Find the length of the curve
y = x^2 - (1/8)lnx, 1 ≤ x ≤ e
∫[1,e] √(1 + (2x - 1/(8x)))^2 dx
how would you simplify that to a fraction
In google paste:
Find the length of the curve calculator
When you see list of results click on:
Arc Length Calculator for Curve - eMathHelp
When page be open in rectangle
Enter a function
paste:
x^2 - (1/8)lnx
Enter a lower limit: 1
Enter an upper limit: e
click on CALCULATE and follow steps
To find the length of the curve, we will use the arc length formula for a function y = f(x).
The arc length formula is given by:
L = ∫√(1 + (dy/dx)^2) dx
Let's start by finding dy/dx.
Given: y = x^2 - (1/8)lnx
To find dy/dx, we need to differentiate y with respect to x.
dy/dx = d/dx(x^2) - d/dx((1/8)lnx)
Differentiating x^2 with respect to x, we get:
d/dx(x^2) = 2x
Differentiating (1/8)lnx with respect to x using the chain rule, we get:
d/dx((1/8)lnx) = (1/8) * (1/x) * (dx/dx)
= 1/(8x)
Therefore, dy/dx = 2x - 1/(8x)
Next, we can substitute dy/dx into the arc length formula:
L = ∫√(1 + (dy/dx)^2) dx
= ∫√(1 + (2x - 1/(8x))^2) dx
Now, we can proceed to evaluate this integral over the given interval of x, which is from 1 to e.
L = ∫(1 to e) √(1 + (2x - 1/(8x))^2) dx
Since the integral might be difficult to solve analytically, it may be more convenient to use numerical methods or computer software to evaluate this integral and find the length of the curve.