Related Rates Problem Involving Airplanes












4












$begingroup$


I took a test yesterday, and would like to know how to answer this specific question on the exam:



One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.



I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.



$dy/dt = 240$ mi/hr



$dx/dt = 300$ mi/hr






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
    $endgroup$
    – Pranav Marathe
    Jan 29 '15 at 23:31






  • 1




    $begingroup$
    To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
    $endgroup$
    – apnorton
    Jan 29 '15 at 23:46






  • 1




    $begingroup$
    @anorton Thank you for your help. I added some background information.
    $endgroup$
    – Audrey
    Jan 30 '15 at 0:03










  • $begingroup$
    @Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
    $endgroup$
    – Loreno Heer
    Jan 30 '15 at 0:22
















4












$begingroup$


I took a test yesterday, and would like to know how to answer this specific question on the exam:



One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.



I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.



$dy/dt = 240$ mi/hr



$dx/dt = 300$ mi/hr






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
    $endgroup$
    – Pranav Marathe
    Jan 29 '15 at 23:31






  • 1




    $begingroup$
    To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
    $endgroup$
    – apnorton
    Jan 29 '15 at 23:46






  • 1




    $begingroup$
    @anorton Thank you for your help. I added some background information.
    $endgroup$
    – Audrey
    Jan 30 '15 at 0:03










  • $begingroup$
    @Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
    $endgroup$
    – Loreno Heer
    Jan 30 '15 at 0:22














4












4








4


1



$begingroup$


I took a test yesterday, and would like to know how to answer this specific question on the exam:



One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.



I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.



$dy/dt = 240$ mi/hr



$dx/dt = 300$ mi/hr






calculus







share|cite|improve this question











$endgroup$




I took a test yesterday, and would like to know how to answer this specific question on the exam:



One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.



I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.



$dy/dt = 240$ mi/hr



$dx/dt = 300$ mi/hr






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 '15 at 0:03







Audrey

















asked Jan 29 '15 at 23:21









AudreyAudrey

448




448








  • 1




    $begingroup$
    Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
    $endgroup$
    – Pranav Marathe
    Jan 29 '15 at 23:31






  • 1




    $begingroup$
    To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
    $endgroup$
    – apnorton
    Jan 29 '15 at 23:46






  • 1




    $begingroup$
    @anorton Thank you for your help. I added some background information.
    $endgroup$
    – Audrey
    Jan 30 '15 at 0:03










  • $begingroup$
    @Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
    $endgroup$
    – Loreno Heer
    Jan 30 '15 at 0:22














  • 1




    $begingroup$
    Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
    $endgroup$
    – Pranav Marathe
    Jan 29 '15 at 23:31






  • 1




    $begingroup$
    To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
    $endgroup$
    – apnorton
    Jan 29 '15 at 23:46






  • 1




    $begingroup$
    @anorton Thank you for your help. I added some background information.
    $endgroup$
    – Audrey
    Jan 30 '15 at 0:03










  • $begingroup$
    @Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
    $endgroup$
    – Loreno Heer
    Jan 30 '15 at 0:22








1




1




$begingroup$
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
$endgroup$
– Pranav Marathe
Jan 29 '15 at 23:31




$begingroup$
Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time.
$endgroup$
– Pranav Marathe
Jan 29 '15 at 23:31




1




1




$begingroup$
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 29 '15 at 23:46




$begingroup$
To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post.
$endgroup$
– apnorton
Jan 29 '15 at 23:46




1




1




$begingroup$
@anorton Thank you for your help. I added some background information.
$endgroup$
– Audrey
Jan 30 '15 at 0:03




$begingroup$
@anorton Thank you for your help. I added some background information.
$endgroup$
– Audrey
Jan 30 '15 at 0:03












$begingroup$
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
$endgroup$
– Loreno Heer
Jan 30 '15 at 0:22




$begingroup$
@Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2...
$endgroup$
– Loreno Heer
Jan 30 '15 at 0:22










2 Answers
2






active

oldest

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0












$begingroup$

Paths of the planes



$p_1(t) = (t cdot 300, 0)$



$p_2(t) = (0,(t-1/6) cdot (240))$



$s(t) = p_1(t) - p_2(t)$



$left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You have $x^2 + y^2 = z^2$
    your goal is to find $frac {dz}{dt}$



    $2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$



    Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$



    $x$ is the distance the west-bound airplane travels in 30 minutes.



    $y$ is the distance the south-bound airplane travels in 20 minutes.



    Use the Pythagorean theorem above to find $z.$



    distance = speed $times$ time



    You will need to convert units because the speed is in mph, and time is in minutes.



    And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.






    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












      $begingroup$

      Paths of the planes



      $p_1(t) = (t cdot 300, 0)$



      $p_2(t) = (0,(t-1/6) cdot (240))$



      $s(t) = p_1(t) - p_2(t)$



      $left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Paths of the planes



        $p_1(t) = (t cdot 300, 0)$



        $p_2(t) = (0,(t-1/6) cdot (240))$



        $s(t) = p_1(t) - p_2(t)$



        $left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Paths of the planes



          $p_1(t) = (t cdot 300, 0)$



          $p_2(t) = (0,(t-1/6) cdot (240))$



          $s(t) = p_1(t) - p_2(t)$



          $left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$






          share|cite|improve this answer











          $endgroup$



          Paths of the planes



          $p_1(t) = (t cdot 300, 0)$



          $p_2(t) = (0,(t-1/6) cdot (240))$



          $s(t) = p_1(t) - p_2(t)$



          $left.left(frac{d}{dt} |s(t)| right) rightvert_{t=.5} = ...$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 29 '15 at 23:43

























          answered Jan 29 '15 at 23:31









          Loreno HeerLoreno Heer

          3,34411534




          3,34411534























              0












              $begingroup$

              You have $x^2 + y^2 = z^2$
              your goal is to find $frac {dz}{dt}$



              $2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$



              Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$



              $x$ is the distance the west-bound airplane travels in 30 minutes.



              $y$ is the distance the south-bound airplane travels in 20 minutes.



              Use the Pythagorean theorem above to find $z.$



              distance = speed $times$ time



              You will need to convert units because the speed is in mph, and time is in minutes.



              And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You have $x^2 + y^2 = z^2$
                your goal is to find $frac {dz}{dt}$



                $2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$



                Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$



                $x$ is the distance the west-bound airplane travels in 30 minutes.



                $y$ is the distance the south-bound airplane travels in 20 minutes.



                Use the Pythagorean theorem above to find $z.$



                distance = speed $times$ time



                You will need to convert units because the speed is in mph, and time is in minutes.



                And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You have $x^2 + y^2 = z^2$
                  your goal is to find $frac {dz}{dt}$



                  $2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$



                  Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$



                  $x$ is the distance the west-bound airplane travels in 30 minutes.



                  $y$ is the distance the south-bound airplane travels in 20 minutes.



                  Use the Pythagorean theorem above to find $z.$



                  distance = speed $times$ time



                  You will need to convert units because the speed is in mph, and time is in minutes.



                  And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.






                  share|cite|improve this answer











                  $endgroup$



                  You have $x^2 + y^2 = z^2$
                  your goal is to find $frac {dz}{dt}$



                  $2zfrac {dz}{dt} = 2xfrac {dx}{dt} + 2yfrac {dy}{dt}$



                  Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$



                  $x$ is the distance the west-bound airplane travels in 30 minutes.



                  $y$ is the distance the south-bound airplane travels in 20 minutes.



                  Use the Pythagorean theorem above to find $z.$



                  distance = speed $times$ time



                  You will need to convert units because the speed is in mph, and time is in minutes.



                  And $frac {dx}{dt}, frac {dy}{dt}$ are the speeds of the planes.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 15 '18 at 22:48

























                  answered Feb 15 '18 at 22:42









                  Doug MDoug M

                  45.2k31954




                  45.2k31954






























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