end{align}
          Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.






          share|cite|improve this answer









          $endgroup$



          Place the town in the origin. Now let the first axis point due east, and the second axis due north. Now the trains can be parametrized by:
          begin{align}
          vec{r_1}(t) &= langle 0, -70+140trangle\
          vec{r_2}(t) &= langle -90-150t,0rangle .
          end{align}



          Hence the distance between the trains is:
          begin{align}
          d(t) = sqrt{(vec{r_2}(t)-vec{r_1}(t))^2}= 10 sqrt{421t^2+74t+130}.
          end{align}
          Now all we need to find is $d'(1/2)$, which after some calculations is 150 mph.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 8:26









          MartinMartin

          1,7541412




          1,7541412























              0












              $begingroup$

              The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.



              The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.



              The distance between the two trains is (by Pythagoras):
              $$d=sqrt{(70-140t)^2+(90+150t)^2}.$$
              The distance is changing at time $t=0.5$ hours at the rate:
              $$d'(0.5)=frac{-2(70-140cdot 0.5)cdot (-140)+2(90+150cdot 0.5)cdot 150}{2sqrt{(70-140cdot 0.5)^2+(90+150cdot 0.5)^2}}=150 text{mph}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.



                The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.



                The distance between the two trains is (by Pythagoras):
                $$d=sqrt{(70-140t)^2+(90+150t)^2}.$$
                The distance is changing at time $t=0.5$ hours at the rate:
                $$d'(0.5)=frac{-2(70-140cdot 0.5)cdot (-140)+2(90+150cdot 0.5)cdot 150}{2sqrt{(70-140cdot 0.5)^2+(90+150cdot 0.5)^2}}=150 text{mph}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.



                  The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.



                  The distance between the two trains is (by Pythagoras):
                  $$d=sqrt{(70-140t)^2+(90+150t)^2}.$$
                  The distance is changing at time $t=0.5$ hours at the rate:
                  $$d'(0.5)=frac{-2(70-140cdot 0.5)cdot (-140)+2(90+150cdot 0.5)cdot 150}{2sqrt{(70-140cdot 0.5)^2+(90+150cdot 0.5)^2}}=150 text{mph}.$$






                  share|cite|improve this answer









                  $endgroup$



                  The first train is $70$ miles south and moving north at $140$ mph, so after $t$ hours it will be $70-140t$ miles from the Traveler's Town.



                  The second train is $90$ miles west and moving west at $150$ mph, so after $t$ hours it will be $90+150t$ miles from the Traveler's Town.



                  The distance between the two trains is (by Pythagoras):
                  $$d=sqrt{(70-140t)^2+(90+150t)^2}.$$
                  The distance is changing at time $t=0.5$ hours at the rate:
                  $$d'(0.5)=frac{-2(70-140cdot 0.5)cdot (-140)+2(90+150cdot 0.5)cdot 150}{2sqrt{(70-140cdot 0.5)^2+(90+150cdot 0.5)^2}}=150 text{mph}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 15:52









                  farruhotafarruhota

                  19.7k2738




                  19.7k2738






























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