Asked by Pegs
Can someone please help me find the arc length of this curve?
x = 3y^(4/3) - 3/32y^(2/3)
y is between -64 and 64
x = 3y^(4/3) - 3/32y^(2/3)
y is between -64 and 64
Answers
Answered by
Reiny
if y = f(x)
then length of y from a to b is
[integral from a to b] √(1 + (dy/dx)^2) dx
You have a nasty equation which is not a function.
You can do two things
1. find dy/dx implicity or
2. take the inverse to get it into y = ...
In that case you will have to change the "y is between -64 to 64" to "x is between ...."
I am also not sure about the last term,
is it (-3/32)y^(2/3) or -3/(32y^(2/3)) ?
then length of y from a to b is
[integral from a to b] √(1 + (dy/dx)^2) dx
You have a nasty equation which is not a function.
You can do two things
1. find dy/dx implicity or
2. take the inverse to get it into y = ...
In that case you will have to change the "y is between -64 to 64" to "x is between ...."
I am also not sure about the last term,
is it (-3/32)y^(2/3) or -3/(32y^(2/3)) ?
Answered by
Pegs
Oh wow...so I need to find it implicitly first??? How confusing. I was just trying to figure out the arc length using the first equation and I assumed it didn't matter that the x and y terms were switched
Yes, the last term is (-3/32)y^(2/3), the fraction before y is just a constant
Yes, the last term is (-3/32)y^(2/3), the fraction before y is just a constant
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