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Vector Valued Functions Calculator

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Example
Created on 2024-06-20Asked by Mila Sanchez (Solvelet student)
Find the arc length of the vector function r(t)=cos(t),sin(t),t\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle for 0t2π0 \leq t \leq 2\pi.

Solution

To find the arc length of the vector function r(t)=cos(t),sin(t),t\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle for 0t2π0 \leq t \leq 2\pi: 1. **Find the derivative of r(t)\mathbf{r}(t):** r(t)=sin(t),cos(t),1. \mathbf{r}'(t) = \left\langle -\sin(t), \cos(t), 1 \right\rangle. 2. **Compute the magnitude of r(t)\mathbf{r}'(t):** r(t)=(sin(t))2+(cos(t))2+12=sin2(t)+cos2(t)+1=1+1=2. \|\mathbf{r}'(t)\| = \sqrt{(-\sin(t))^2 + (\cos(t))^2 + 1^2} = \sqrt{\sin^2(t) + \cos^2(t) + 1} = \sqrt{1 + 1} = \sqrt{2}. 3. **Integrate the magnitude over the given interval:** L=02πr(t)dt=02π2dt=202πdt=22π=2π2. L = \int_{0}^{2\pi} \|\mathbf{r}'(t)\| \, dt = \int_{0}^{2\pi} \sqrt{2} \, dt = \sqrt{2} \int_{0}^{2\pi} dt = \sqrt{2} \cdot 2\pi = 2\pi\sqrt{2}. 4. **Result:** The arc length of the vector function r(t)=cos(t),sin(t),t\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle for 0t2π0 \leq t \leq 2\pi is 2π22\pi\sqrt{2}. Solved on Solvelet with Basic AI Model
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