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
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)) ?
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
To find the arc length of a curve, you will need to use the formula:
L = ∫[a,b] √[1 + (dy/dx)^2] dx
In this case, you need to find dy/dx and then evaluate the integral over the given range.
Let's start by finding dy/dx. We can use the Chain Rule to differentiate the equation:
Given:
x = 3y^(4/3) - 3/32y^(2/3)
Differentiating both sides with respect to x:
1 = 12y^(1/3)(dy/dx) - 3/16y^(-1/3)(dy/dx)
Now, solve for dy/dx by isolating the term:
dy/dx = (1 + 3/16y^(-1/3)) / (12y^(1/3))
Next, we will substitute this value into the arc length formula. But before that, let's simplify the expression:
dy/dx = (16y^(-1/3) + 3) / (48y^(1/3))
Now you can plug dy/dx into the arc length formula:
L = ∫[a,b] √[1 + ((16y^(-1/3) + 3) / (48y^(1/3)))^2] dx
Where [a, b] is the given range for y, which is -64 to 64.
To evaluate this integral, you will need to use numerical methods or a computer software capable of solving definite integrals.