Find the exact length of the curve.
x=(y^4)/8+1/(4y^2), 1 ≤ y ≤ 2.
y' = y^3/2 - 1/(2y^3)
1+y'^2 = (y^6+1)^2/4y^6
so, the arc length is
s = 1/2 ∫[1,2] y^3 + 1/y^3 dy = 33/16
Well, I could try to measure it with a ruler, but I'm afraid my arms are too short.
To find the exact length of the curve, we will use the arc length formula:
L = ∫ sqrt(1 + (dx/dy)^2) dy
First, let's find dy/dx by taking the derivative of x with respect to y:
x = (y^4)/8 + 1/(4y^2)
To simplify, we can rewrite the equation as:
8x = y^4 + 2/y^2
Differentiating both sides with respect to y:
8(dx/dy) = 4y^3 - 4/y^3
Now solve for (dx/dy):
(dx/dy) = (4y^3 - 4/y^3) / 8
Simplifying further:
(dx/dy) = (y^6 - 1) / (2y^3)
Now, substitute this expression into the arc length formula:
L = ∫ sqrt(1 + (dx/dy)^2) dy
L = ∫ sqrt(1 + ((y^6 - 1) / (2y^3))^2) dy
To evaluate this integral, we need to simplify the expression under the square root:
sqrt(1 + ((y^6 - 1) / (2y^3))^2) = sqrt(1 + (y^12 - 2y^6 + 1) / (4y^6)) = sqrt((4y^6 + y^12 - 2y^6 + 1) / (4y^6))
= sqrt((y^12 + 2y^6 + 1) / 4y^6) = sqrt((y^6 + 1)^2 / (2y^6)) = (y^6 + 1) / sqrt(2y^6)
Now we can rewrite the arc length integral:
L = ∫ (y^6 + 1) / sqrt(2y^6) dy, from y = 1 to y = 2
To evaluate this integral, we can make a substitution:
Let u = y^6 + 1
Then du = 6y^5 dy
Solving for dy, we have dy = du / (6y^5)
After substituting and simplifying, the integral becomes:
L = (1/6) ∫ sqrt(u) / sqrt(2y^6) du, from u = 2 to u = 65
Finally, we can evaluate the definite integral:
L = (1/6) ∫ sqrt(u) / sqrt(2y^6) du, from u = 2 to u = 65
= (1/6) ∫ sqrt(u/2y^6) du, from u = 2 to u = 65
= (1/6) ∫ sqrt(u) / (y^3 * sqrt(2)) du, from u = 2 to u = 65
Now, by substituting u back in terms of y, we can continue evaluating the integral to find the exact length of the curve. Unfortunately, due to the complexity of the expression, it is not possible to provide an exact value without using numerical methods.
To find the exact length of the curve, we can use the arc length formula:
L = ∫[a,b] √(1+(dy/dx)^2) dx
Given the equation of the curve in terms of y, we need to express dy/dx in terms of y to calculate the integral.
First, let's find dy/dx.
x = (y^4)/8 + 1/(4y^2)
Differentiating both sides with respect to y:
dx/dy = (4y^3)/8 - (1/(4y^3)) * (-2/(y^3))
Simplifying dx/dy:
dx/dy = y^3 - (1/(2y^6))
Next, to find dy/dx, we take the reciprocal of dx/dy:
dy/dx = 1/dx/dy
dy/dx = (2y^6)/(y^3 - (1/(2y^6)))
Now, let's substitute this expression into the arc length formula:
L = ∫[a,b] √(1+(dy/dx)^2) dx
L = ∫[a,b] √(1+((2y^6)/(y^3 - (1/(2y^6))))^2) dx
Since the limits of integration are given as 1 ≤ y ≤ 2, we need to change the variable from dx to dy. To do this, we use the fact that dx = (dx/dy) * dy:
L = ∫[1,2] √(1+((2y^6)/(y^3 - (1/(2y^6))))^2) * (dx/dy) dy
Now, we can simplify and solve this integral numerically using appropriate software or calculators.