Find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.
1. y = (1/3)(x^2 + 2)^(3/2), from x = 0 to x = 3
and
2. x = (y^3)/3 + 1/(4y), from y = 1 to y = 3
[HINT: 1 + (dx/dy)^2 is a perfect square]
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To find the exact length of the curve using antidifferentiation, we can use the formula for arc length:
L = ∫(a to b) √(1 + (dy/dx)^2) dx
Let's start with the first equation:
1. y = (1/3)(x^2 + 2)^(3/2), from x = 0 to x = 3
First, we need to find dy/dx by taking the derivative of y with respect to x:
dy/dx = d/dx[(1/3)(x^2 + 2)^(3/2)]
= (1/3) * (3/2) * (x^2 + 2)^(1/2) * 2x
= (x(x^2 + 2)^(1/2))/√(x^2 + 2)
Next, we can substitute this into the arc length formula:
L = ∫(0 to 3) √(1 + [(x(x^2 + 2)^(1/2))/√(x^2 + 2)]^2) dx
= ∫(0 to 3) √(1 + (x^2(x^2 + 2))/(x^2 + 2)) dx
= ∫(0 to 3) √(1 + x^2) dx
Now we need to simplify the integrand algebraically before finding an antiderivative:
√(1 + x^2) is already in a suitable form, so we can proceed to find the antiderivative:
L = ∫(0 to 3) √(1 + x^2) dx
= [1/2 * (x * √(1 + x^2) + ln(x + √(1 + x^2))) ](0 to 3)
= (1/2 * (3 * √(1 + 3^2) + ln(3 + √(1 + 3^2)))) - (1/2 * (0 * √(1 + 0^2) + ln(0 + √(1 + 0^2))))
Simplifying further:
L = (1/2 * (3 * √10 + ln(3 + √10))) - (1/2 * ln(1))
= (3/2 * √10 + 1/2 * ln(3 + √10))
Thus, the exact length of the curve is (3/2 * √10 + 1/2 * ln(3 + √10)).
Now, let's move to the second equation:
2. x = (y^3)/3 + 1/(4y), from y = 1 to y = 3
Following the same steps as above, we need to find dx/dy by taking the derivative of x with respect to y:
dx/dy = d/dy[(y^3)/3 + 1/(4y)]
= (3y^2)/3 - (1/(4y^2))
= y^2 - 1/(4y^2)
Substituting this into the arc length formula:
L = ∫(1 to 3) √(1 + (y^2 - 1/(4y^2))^2) dy
= ∫(1 to 3) √(1 + (y^4 - 2 + 1/(16y^4))) dy
= ∫(1 to 3) √(y^4 + (16y^4 - 32 + 1)/(16y^4)) dy
= ∫(1 to 3) √((17y^4 - 31)/(16y^4)) dy
We can simplify the integrand algebraically further:
L = ∫(1 to 3) √(17y^4 - 31)/(4y^2) dy
Unfortunately, the integrand cannot be simplified further. We would then need to evaluate this integral numerically using numerical integration techniques or available software.