Finding the circumference of a double cycloid












1












$begingroup$


So I have the following problem:



Find the the circumference of a melon, if the boundary line of the melon is a double cycloid and following equation describe the boundary line of the melon.



{x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} and



{x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))}



I have to give the following:



1) integrand for the one side of the melon: $$ g(t)= $$
2)indefinite integral for the half of the melon's circumference: $$ C(t)= ... + C $$
3) the circumference of the melon: $$ C=$$



I don't know how to find the upper and lower bound. So it would be very helpful, if somebody could give me the equations to draw a graph in geogebra or in wolframalpha.



What I have found out:
1) integrand for the one side of the melon: $$ g(t)=sqrt{50-50cos(t)} $$



3) the circumference of the melon: it should be $$ C=80$$, but if I take t from $ 0$ to $ 2pi $ $ g(t)=sqrt{50-50cos(t)} $ the answer comes 0. What is wrong?






calculus integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So I have the following problem:



    Find the the circumference of a melon, if the boundary line of the melon is a double cycloid and following equation describe the boundary line of the melon.



    {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} and



    {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))}



    I have to give the following:



    1) integrand for the one side of the melon: $$ g(t)= $$
    2)indefinite integral for the half of the melon's circumference: $$ C(t)= ... + C $$
    3) the circumference of the melon: $$ C=$$



    I don't know how to find the upper and lower bound. So it would be very helpful, if somebody could give me the equations to draw a graph in geogebra or in wolframalpha.



    What I have found out:
    1) integrand for the one side of the melon: $$ g(t)=sqrt{50-50cos(t)} $$



    3) the circumference of the melon: it should be $$ C=80$$, but if I take t from $ 0$ to $ 2pi $ $ g(t)=sqrt{50-50cos(t)} $ the answer comes 0. What is wrong?






    calculus integration definite-integrals indefinite-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So I have the following problem:



      Find the the circumference of a melon, if the boundary line of the melon is a double cycloid and following equation describe the boundary line of the melon.



      {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} and



      {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))}



      I have to give the following:



      1) integrand for the one side of the melon: $$ g(t)= $$
      2)indefinite integral for the half of the melon's circumference: $$ C(t)= ... + C $$
      3) the circumference of the melon: $$ C=$$



      I don't know how to find the upper and lower bound. So it would be very helpful, if somebody could give me the equations to draw a graph in geogebra or in wolframalpha.



      What I have found out:
      1) integrand for the one side of the melon: $$ g(t)=sqrt{50-50cos(t)} $$



      3) the circumference of the melon: it should be $$ C=80$$, but if I take t from $ 0$ to $ 2pi $ $ g(t)=sqrt{50-50cos(t)} $ the answer comes 0. What is wrong?






      calculus integration definite-integrals indefinite-integrals







      share|cite|improve this question











      $endgroup$




      So I have the following problem:



      Find the the circumference of a melon, if the boundary line of the melon is a double cycloid and following equation describe the boundary line of the melon.



      {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} and



      {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))}



      I have to give the following:



      1) integrand for the one side of the melon: $$ g(t)= $$
      2)indefinite integral for the half of the melon's circumference: $$ C(t)= ... + C $$
      3) the circumference of the melon: $$ C=$$



      I don't know how to find the upper and lower bound. So it would be very helpful, if somebody could give me the equations to draw a graph in geogebra or in wolframalpha.



      What I have found out:
      1) integrand for the one side of the melon: $$ g(t)=sqrt{50-50cos(t)} $$



      3) the circumference of the melon: it should be $$ C=80$$, but if I take t from $ 0$ to $ 2pi $ $ g(t)=sqrt{50-50cos(t)} $ the answer comes 0. What is wrong?






      calculus integration definite-integrals indefinite-integrals




      calculus integration definite-integrals indefinite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 10:18







      Student123

















      asked Dec 3 '18 at 14:16









      Student123Student123

      536




      536






















          1 Answer
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          oldest

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          1












          $begingroup$

          This might help, so I am going to put it here.
          Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=sqrt{left(frac{dx}{dt}right)^{2}+left(frac{dy}{dt}right)^{2}}dt=sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=rcdotsqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(theta)=1-2sin^{2}(theta/2)$ Factoring the radical gets $sqrt{2(1-cos(t)).}$ Applying the identity gets $sqrt{2(1-(1-2sin^{2}(t/2))}=sqrt{4sin^{2}(t/2)}=2sin(frac{t}{2})$. $$f(t)=rcdot2sin(frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$int_{0}^{2pi}rcdot2sin(frac{t}{2})dt=-rcdot4cos(frac{t}{2})Big|_{0}^{2pi}=8r$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            As far as i can tell, no. Half of the circumference is what I show in the answer.
            $endgroup$
            – Narlin
            Dec 3 '18 at 15:14










          • $begingroup$
            where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
            $endgroup$
            – Student123
            Dec 6 '18 at 10:50










          • $begingroup$
            sosmath.com/trig/douangl/douangl.html about half way down the page.
            $endgroup$
            – Narlin
            Dec 6 '18 at 20:55











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          This might help, so I am going to put it here.
          Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=sqrt{left(frac{dx}{dt}right)^{2}+left(frac{dy}{dt}right)^{2}}dt=sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=rcdotsqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(theta)=1-2sin^{2}(theta/2)$ Factoring the radical gets $sqrt{2(1-cos(t)).}$ Applying the identity gets $sqrt{2(1-(1-2sin^{2}(t/2))}=sqrt{4sin^{2}(t/2)}=2sin(frac{t}{2})$. $$f(t)=rcdot2sin(frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$int_{0}^{2pi}rcdot2sin(frac{t}{2})dt=-rcdot4cos(frac{t}{2})Big|_{0}^{2pi}=8r$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            As far as i can tell, no. Half of the circumference is what I show in the answer.
            $endgroup$
            – Narlin
            Dec 3 '18 at 15:14










          • $begingroup$
            where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
            $endgroup$
            – Student123
            Dec 6 '18 at 10:50










          • $begingroup$
            sosmath.com/trig/douangl/douangl.html about half way down the page.
            $endgroup$
            – Narlin
            Dec 6 '18 at 20:55
















          1












          $begingroup$

          This might help, so I am going to put it here.
          Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=sqrt{left(frac{dx}{dt}right)^{2}+left(frac{dy}{dt}right)^{2}}dt=sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=rcdotsqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(theta)=1-2sin^{2}(theta/2)$ Factoring the radical gets $sqrt{2(1-cos(t)).}$ Applying the identity gets $sqrt{2(1-(1-2sin^{2}(t/2))}=sqrt{4sin^{2}(t/2)}=2sin(frac{t}{2})$. $$f(t)=rcdot2sin(frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$int_{0}^{2pi}rcdot2sin(frac{t}{2})dt=-rcdot4cos(frac{t}{2})Big|_{0}^{2pi}=8r$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            As far as i can tell, no. Half of the circumference is what I show in the answer.
            $endgroup$
            – Narlin
            Dec 3 '18 at 15:14










          • $begingroup$
            where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
            $endgroup$
            – Student123
            Dec 6 '18 at 10:50










          • $begingroup$
            sosmath.com/trig/douangl/douangl.html about half way down the page.
            $endgroup$
            – Narlin
            Dec 6 '18 at 20:55














          1












          1








          1





          $begingroup$

          This might help, so I am going to put it here.
          Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=sqrt{left(frac{dx}{dt}right)^{2}+left(frac{dy}{dt}right)^{2}}dt=sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=rcdotsqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(theta)=1-2sin^{2}(theta/2)$ Factoring the radical gets $sqrt{2(1-cos(t)).}$ Applying the identity gets $sqrt{2(1-(1-2sin^{2}(t/2))}=sqrt{4sin^{2}(t/2)}=2sin(frac{t}{2})$. $$f(t)=rcdot2sin(frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$int_{0}^{2pi}rcdot2sin(frac{t}{2})dt=-rcdot4cos(frac{t}{2})Big|_{0}^{2pi}=8r$$






          share|cite|improve this answer









          $endgroup$



          This might help, so I am going to put it here.
          Calculate the length of a cycloid curve with a radius of 5 inches. The cycloid parametric equations are $x(t)=r(t-sin(t))$ and $y(t)=r(1-cos(t))$. $$f(t)=sqrt{left(frac{dx}{dt}right)^{2}+left(frac{dy}{dt}right)^{2}}dt=sqrt{r^{2}(t-sin(t))^{2}+r^{2}(1-cos(t))^{2}}dt$$ $$=rcdotsqrt{2-2cos(t)}dt$$ Normally, I would stop at this point and integrate, but this can be beneficially simplified using a trig identity. $cos(theta)=1-2sin^{2}(theta/2)$ Factoring the radical gets $sqrt{2(1-cos(t)).}$ Applying the identity gets $sqrt{2(1-(1-2sin^{2}(t/2))}=sqrt{4sin^{2}(t/2)}=2sin(frac{t}{2})$. $$f(t)=rcdot2sin(frac{t}{2})dt$$ This last eqn is a great deal easier to integrate. $$int_{0}^{2pi}rcdot2sin(frac{t}{2})dt=-rcdot4cos(frac{t}{2})Big|_{0}^{2pi}=8r$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 14:51









          NarlinNarlin

          524310




          524310








          • 1




            $begingroup$
            As far as i can tell, no. Half of the circumference is what I show in the answer.
            $endgroup$
            – Narlin
            Dec 3 '18 at 15:14










          • $begingroup$
            where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
            $endgroup$
            – Student123
            Dec 6 '18 at 10:50










          • $begingroup$
            sosmath.com/trig/douangl/douangl.html about half way down the page.
            $endgroup$
            – Narlin
            Dec 6 '18 at 20:55














          • 1




            $begingroup$
            As far as i can tell, no. Half of the circumference is what I show in the answer.
            $endgroup$
            – Narlin
            Dec 3 '18 at 15:14










          • $begingroup$
            where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
            $endgroup$
            – Student123
            Dec 6 '18 at 10:50










          • $begingroup$
            sosmath.com/trig/douangl/douangl.html about half way down the page.
            $endgroup$
            – Narlin
            Dec 6 '18 at 20:55








          1




          1




          $begingroup$
          As far as i can tell, no. Half of the circumference is what I show in the answer.
          $endgroup$
          – Narlin
          Dec 3 '18 at 15:14




          $begingroup$
          As far as i can tell, no. Half of the circumference is what I show in the answer.
          $endgroup$
          – Narlin
          Dec 3 '18 at 15:14












          $begingroup$
          where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
          $endgroup$
          – Student123
          Dec 6 '18 at 10:50




          $begingroup$
          where do you get the $cos(theta)=1-2sin^{2}(theta/2)$ from?
          $endgroup$
          – Student123
          Dec 6 '18 at 10:50












          $begingroup$
          sosmath.com/trig/douangl/douangl.html about half way down the page.
          $endgroup$
          – Narlin
          Dec 6 '18 at 20:55




          $begingroup$
          sosmath.com/trig/douangl/douangl.html about half way down the page.
          $endgroup$
          – Narlin
          Dec 6 '18 at 20:55


















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