Consider the vector field $\vec{P}=\langle 7 x+6 y, 6 x+4 y\rangle$.
Is this vector field Conservative? Conservative $\checkmark \sigma$
Use method of your choice to evaluate $\int_{C} \vec{F} \cdot d \vec{r}$ along the curve $C_{:} \vec{r}(t)=t^{2} \vec{i}+t^{3} \vec{j}$, for $0 \leq t \leq 1$.
\[
\int_{C} \vec{F} \cdot d \vec{r}=
\]
Answer
Finally, we evaluate the integral \(\int_{0}^{1} \vec{F} \cdot d\vec{r} = \int_{0}^{1} (14t^3 + 30t^4 + 12t^5) dt = \left[\frac{14}{4}t^4 + \frac{30}{5}t^5 + \frac{12}{6}t^6\right]_{0}^{1} = \frac{14}{4} + \frac{30}{5} + \frac{12}{6} = \boxed{11}\).
Step 1 ：First, we need to check if the vector field is conservative. A vector field \(\vec{P} = \langle P_1, P_2 \rangle\) is conservative if \(\frac{\partial P_2}{\partial x} = \frac{\partial P_1}{\partial y}\).
Step 2 ：For \(\vec{P} = \langle 7x + 6y, 6x + 4y \rangle\), we have \(\frac{\partial P_2}{\partial x} = 6\) and \(\frac{\partial P_1}{\partial y} = 6\). Since these are equal, the vector field is conservative.
Step 3 ：Next, we need to evaluate the line integral \(\int_{C} \vec{F} \cdot d\vec{r}\).
Step 4 ：Given the curve \(C: \vec{r}(t) = t^2\vec{i} + t^3\vec{j}\), for \(0 \leq t \leq 1\), we can find \(d\vec{r} = 2t dt\vec{i} + 3t^2 dt\vec{j}\).
Step 5 ：Substitute \(\vec{r}(t)\) into \(\vec{P}\) to get \(\vec{F} = \langle 7t^2 + 6t^3, 6t^2 + 4t^3 \rangle\).
Step 6 ：Then, \(\vec{F} \cdot d\vec{r} = (7t^2 + 6t^3)2t dt + (6t^2 + 4t^3)3t^2 dt = 14t^3 dt + 12t^4 dt + 18t^4 dt + 12t^5 dt = 14t^3 dt + 30t^4 dt + 12t^5 dt\).
Step 7 ：Finally, we evaluate the integral \(\int_{0}^{1} \vec{F} \cdot d\vec{r} = \int_{0}^{1} (14t^3 + 30t^4 + 12t^5) dt = \left[\frac{14}{4}t^4 + \frac{30}{5}t^5 + \frac{12}{6}t^6\right]_{0}^{1} = \frac{14}{4} + \frac{30}{5} + \frac{12}{6} = \boxed{11}\).
