Calculus Help Finding the Arc Length












1












$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







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  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38
















1












$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38














1












1








1





$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







share|cite|improve this question











$endgroup$




So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length




calculus integration definite-integrals arc-length






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 7:20









Martin Rosenau

1,1661310




1,1661310










asked Dec 17 '18 at 5:53









krauser126krauser126

474




474












  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38


















  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38
















$begingroup$
$ln3+1$ comes from what Robert Z suggests
$endgroup$
– Sameer Baheti
Dec 17 '18 at 6:38




$begingroup$
$ln3+1$ comes from what Robert Z suggests
$endgroup$
– Sameer Baheti
Dec 17 '18 at 6:38










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52


















1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52
















1












1








1





$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$



You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 6:36

























answered Dec 17 '18 at 6:15









Robert ZRobert Z

98.6k1068139




98.6k1068139












  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52




















  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52


















$begingroup$
So what is the final result?
$endgroup$
– Robert Z
Dec 17 '18 at 6:45




$begingroup$
So what is the final result?
$endgroup$
– Robert Z
Dec 17 '18 at 6:45












$begingroup$
Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
$endgroup$
– krauser126
Dec 17 '18 at 6:48




$begingroup$
Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
$endgroup$
– krauser126
Dec 17 '18 at 6:48












$begingroup$
"completing the square doesn't work here" why?
$endgroup$
– Robert Z
Dec 17 '18 at 6:49




$begingroup$
"completing the square doesn't work here" why?
$endgroup$
– Robert Z
Dec 17 '18 at 6:49












$begingroup$
Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
$endgroup$
– krauser126
Dec 17 '18 at 6:50




$begingroup$
Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
$endgroup$
– krauser126
Dec 17 '18 at 6:50












$begingroup$
@krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
$endgroup$
– Dylan
Dec 17 '18 at 6:52






$begingroup$
@krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
$endgroup$
– Dylan
Dec 17 '18 at 6:52




















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