Find the arc length on the interval for t between 0 and 1 inclusive for the curve described with the parametric equations: x = 1 + 3t^2, y = 2t^3 + 4. (20 points)
A) 2 times the square root of 2 minus 1
B) 4 times the square root of 2 minus 2
C) 2 times the square root of 2 minus 2
D) the square root of 2 minus 2
s = ∫[0,1] √(36t^2 + 36t^4) dt
= ∫[0,1] 6t√(1 + t^2) dt
Now let u = 1+t^2 and du = 2t dt, giving
s = ∫[1,2] 3√u du
now finish it off
Is the integral from 1 to 2 or 2 to 1
come on, think a bit. You had
∫ from 0 to 1 dt
u = 1+t^2, so now you have
∫ from 1 to 2 du
To find the arc length of a curve described by parametric equations, we need to use the formula:
L = ∫[a,b] √(dx/dt)^2 + (dy/dt)^2 dt
where a and b are the limits of the parameter t.
In this case, the parametric equations are:
x = 1 + 3t^2
y = 2t^3 + 4
First, we need to find the derivatives of x and y with respect to t:
dx/dt = d/dt (1 + 3t^2) = 6t
dy/dt = d/dt (2t^3 + 4) = 6t^2
Now we can substitute these derivatives into the arc length formula and compute the integral:
L = ∫[0,1] √(6t)^2 + (6t^2)^2 dt
= ∫[0,1] √(36t^2 + 36t^4) dt
= ∫[0,1] √36t^2(1 + t^2) dt
= ∫[0,1] 6t√(1 + t^2) dt
To solve this integral, we need to make a trigonometric substitution:
Let t = tan(θ), then dt = sec^2(θ) dθ
Now we need to find the limits of integration for θ when t is between 0 and 1. Since t = tan(θ), we have:
θ = arctan(t)
So when t = 0, θ = arctan(0) = 0, and when t = 1, θ = arctan(1) = π/4.
Substituting the trigonometric substitution and the limits of integration, we have:
L = ∫[0,π/4] 6tan(θ)√(1 + tan^2(θ)) sec^2(θ) dθ
= 6∫[0,π/4] tan(θ) sec(θ) dθ
= 6∫[0,π/4] sin(θ)/cos(θ) (1/cos(θ)) dθ
= 6∫[0,π/4] sin(θ)/cos^2(θ) dθ
Now we can use a u-substitution: Let u = cos(θ), then du = -sin(θ) dθ.
Changing the limits of integration accordingly, when θ = 0, u = cos(0) = 1, and when θ = π/4, u = cos(π/4) = 1/√2.
Substituting the u-substitution and the new limits of integration, we have:
L = 6∫[1,1/√2] (1/u^2)(-du)
= -6∫[1,1/√2] 1/u^2 du
= -6[-1/u]∣[1,1/√2]
= -6[(1/1) - (1/(1/√2))]
= -6[1 - √2]
= 6(√2 - 1)
Therefore, the arc length on the interval for t between 0 and 1 inclusive is 6(√2 - 1).
So the correct answer is D) the square root of 2 minus 2.