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
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
$endgroup$
add a comment |
$begingroup$
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
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
$endgroup$
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
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
edited Dec 26 '18 at 2:29
Key Flex
8,58571233
8,58571233
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
asked Dec 25 '18 at 15:19
SoumeeSoumee
670614
670614
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
add a comment |
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
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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*}
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$
Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
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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
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– Satish Ramanathan
Dec 25 '18 at 16:07
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@SatishRamanathan I didn't find anything wrong with your solution. But, I wonder how this question is yielding to two different answers.
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– 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
add a comment |
$begingroup$
$$x = 2cos t -cos 2t$$
$$y = 2sin t -sin 2t$$
$$frac{dx}{dt} = -2sin t +2sin 2t$$
$$frac{dy}{dt} = 2cos t -2cos 2t$$
$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
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The answer in the book is $frac{3}{2}$
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– Soumee
Dec 25 '18 at 15:49
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It seem its (1,2) of cardioid, and it is negative value. wolframalpha.com/input/?i=x%3D2cost-cos2t,+y%3D2sint-sin2t,
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– Takahiro Waki
Dec 25 '18 at 16:18
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@TakahiroWaki Yes, you are right. The answer is $frac{-3}{2}$.
$endgroup$
– Soumee
Dec 26 '18 at 6:20
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$
I get
$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
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active
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$begingroup$
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*}
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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*}
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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*}
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
$endgroup$
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*}
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
edited Dec 26 '18 at 2:23
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
answered Dec 26 '18 at 2:12
RócherzRócherz
3,0263823
3,0263823
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$
Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$
Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$
Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
$endgroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=dfrac{2cos t-2cos2t}{-2sin t+2sin2t}=dfrac{cos2t-cos t}{sin t-sin2t}$$
$$dfrac{d^2y}{dx^2}=dfrac{(-2sin2t+sin t)(sin t-sin2t)-(cos2t-cos t)(cos t-2cos2t)}{(sin t-sin2t)^2-2(sin t-sin2t)}$$
Now $$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=dfrac{(1)(1-0)-(-1-0)(0-2(-1))}{-2(1-0)^3}=-dfrac32$$
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
edited Dec 25 '18 at 16:13
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
answered Dec 25 '18 at 15:59
Key FlexKey Flex
8,58571233
8,58571233
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$$x = 2cos t -cos 2t$$
$$y = 2sin t -sin 2t$$
$$frac{dx}{dt} = -2sin t +2sin 2t$$
$$frac{dy}{dt} = 2cos t -2cos 2t$$
$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$x = 2cos t -cos 2t$$
$$y = 2sin t -sin 2t$$
$$frac{dx}{dt} = -2sin t +2sin 2t$$
$$frac{dy}{dt} = 2cos t -2cos 2t$$
$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$x = 2cos t -cos 2t$$
$$y = 2sin t -sin 2t$$
$$frac{dx}{dt} = -2sin t +2sin 2t$$
$$frac{dy}{dt} = 2cos t -2cos 2t$$
$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$$x = 2cos t -cos 2t$$
$$y = 2sin t -sin 2t$$
$$frac{dx}{dt} = -2sin t +2sin 2t$$
$$frac{dy}{dt} = 2cos t -2cos 2t$$
$$dfrac{d^2y}{dx^2}=dfrac{(+sin 2t-sin t)(-sin t+2sin 2t)-(-cos 2t+cos t)(-cos t+2cos 2t)}{(-sin t+sin 2t)^2}cdotfrac{dt}{dx}$$
$$frac{d^2y}{dx^2}_{[frac{pi}{2}]} = -frac{3}{2}$$
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
edited Dec 26 '18 at 1:29
Rócherz
3,0263823
3,0263823
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
answered Dec 25 '18 at 15:46
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$
I get
$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$
I get
$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$
I get
$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
$$dfrac{dy}{dx}=dfrac{dfrac{dy}{dt}}{dfrac{dx}{dt}}=frac{y'_t}{x'_t}$$
$$dfrac{d^2y}{dx^2}=frac{left(frac{y'_t}{x'_t}right)'_t}{x'_t}=
frac{x'_ty''_{tt}-x''_{tt}y'_t}{(x'_t)^3}$$
I get
$$dfrac{d^2y}{dx^2}bigg|_{t=frac{pi}{2}}=-frac{3}{2}$$
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
answered Dec 25 '18 at 16:49
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
add a comment |
add a comment |
$begingroup$
Alternatively:
$$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
$endgroup$
Alternatively:
$$xleft(frac{pi}{2}right)=1; yleft(frac{pi}{2}right)=2;\
x=2cos t-cos 2t Rightarrow cos^2t-2cos t+x-1=0 Rightarrow cos t=frac12-frac{sqrt{3-2x}}2;\
x^2+y^2=5-4cos t=5-2+2sqrt{3-2x};\
2x+2yy'=frac{-2}{sqrt{3-2x}} Rightarrow y'(1)=-1;\
2+2y'^2+2yy''=frac{-2}{sqrt{(3-2x)^2}} Rightarrow y''(1)=frac{-6}4=-frac32.$$
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
answered Dec 25 '18 at 18:47
farruhotafarruhota
21.9k2842
21.9k2842
add a comment |
add a comment |
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$begingroup$
Please use MathJax to type your problem. Don't give us the hand-written solutions.
$endgroup$
– jayant98
Dec 25 '18 at 15:24