If $x=2cos t-cos 2t $, $y=2sin t -sin 2t$, find the value of $frac{d^2y}{dx^2}$ when $t= pi /2$












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Question $boldsymbol 7$. If $x=2cos t-cos 2t $, $y=2sin t -sin 2t$, find the value of $frac{d^2y}{dx^2}$ when $t= pi /2$



Solution:



begin{align*}
y &= 2sin t -sin 2t \
frac{dy}{dt} &= 2cos t -2cos 2t \
frac{dx}{dt} &=-2sin t +2sin 2t \
frac{dy}{dx} &=frac{dy/dt}{dx/dt} =frac{cos t-cos 2t}{-sin t +sin 2t} \
frac{d^2y}{dx^2} &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) (-sin t -2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{(sin t -sin 2t) (sin t +2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{sin^2 t +2sin t sin 2t -sin t sin 2t -2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{1 +2(cos^2 2t -sin^2 2t) +sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx}
end{align*}



Have I done any mistake so far?






calculus







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  • 2




    $begingroup$
    Please use MathJax to type your problem. Don't give us the hand-written solutions.
    $endgroup$
    – jayant98
    Dec 25 '18 at 15:24
















0












$begingroup$


Question $boldsymbol 7$. If $x=2cos t-cos 2t $, $y=2sin t -sin 2t$, find the value of $frac{d^2y}{dx^2}$ when $t= pi /2$



Solution:



begin{align*}
y &= 2sin t -sin 2t \
frac{dy}{dt} &= 2cos t -2cos 2t \
frac{dx}{dt} &=-2sin t +2sin 2t \
frac{dy}{dx} &=frac{dy/dt}{dx/dt} =frac{cos t-cos 2t}{-sin t +sin 2t} \
frac{d^2y}{dx^2} &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) (-sin t -2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{(sin t -sin 2t) (sin t +2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{sin^2 t +2sin t sin 2t -sin t sin 2t -2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{1 +2(cos^2 2t -sin^2 2t) +sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx}
end{align*}



Have I done any mistake so far?






calculus







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  • 2




    $begingroup$
    Please use MathJax to type your problem. Don't give us the hand-written solutions.
    $endgroup$
    – jayant98
    Dec 25 '18 at 15:24














0












0








0


1



$begingroup$


Question $boldsymbol 7$. If $x=2cos t-cos 2t $, $y=2sin t -sin 2t$, find the value of $frac{d^2y}{dx^2}$ when $t= pi /2$



Solution:



begin{align*}
y &= 2sin t -sin 2t \
frac{dy}{dt} &= 2cos t -2cos 2t \
frac{dx}{dt} &=-2sin t +2sin 2t \
frac{dy}{dx} &=frac{dy/dt}{dx/dt} =frac{cos t-cos 2t}{-sin t +sin 2t} \
frac{d^2y}{dx^2} &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) (-sin t -2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{(sin t -sin 2t) (sin t +2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{sin^2 t +2sin t sin 2t -sin t sin 2t -2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{1 +2(cos^2 2t -sin^2 2t) +sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx}
end{align*}



Have I done any mistake so far?






calculus







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$endgroup$




Question $boldsymbol 7$. If $x=2cos t-cos 2t $, $y=2sin t -sin 2t$, find the value of $frac{d^2y}{dx^2}$ when $t= pi /2$



Solution:



begin{align*}
y &= 2sin t -sin 2t \
frac{dy}{dt} &= 2cos t -2cos 2t \
frac{dx}{dt} &=-2sin t +2sin 2t \
frac{dy}{dx} &=frac{dy/dt}{dx/dt} =frac{cos t-cos 2t}{-sin t +sin 2t} \
frac{d^2y}{dx^2} &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) (-sin t -2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{(sin t -sin 2t) (sin t +2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{sin^2 t +2sin t sin 2t -sin t sin 2t -2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{1 +2(cos^2 2t -sin^2 2t) +sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx}
end{align*}



Have I done any mistake so far?






calculus




calculus






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edited Dec 26 '18 at 2:29









Key Flex

8,10761233




8,10761233










asked Dec 25 '18 at 15:19









SoumeeSoumee

668614




668614








  • 2




    $begingroup$
    Please use MathJax to type your problem. Don't give us the hand-written solutions.
    $endgroup$
    – jayant98
    Dec 25 '18 at 15:24














  • 2




    $begingroup$
    Please use MathJax to type your problem. Don't give us the hand-written solutions.
    $endgroup$
    – jayant98
    Dec 25 '18 at 15:24








2




2




$begingroup$
Please use MathJax to type your problem. Don't give us the hand-written solutions.
$endgroup$
– jayant98
Dec 25 '18 at 15:24




$begingroup$
Please use MathJax to type your problem. Don't give us the hand-written solutions.
$endgroup$
– jayant98
Dec 25 '18 at 15:24










5 Answers
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Mistake: $displaystyle frac{d}{dt} (cos t -cos 2t) =-sin t color{red}{boldsymbol +}2sin 2t$.



begin{align*}
frac{d^2y}{dx^2}
&=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&=frac{(sin 2t -sin t) (-sin t color{red}{boldsymbol +}2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{(sin t -sin 2t) (sin t color{red}{boldsymbol -}2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{sin^2 t color{red}{boldsymbol -}2sin t sin 2t -sin t sin 2t color{red}{boldsymbol +}2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{1 +2(cos^2 2t color{red}{boldsymbol +}sin^2 2t) color{red}{boldsymbol{-3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= frac{color{red}{boldsymbol{3 -3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
&= ldots
end{align*}






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  • $begingroup$
    Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
    $endgroup$
    – Soumee
    Dec 26 '18 at 6:17



















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$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$



Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$






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  • $begingroup$
    could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
    $endgroup$
    – Satish Ramanathan
    Dec 25 '18 at 16:07










  • $begingroup$
    @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
    $endgroup$
    – Key Flex
    Dec 25 '18 at 16:22






  • 1




    $begingroup$
    You are right in your answer. I made a mistake. I am upvoting yours.
    $endgroup$
    – Satish Ramanathan
    Dec 25 '18 at 16:40



















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$$x = 2cos t -cos 2t$$



$$y = 2sin t -sin 2t$$



$$frac{dx}{dt} = -2sin t +2sin 2t$$



$$frac{dy}{dt} = 2cos t -2cos 2t$$



$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$






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  • $begingroup$
    The answer in the book is $frac{3}{2}$
    $endgroup$
    – Soumee
    Dec 25 '18 at 15:49










  • $begingroup$
    It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
    $endgroup$
    – Takahiro Waki
    Dec 25 '18 at 16:18










  • $begingroup$
    @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
    $endgroup$
    – Soumee
    Dec 26 '18 at 6:20





















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$begingroup$

$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$



I get



$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$






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    1












    $begingroup$

    Alternatively:
    $$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
    x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
    x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
    2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
    2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$






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      5 Answers
      5






      active

      oldest

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      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Mistake: $displaystyle frac{d}{dt} (cos t -cos 2t) =-sin t color{red}{boldsymbol +}2sin 2t$.



      begin{align*}
      frac{d^2y}{dx^2}
      &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) (-sin t color{red}{boldsymbol +}2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{(sin t -sin 2t) (sin t color{red}{boldsymbol -}2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{sin^2 t color{red}{boldsymbol -}2sin t sin 2t -sin t sin 2t color{red}{boldsymbol +}2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{1 +2(cos^2 2t color{red}{boldsymbol +}sin^2 2t) color{red}{boldsymbol{-3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{color{red}{boldsymbol{3 -3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= ldots
      end{align*}






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:17
















      1












      $begingroup$

      Mistake: $displaystyle frac{d}{dt} (cos t -cos 2t) =-sin t color{red}{boldsymbol +}2sin 2t$.



      begin{align*}
      frac{d^2y}{dx^2}
      &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) (-sin t color{red}{boldsymbol +}2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{(sin t -sin 2t) (sin t color{red}{boldsymbol -}2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{sin^2 t color{red}{boldsymbol -}2sin t sin 2t -sin t sin 2t color{red}{boldsymbol +}2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{1 +2(cos^2 2t color{red}{boldsymbol +}sin^2 2t) color{red}{boldsymbol{-3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{color{red}{boldsymbol{3 -3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= ldots
      end{align*}






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:17














      1












      1








      1





      $begingroup$

      Mistake: $displaystyle frac{d}{dt} (cos t -cos 2t) =-sin t color{red}{boldsymbol +}2sin 2t$.



      begin{align*}
      frac{d^2y}{dx^2}
      &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) (-sin t color{red}{boldsymbol +}2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{(sin t -sin 2t) (sin t color{red}{boldsymbol -}2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{sin^2 t color{red}{boldsymbol -}2sin t sin 2t -sin t sin 2t color{red}{boldsymbol +}2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{1 +2(cos^2 2t color{red}{boldsymbol +}sin^2 2t) color{red}{boldsymbol{-3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{color{red}{boldsymbol{3 -3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= ldots
      end{align*}






      share|cite|improve this answer











      $endgroup$



      Mistake: $displaystyle frac{d}{dt} (cos t -cos 2t) =-sin t color{red}{boldsymbol +}2sin 2t$.



      begin{align*}
      frac{d^2y}{dx^2}
      &=frac{d}{dt} left(frac{cos t-cos 2t}{-sin t +sin 2t}right) cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) frac{d}{dt}(cos t -cos 2t) -(cos t -cos 2t) frac{d}{dt}(sin 2t -sin t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &=frac{(sin 2t -sin t) (-sin t color{red}{boldsymbol +}2sin 2t) -(cos t -cos 2t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{(sin t -sin 2t) (sin t color{red}{boldsymbol -}2sin 2t) +(cos 2t -cos t) (2cos 2t -cos t)}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{sin^2 t color{red}{boldsymbol -}2sin t sin 2t -sin t sin 2t color{red}{boldsymbol +}2sin^2 2t +ldots \ ldots +2cos^2 2t -cos t cos 2t -2cos t cos 2t +cos^2 t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{1 +2(cos^2 2t color{red}{boldsymbol +}sin^2 2t) color{red}{boldsymbol{-3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= frac{color{red}{boldsymbol{3 -3}}sin t sin 2t -3cos t cos 2t}{(sin 2t -sin t)^2} cdot frac{dt}{dx} \
      &= ldots
      end{align*}







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 26 '18 at 2:23

























      answered Dec 26 '18 at 2:12









      RócherzRócherz

      2,7762721




      2,7762721












      • $begingroup$
        Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:17


















      • $begingroup$
        Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:17
















      $begingroup$
      Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
      $endgroup$
      – Soumee
      Dec 26 '18 at 6:17




      $begingroup$
      Thanks for specifically pointing out my mistake. Though I would equally thank Key Flex for editing the question. Since I have to mark just one answer as the correct answer I would like to mark Rocherz's answer as the correct one as I really wanted to know what mistake I had done. I am thankful to others as well for contributing to the solution of the problem.
      $endgroup$
      – Soumee
      Dec 26 '18 at 6:17











      2












      $begingroup$

      $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
      $$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$



      Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:07










      • $begingroup$
        @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
        $endgroup$
        – Key Flex
        Dec 25 '18 at 16:22






      • 1




        $begingroup$
        You are right in your answer. I made a mistake. I am upvoting yours.
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:40
















      2












      $begingroup$

      $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
      $$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$



      Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:07










      • $begingroup$
        @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
        $endgroup$
        – Key Flex
        Dec 25 '18 at 16:22






      • 1




        $begingroup$
        You are right in your answer. I made a mistake. I am upvoting yours.
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:40














      2












      2








      2





      $begingroup$

      $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
      $$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$



      Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$






      share|cite|improve this answer











      $endgroup$



      $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
      $$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$



      Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 25 '18 at 16:13

























      answered Dec 25 '18 at 15:59









      Key FlexKey Flex

      8,10761233




      8,10761233












      • $begingroup$
        could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:07










      • $begingroup$
        @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
        $endgroup$
        – Key Flex
        Dec 25 '18 at 16:22






      • 1




        $begingroup$
        You are right in your answer. I made a mistake. I am upvoting yours.
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:40


















      • $begingroup$
        could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:07










      • $begingroup$
        @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
        $endgroup$
        – Key Flex
        Dec 25 '18 at 16:22






      • 1




        $begingroup$
        You are right in your answer. I made a mistake. I am upvoting yours.
        $endgroup$
        – Satish Ramanathan
        Dec 25 '18 at 16:40
















      $begingroup$
      could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
      $endgroup$
      – Satish Ramanathan
      Dec 25 '18 at 16:07




      $begingroup$
      could you tell me if there is conceptually anything wrong and I cannot find anything wrong unless you tell me if something is basically wrong
      $endgroup$
      – Satish Ramanathan
      Dec 25 '18 at 16:07












      $begingroup$
      @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
      $endgroup$
      – Key Flex
      Dec 25 '18 at 16:22




      $begingroup$
      @SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
      $endgroup$
      – Key Flex
      Dec 25 '18 at 16:22




      1




      1




      $begingroup$
      You are right in your answer. I made a mistake. I am upvoting yours.
      $endgroup$
      – Satish Ramanathan
      Dec 25 '18 at 16:40




      $begingroup$
      You are right in your answer. I made a mistake. I am upvoting yours.
      $endgroup$
      – Satish Ramanathan
      Dec 25 '18 at 16:40











      2












      $begingroup$

      $$x = 2cos t -cos 2t$$



      $$y = 2sin t -sin 2t$$



      $$frac{dx}{dt} = -2sin t +2sin 2t$$



      $$frac{dy}{dt} = 2cos t -2cos 2t$$



      $$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
      $$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        The answer in the book is $frac{3}{2}$
        $endgroup$
        – Soumee
        Dec 25 '18 at 15:49










      • $begingroup$
        It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
        $endgroup$
        – Takahiro Waki
        Dec 25 '18 at 16:18










      • $begingroup$
        @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:20


















      2












      $begingroup$

      $$x = 2cos t -cos 2t$$



      $$y = 2sin t -sin 2t$$



      $$frac{dx}{dt} = -2sin t +2sin 2t$$



      $$frac{dy}{dt} = 2cos t -2cos 2t$$



      $$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
      $$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        The answer in the book is $frac{3}{2}$
        $endgroup$
        – Soumee
        Dec 25 '18 at 15:49










      • $begingroup$
        It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
        $endgroup$
        – Takahiro Waki
        Dec 25 '18 at 16:18










      • $begingroup$
        @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:20
















      2












      2








      2





      $begingroup$

      $$x = 2cos t -cos 2t$$



      $$y = 2sin t -sin 2t$$



      $$frac{dx}{dt} = -2sin t +2sin 2t$$



      $$frac{dy}{dt} = 2cos t -2cos 2t$$



      $$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
      $$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$






      share|cite|improve this answer











      $endgroup$



      $$x = 2cos t -cos 2t$$



      $$y = 2sin t -sin 2t$$



      $$frac{dx}{dt} = -2sin t +2sin 2t$$



      $$frac{dy}{dt} = 2cos t -2cos 2t$$



      $$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
      $$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 26 '18 at 1:29









      Rócherz

      2,7762721




      2,7762721










      answered Dec 25 '18 at 15:46









      Satish RamanathanSatish Ramanathan

      9,92531323




      9,92531323












      • $begingroup$
        The answer in the book is $frac{3}{2}$
        $endgroup$
        – Soumee
        Dec 25 '18 at 15:49










      • $begingroup$
        It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
        $endgroup$
        – Takahiro Waki
        Dec 25 '18 at 16:18










      • $begingroup$
        @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:20




















      • $begingroup$
        The answer in the book is $frac{3}{2}$
        $endgroup$
        – Soumee
        Dec 25 '18 at 15:49










      • $begingroup$
        It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
        $endgroup$
        – Takahiro Waki
        Dec 25 '18 at 16:18










      • $begingroup$
        @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
        $endgroup$
        – Soumee
        Dec 26 '18 at 6:20


















      $begingroup$
      The answer in the book is $frac{3}{2}$
      $endgroup$
      – Soumee
      Dec 25 '18 at 15:49




      $begingroup$
      The answer in the book is $frac{3}{2}$
      $endgroup$
      – Soumee
      Dec 25 '18 at 15:49












      $begingroup$
      It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
      $endgroup$
      – Takahiro Waki
      Dec 25 '18 at 16:18




      $begingroup$
      It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
      $endgroup$
      – Takahiro Waki
      Dec 25 '18 at 16:18












      $begingroup$
      @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
      $endgroup$
      – Soumee
      Dec 26 '18 at 6:20






      $begingroup$
      @TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
      $endgroup$
      – Soumee
      Dec 26 '18 at 6:20













      1












      $begingroup$

      $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
      $$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
      frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$



      I get



      $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
        $$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
        frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$



        I get



        $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
          $$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
          frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$



          I get



          $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$






          share|cite|improve this answer









          $endgroup$



          $$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
          $$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
          frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$



          I get



          $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 16:49









          Aleksas DomarkasAleksas Domarkas

          1,24916




          1,24916























              1












              $begingroup$

              Alternatively:
              $$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
              x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
              x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
              2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
              2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Alternatively:
                $$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
                x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
                x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
                2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
                2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Alternatively:
                  $$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
                  x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
                  x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
                  2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
                  2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$






                  share|cite|improve this answer









                  $endgroup$



                  Alternatively:
                  $$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
                  x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
                  x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
                  2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
                  2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 18:47









                  farruhotafarruhota

                  20.2k2738




                  20.2k2738






























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