Use calculus to find length of y=-mx+b, then show that the answer agrees with the answer when using...












0












$begingroup$


First question, sorry for poor formatting. This question is from my Calculus 2 class, and I am pretty sure I am supposed to be using arc length formula for the question.



Exact words:





  1. Consider the line segment shown



    *shows simple right triangle with $y=-mx+b$ over hypotenuse, and the $90$ degree corner sitting at point $(0,0)$



    a. Use calculus to find its length (show your calculation of the integral involved)



    b. Now show that your answer to part a agrees with what you get when you simply use the Pythagorean Theorem (or distance formula)




So I think I got through the first part:



Taking the derivative of $y=-mx+b$ and putting it into the arc length formula I got $(b-a)sqrt{1+m^2}$



The second part is where I am lost... I feel like I am missing something pretty easy but I don't understand how I would make the Pythagorean theorem or distance formula to agree with my answer. My answer to part a might just be wrong too but I am not sure how else to do it.



There are a lot more parts to the problem, if I can get past this hopefully it can show me how to approach the other problems.



Appreciate any help I can get, thanks!






calculus arc-length







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  • 2




    $begingroup$
    What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
    $endgroup$
    – platty
    Nov 28 '18 at 6:27










  • $begingroup$
    desmos.com/calculator/zcn4cv4qj6
    $endgroup$
    – Mason
    Nov 28 '18 at 6:28
















0












$begingroup$


First question, sorry for poor formatting. This question is from my Calculus 2 class, and I am pretty sure I am supposed to be using arc length formula for the question.



Exact words:





  1. Consider the line segment shown



    *shows simple right triangle with $y=-mx+b$ over hypotenuse, and the $90$ degree corner sitting at point $(0,0)$



    a. Use calculus to find its length (show your calculation of the integral involved)



    b. Now show that your answer to part a agrees with what you get when you simply use the Pythagorean Theorem (or distance formula)




So I think I got through the first part:



Taking the derivative of $y=-mx+b$ and putting it into the arc length formula I got $(b-a)sqrt{1+m^2}$



The second part is where I am lost... I feel like I am missing something pretty easy but I don't understand how I would make the Pythagorean theorem or distance formula to agree with my answer. My answer to part a might just be wrong too but I am not sure how else to do it.



There are a lot more parts to the problem, if I can get past this hopefully it can show me how to approach the other problems.



Appreciate any help I can get, thanks!






calculus arc-length







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
    $endgroup$
    – platty
    Nov 28 '18 at 6:27










  • $begingroup$
    desmos.com/calculator/zcn4cv4qj6
    $endgroup$
    – Mason
    Nov 28 '18 at 6:28














0












0








0





$begingroup$


First question, sorry for poor formatting. This question is from my Calculus 2 class, and I am pretty sure I am supposed to be using arc length formula for the question.



Exact words:





  1. Consider the line segment shown



    *shows simple right triangle with $y=-mx+b$ over hypotenuse, and the $90$ degree corner sitting at point $(0,0)$



    a. Use calculus to find its length (show your calculation of the integral involved)



    b. Now show that your answer to part a agrees with what you get when you simply use the Pythagorean Theorem (or distance formula)




So I think I got through the first part:



Taking the derivative of $y=-mx+b$ and putting it into the arc length formula I got $(b-a)sqrt{1+m^2}$



The second part is where I am lost... I feel like I am missing something pretty easy but I don't understand how I would make the Pythagorean theorem or distance formula to agree with my answer. My answer to part a might just be wrong too but I am not sure how else to do it.



There are a lot more parts to the problem, if I can get past this hopefully it can show me how to approach the other problems.



Appreciate any help I can get, thanks!






calculus arc-length







share|cite|improve this question











$endgroup$




First question, sorry for poor formatting. This question is from my Calculus 2 class, and I am pretty sure I am supposed to be using arc length formula for the question.



Exact words:





  1. Consider the line segment shown



    *shows simple right triangle with $y=-mx+b$ over hypotenuse, and the $90$ degree corner sitting at point $(0,0)$



    a. Use calculus to find its length (show your calculation of the integral involved)



    b. Now show that your answer to part a agrees with what you get when you simply use the Pythagorean Theorem (or distance formula)




So I think I got through the first part:



Taking the derivative of $y=-mx+b$ and putting it into the arc length formula I got $(b-a)sqrt{1+m^2}$



The second part is where I am lost... I feel like I am missing something pretty easy but I don't understand how I would make the Pythagorean theorem or distance formula to agree with my answer. My answer to part a might just be wrong too but I am not sure how else to do it.



There are a lot more parts to the problem, if I can get past this hopefully it can show me how to approach the other problems.



Appreciate any help I can get, thanks!






calculus arc-length




calculus arc-length






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share|cite|improve this question













share|cite|improve this question




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edited Nov 28 '18 at 6:31









Mason

1,9551530




1,9551530










asked Nov 28 '18 at 6:23









Tyler ETyler E

1




1








  • 2




    $begingroup$
    What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
    $endgroup$
    – platty
    Nov 28 '18 at 6:27










  • $begingroup$
    desmos.com/calculator/zcn4cv4qj6
    $endgroup$
    – Mason
    Nov 28 '18 at 6:28














  • 2




    $begingroup$
    What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
    $endgroup$
    – platty
    Nov 28 '18 at 6:27










  • $begingroup$
    desmos.com/calculator/zcn4cv4qj6
    $endgroup$
    – Mason
    Nov 28 '18 at 6:28








2




2




$begingroup$
What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
$endgroup$
– platty
Nov 28 '18 at 6:27




$begingroup$
What is $a$ in your answer? You said that the diagram has a right triangle, so you should be able to use the Pythagorean theorem if you can find the side lengths. Also, attaching an image might be helpful.
$endgroup$
– platty
Nov 28 '18 at 6:27












$begingroup$
desmos.com/calculator/zcn4cv4qj6
$endgroup$
– Mason
Nov 28 '18 at 6:28




$begingroup$
desmos.com/calculator/zcn4cv4qj6
$endgroup$
– Mason
Nov 28 '18 at 6:28










2 Answers
2






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1












$begingroup$

The line $y=-mx+b$ looks like this



enter image description here



So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length




$sqrt{b^2+(b/m)^2} = bsqrt{1+m^{-2}}$




But also we can compute the area of the arc length
given by $int_0^{frac{b}{m}}sqrt{1+f'left(tright)^2}dt$ where $f(x)=-mx+b$.



$int_0^{b/m}sqrt{1+f'left(tright)^2}dt$




$=int_0^{b/m} sqrt{1+m^2}dt= frac{b}{m}sqrt{1+m^2}$.




So we get the same result as when we employed Pythagorean theorem.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint:when $x=0, y=b $ and when $y=0,x=frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.



    But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      The line $y=-mx+b$ looks like this



      enter image description here



      So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length




      $sqrt{b^2+(b/m)^2} = bsqrt{1+m^{-2}}$




      But also we can compute the area of the arc length
      given by $int_0^{frac{b}{m}}sqrt{1+f'left(tright)^2}dt$ where $f(x)=-mx+b$.



      $int_0^{b/m}sqrt{1+f'left(tright)^2}dt$




      $=int_0^{b/m} sqrt{1+m^2}dt= frac{b}{m}sqrt{1+m^2}$.




      So we get the same result as when we employed Pythagorean theorem.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        The line $y=-mx+b$ looks like this



        enter image description here



        So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length




        $sqrt{b^2+(b/m)^2} = bsqrt{1+m^{-2}}$




        But also we can compute the area of the arc length
        given by $int_0^{frac{b}{m}}sqrt{1+f'left(tright)^2}dt$ where $f(x)=-mx+b$.



        $int_0^{b/m}sqrt{1+f'left(tright)^2}dt$




        $=int_0^{b/m} sqrt{1+m^2}dt= frac{b}{m}sqrt{1+m^2}$.




        So we get the same result as when we employed Pythagorean theorem.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          The line $y=-mx+b$ looks like this



          enter image description here



          So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length




          $sqrt{b^2+(b/m)^2} = bsqrt{1+m^{-2}}$




          But also we can compute the area of the arc length
          given by $int_0^{frac{b}{m}}sqrt{1+f'left(tright)^2}dt$ where $f(x)=-mx+b$.



          $int_0^{b/m}sqrt{1+f'left(tright)^2}dt$




          $=int_0^{b/m} sqrt{1+m^2}dt= frac{b}{m}sqrt{1+m^2}$.




          So we get the same result as when we employed Pythagorean theorem.






          share|cite|improve this answer











          $endgroup$



          The line $y=-mx+b$ looks like this



          enter image description here



          So if you have a right triangle with lengths $b$, and $b/m$ the hypotenuse should be (by Pythagorean theorem) length




          $sqrt{b^2+(b/m)^2} = bsqrt{1+m^{-2}}$




          But also we can compute the area of the arc length
          given by $int_0^{frac{b}{m}}sqrt{1+f'left(tright)^2}dt$ where $f(x)=-mx+b$.



          $int_0^{b/m}sqrt{1+f'left(tright)^2}dt$




          $=int_0^{b/m} sqrt{1+m^2}dt= frac{b}{m}sqrt{1+m^2}$.




          So we get the same result as when we employed Pythagorean theorem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 6:44

























          answered Nov 28 '18 at 6:37









          MasonMason

          1,9551530




          1,9551530























              0












              $begingroup$

              Hint:when $x=0, y=b $ and when $y=0,x=frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.



              But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Hint:when $x=0, y=b $ and when $y=0,x=frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.



                But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint:when $x=0, y=b $ and when $y=0,x=frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.



                  But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.






                  share|cite|improve this answer











                  $endgroup$



                  Hint:when $x=0, y=b $ and when $y=0,x=frac bm.$Now apply Pythagorean theorem to the right angled triangle constructed using the points $(0,0), (0,b)$ and $(b/m, 0)$.



                  But this will give you an answer different from the one you wrote above. Also you haven't mentioned anything about $a$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 '18 at 6:37

























                  answered Nov 28 '18 at 6:28









                  Thomas ShelbyThomas Shelby

                  2,710421




                  2,710421






























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