Evaluate the line integral of the vector field $F(x,y)=(2x,3y)$ along the curve $C$ given by the vector function $r\left(t\right)=(t,t^2)$, where $t$ goes from $0$ to $1$.

• A. $3$
• B. $2$
• C. $4$
• D. $5$

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the formula for a line integral of a vector field along a curve. The line integral of a vector field F(x, y) = (P(x, y), Q(x, y)) along a curve C parameterized by r(t) = (x(t), y(t)) is given by: ∫_C F · dr = ∫_a^b [P(x(t), y(t)) dx/dt + Q(x(t), y(t)) dy/dt] dt.
2. Step 2: Identify the components of the vector field F(x, y) = (2x, 3y) and the parameterization of the curve r(t) = (t, t^2). Here, P(x, y) = 2x and Q(x, y) = 3y. The parameterization gives x(t) = t and y(t) = t^2.
3. Step 3: Compute the derivatives dx/dt and dy/dt from the parameterization. Since x(t) = t, dx/dt = 1. Similarly, since y(t) = t^2, dy/dt = 2t.
4. Step 4: Substitute x(t), y(t), dx/dt, and dy/dt into the line integral formula. The integral becomes: ∫_0^1 [2(t) * 1 + 3(t^2) * 2t] dt.
5. Step 5: Simplify the integrand to get: ∫_0^1 [2t + 6t^3] dt. Then, split the integral into two parts: ∫_0^1 2t dt + ∫_0^1 6t^3 dt. Evaluate each integral separately to find the final result.