Let's solve the problem step by step: **Given:** - Vector field: \(\vec{F}(x, y, z) = (2y, 2xz, 4y)\) - Path: \(\vec{r}(t) = (\cos(2t), 3\sin(2t), 5t)\) - From \(t = 0\) to \(t = \frac{\pi}{2}\) We are to compute the **work** done by the force as the object moves along the path. ### Step 1: Work Integral The work is given by: \[ W = \int_C \vec{F} \cdot d\vec{r} \] where \(d\vec{r} = \vec{r}'(t) dt\). ### Step 2: Compute \(\vec{r}'(t)\) \[ \vec{r}(t) = (\cos(2t), 3\sin(2t), 5t) \] \[ \vec{r}'(t) = (-2\sin(2t), 6\cos(2t), 5) \] ### Step 3: Plug Path into the Vector Field \[ \vec{F}(\vec{r}(t)) = (2y, 2xz, 4y) \] So, - \(x = \cos(2t)\) - \(y = 3\sin(2t)\) - \(z = 5t\) Thus: - First component: \(2y = 2 \cdot 3\sin(2t) = 6\sin(2t)\) - Second component: \(2xz = 2 \cdot \cos(2t) \cdot 5t = 10t\cos(2t)\) - Third component: \(4y = 4 \cdot 3\sin(2t) = 12\sin(2t)\) So: \[ \vec{F}(\vec{r}(t)) = (6\sin(2t), 10t\cos(2t), 12\sin(2t)) \] ### Step 4: Dot Product \(\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)\) \[ \begin{align*} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) &= [6\sin(2t)] \cdot [-2\sin(2t)] + [10t\cos(2t)] \cdot [6\cos(2t)] + [12\sin(2t)] \cdot [5] \\ &= -12\sin^2(2t) + 60t\cos^2(2t) + 60\sin(2t) \end{align*} \] ### Step 5: Set Up the Integral \[ W = \int_{0}^{\frac{\pi}{2}} \left(-12\sin^2(2t) + 60t\cos^2(2t) + 60\sin(2t)\right) dt \] ### Step 6: Integrate Each Term #### First Term: \(-12\int \sin^2(2t) dt\) Use power reduction: \(\sin^2(2t) = \frac{1}{2}(1 - \cos(4t))\) \[ -12 \int_{0}^{\frac{\pi}{2}} \sin^2(2t) dt = -12 \int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - \cos(4t)) dt = -6 \left[t - \frac{\sin(4t)}{4}\right]_{0}^{\frac{\pi}{2}} \] At \(t = \frac{\pi}{2}\): \(t = \frac{\pi}{2}\), \(4t = 2\pi\), \(\sin(2\pi) = 0\) At \(t = 0\): \(t = 0\), \(\sin(0) = 0\) So, \[ -6 \left[ \frac{\pi}{2} - 0 \right] = -3\pi \] #### Second Term: \(60\int t\cos^2(2t) dt\) Let’s use \(\cos^2(2t) = \frac{1}{2}(1 + \cos(4t))\): \[ 60 \int_{0}^{\frac{\pi}{2}} t \cos^2(2t) dt = 60 \int_{0}^{\frac{\pi}{2}} t \cdot \frac{1 + \cos(4t)}{2} dt = 30 \int_{0}^{\frac{\pi}{2}} t dt + 30 \int_{0}^{\frac{\pi}{2}} t \cos(4t) dt \] First part: \[ 30 \int_{0}^{\frac{\pi}{2}} t dt = 30 \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{2}} = 30 \left(\frac{(\frac{\pi}{2})^2}{2}\right) = 30 \cdot \frac{\pi^2}{8} = \frac{15\pi^2}{4} \] Second part (integration by parts, let \(u = t, dv = \cos(4t)dt\)): - \(u = t \to du = dt\) - \(v = \frac{1}{4}\sin(4t)\) \[ \int t \cos(4t) dt = t \cdot \frac{1}{4}\sin(4t) - \int \frac{1}{4}\sin(4t) dt = \frac{t}{4}\sin(4t) + \frac{1}{16}\cos(4t) \] Evaluate from \(0\) to \(\frac{\pi}{2}\): - At \(t = \frac{\pi}{2}\): \(\sin(2\pi) = 0\), \(\cos(2\pi) = 1\) - At \(t = 0\): \(\sin(0) = 0\), \(\cos(0) = 1\) So, \[ \left[\frac{t}{4}\sin(4t) + \frac{1}{16}\cos(4t)\right]_{0}^{\frac{\pi}{2}} = \left[0 + \frac{1}{16}(1)\right] - \left[0 + \frac{1}{16}(1)\right] = 0 \] So, the second part is zero. Total for the second term: \[ \frac{15\pi^2}{4} \] #### Third Term: \(60\int \sin(2t) dt\) \[ 60\int_{0}^{\frac{\pi}{2}} \sin(2t) dt = 60 \left[ -\frac{1}{2}\cos(2t) \right]_{0}^{\frac{\pi}{2}} \] At \(t = \frac{\pi}{2}\): \(\cos(\pi) = -1\) At \(t = 0\): \(\cos(0) = 1\) So, \[ -30\left[ -1 - 1 \right] = -30(-2) = 60 \] ### Step 7: Add All Terms Together \[ W = (-3\pi) + \frac{15\pi^2}{4} + 60 \] --- ### **Final Answer:** \[ \boxed{W = -3\pi + \frac{15\pi^2}{4} + 60} \] This is the exact work done by the force along the given path.