Mth 230 Project Problem 3 Name 1 Given

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Mth 230 Project Problem 3 Name 1 Given Mth 230 Project Problem 3 Name 1 Given Given the vector function $\mathbf{r}(t) = (2, 2\sin(8t), 2\cos(8t))$, perform the following tasks: A) Find the unit tangent vector $\mathbf{T}(t)$. B) Compute the arc length on the interval $[0, t]$. C) Compute the arc length on the interval $[0, 2 \pi]$.

Dr. Jack HW Helper · Accepted Answer

The study of space curves involves understanding key concepts such as tangent vectors, arc length, curvature, osculating and normal planes, and reparameterization based on arc length. Given the vector function $\mathbf{r}(t) = (2, 2\sin(8t), 2\cos(8t))$, our first goal is to find the unit tangent vector $\mathbf{T}(t)$. The tangent vector $\mathbf{r}'(t)$ is obtained by differentiating each component with respect to $t$. Differentiating $x(t) = 2$, $y(t) = 2\sin(8t)$, and $z(t) = 2\cos(8t)$, we get: $$ \mathbf{r}'(t) = \left(0, 16\cos(8t), -16\sin(8t)\right). $$ The magnitude of $\mathbf{r}'(t)$ is: $$ \|\mathbf{r}'(t)\| = \sqrt{0^2 + (16\cos(8t))^2 + (-16\sin(8t))^2} = \sqrt{0 + 256\cos^2(8t) + 256\sin^2(8t)}. $$ Using the fundamental Pythagorean identity $\sin^2 u + \cos^2 u = 1$, the magnitude simplifies to: $$ \|\mathbf{r}'(t)\| = \sqrt{256 (\cos^2(8t) + \sin^2(8t))} = \sqrt{256 \times 1} = 16. $$ The unit tangent vector $\mathbf{T}(t)$ is therefore: $$ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{16} (0, 16\cos(8t), -16\sin(8t)) = (0, \cos(8t), -\sin(8t)). $$ Next, the arc length $s(t)$ from $0$ to $t$ is given by: $$ s(t) = \int_{0}^{t} \|\mathbf{r}'(u)\| du = \int_{0}^{t} 16 du = 16t, $$ since the magnitude is constant. For the interval $[0, t]$, the arc length is thus directly proportional to $t$. On the interval $[0, 2\pi]$, the arc length is: $$ s(2\pi) = 16 \times 2\pi = 32\pi. $$ This indicates that the curve traced out over one period of $2\pi$ has a total length of $32\pi$. The process illustrates how differentiation and basic calculus principles facilitate understanding the geometry of space curves. By obtaining the unit tangent vector, one can analyze the curve's direction, speed, and other properties crucial to the study of differential geometry. In applications such as physics and engineering, these calculations enable the description of particle trajectories, the design of motion paths, a

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Mth 230 Project Problem 3 Name 1 Given

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Mth 230 Project Problem 3 Name 1 Given

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