Answer:
To find the exact length of the curve defined by the equation y = 32x^(3/2) over the interval 0 ≤ x ≤ 2, you can use the formula for the arc length of a curve:
L = ∫[a, b] √(1 + (dy/dx)^2) dx
In this case, a = 0 and b = 2, and you need to find dy/dx (the derivative of y with respect to x).
Given y = 32x^(3/2), we can find dy/dx as follows:
dy/dx = d/dx (32x^(3/2))
Using the power rule for differentiation, we get:
dy/dx = (3/2) * 32x^(3/2 - 1)
dy/dx = (3/2) * 32x^(1/2)
Now, we can substitute this into the formula for arc length:
L = ∫[0, 2] √[1 + ((3/2) * 32x^(1/2))^2] dx
Now, you need to calculate this integral. Let's simplify it first:
L = ∫[0, 2] √[1 + (3/2)^2 * (32x)] dx
L = ∫[0, 2] √[1 + 9 * 32x] dx
L = ∫[0, 2] √[1 + 288x] dx
Now, you can use a substitution to solve this integral. Let u = 1 + 288x, which implies du = 288dx. Solving for dx, we get dx = du/288.
Now, we can rewrite the integral with respect to u:
L = ∫[u(0), u(2)] √u * (1/288) du
u(0) = 1 + 288 * 0 = 1
u(2) = 1 + 288 * 2 = 577
So, the integral becomes:
L = (1/288) ∫[1, 577] √u du
Now, integrate √u with respect to u:
L = (1/288) * [(2/3) * u^(3/2)] | from 1 to 577
L = (2/864) * [577^(3/2) - 1^(3/2)]
L = (1/432) * (577^(3/2) - 1)
Now, you can calculate the numerical value of this expression to find the exact length of the curve over the interval 0 ≤ x ≤ 2.
Explanation: