Find the exact length of the curve given by $x=2t^3$, $y=t^2-2$ for $0\le t\le 5$.

• A. The length is ${\int }_0^5\sqrt{12t^2+4}dt$
• B. The length is ${\int }_0^5\sqrt{4t^6+2t}dt$
• C. The length is ${\int }_0^5\sqrt{9t^4+4t^2}dt$
• D. The length is ${\int }_0^5\sqrt{36t^4+4t^2}dt$

Answer: (D)

Verified step by step guidance

1. Step 1: Recall the formula for the arc length of a curve given parametrically as x = f(t) and y = g(t). The arc length is given by L = ∫_a^b sqrt((dx/dt)^2 + (dy/dt)^2) dt, where dx/dt and dy/dt are the derivatives of x and y with respect to t.
2. Step 2: Compute dx/dt for x = 2t^3. Differentiating with respect to t, we get dx/dt = 6t^2.
3. Step 3: Compute dy/dt for y = t^2 - 2. Differentiating with respect to t, we get dy/dt = 2t.
4. Step 4: Substitute dx/dt and dy/dt into the arc length formula. The integrand becomes sqrt((6t^2)^2 + (2t)^2) = sqrt(36t^4 + 4t^2).
5. Step 5: Set up the integral for the arc length over the interval 0 ≤ t ≤ 5. The arc length is L = ∫_0^5 sqrt(36t^4 + 4t^2) dt.