Mixing
Finding length of curve $y^2 = 64(x+3)^3$ for $0 le x le 3$
$begingroup$
Not getting the right answer for this, can someone point me to where I'm going wrong?
calculus integration algebraic-curves
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
$endgroup$
add a comment |
$begingroup$
Not getting the right answer for this, can someone point me to where I'm going wrong?
calculus integration algebraic-curves
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
$endgroup$
1
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
1
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12
add a comment |
$begingroup$
Not getting the right answer for this, can someone point me to where I'm going wrong?
calculus integration algebraic-curves
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
$endgroup$
Not getting the right answer for this, can someone point me to where I'm going wrong?
calculus integration algebraic-curves
calculus integration algebraic-curves
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
edited Feb 7 '16 at 6:02
Krunal Rindani
143111
143111
asked Feb 7 '16 at 5:39
windy401windy401
63021224
asked Feb 7 '16 at 5:39
windy401windy401
63021224
asked Feb 7 '16 at 5:39
windy401windy401
63021224
63021224
1
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
1
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12
add a comment |
1
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
1
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12
1
1
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
1
1
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You derived $f'$ directly, but I suggest a different method. $(y^2)'=2yy'$, so $2yy'=64cdot 3(x+3)^2$. Then
begin{align}
(y')^2 &= left(frac{64cdot 3(x+3)^2}{2y}right)^2\
&=frac{64^2 3^2 (x+3)^4}{4y^2}\
&=frac{64^2 3^2 (x+3)^4}{4cdot 64(x+3)^3}\
&=144(x+3).
end{align}
Therefore $1+(f'(x))^2 =144x+433$. Of course, what you did is almost correct, except a mistake in calculation.
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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votes
$begingroup$
You derived $f'$ directly, but I suggest a different method. $(y^2)'=2yy'$, so $2yy'=64cdot 3(x+3)^2$. Then
begin{align}
(y')^2 &= left(frac{64cdot 3(x+3)^2}{2y}right)^2\
&=frac{64^2 3^2 (x+3)^4}{4y^2}\
&=frac{64^2 3^2 (x+3)^4}{4cdot 64(x+3)^3}\
&=144(x+3).
end{align}
Therefore $1+(f'(x))^2 =144x+433$. Of course, what you did is almost correct, except a mistake in calculation.
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
$endgroup$
add a comment |
$begingroup$
You derived $f'$ directly, but I suggest a different method. $(y^2)'=2yy'$, so $2yy'=64cdot 3(x+3)^2$. Then
begin{align}
(y')^2 &= left(frac{64cdot 3(x+3)^2}{2y}right)^2\
&=frac{64^2 3^2 (x+3)^4}{4y^2}\
&=frac{64^2 3^2 (x+3)^4}{4cdot 64(x+3)^3}\
&=144(x+3).
end{align}
Therefore $1+(f'(x))^2 =144x+433$. Of course, what you did is almost correct, except a mistake in calculation.
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
$endgroup$
add a comment |
$begingroup$
You derived $f'$ directly, but I suggest a different method. $(y^2)'=2yy'$, so $2yy'=64cdot 3(x+3)^2$. Then
begin{align}
(y')^2 &= left(frac{64cdot 3(x+3)^2}{2y}right)^2\
&=frac{64^2 3^2 (x+3)^4}{4y^2}\
&=frac{64^2 3^2 (x+3)^4}{4cdot 64(x+3)^3}\
&=144(x+3).
end{align}
Therefore $1+(f'(x))^2 =144x+433$. Of course, what you did is almost correct, except a mistake in calculation.
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
$endgroup$
You derived $f'$ directly, but I suggest a different method. $(y^2)'=2yy'$, so $2yy'=64cdot 3(x+3)^2$. Then
begin{align}
(y')^2 &= left(frac{64cdot 3(x+3)^2}{2y}right)^2\
&=frac{64^2 3^2 (x+3)^4}{4y^2}\
&=frac{64^2 3^2 (x+3)^4}{4cdot 64(x+3)^3}\
&=144(x+3).
end{align}
Therefore $1+(f'(x))^2 =144x+433$. Of course, what you did is almost correct, except a mistake in calculation.
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
answered Feb 7 '16 at 6:11
choco_addictedchoco_addicted
8,09261947
8,09261947
add a comment |
add a comment |
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1
$begingroup$
$1+144(x+3)=1+144x+432=144x+433$, not $144x+37$.
$endgroup$
– choco_addicted
Feb 7 '16 at 6:02
1
$begingroup$
It seems that you say that $1+144(x+3)=1+144x+36$ which is not correct.
$endgroup$
– mickep
Feb 7 '16 at 6:02
$begingroup$
Oh ok forgot to switch that part when I fixed another mistake. Thanks for saving me from my late night mistake guys.
$endgroup$
– windy401
Feb 7 '16 at 6:12