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| <meta http-equiv="X-UA-Compatible" content="IE=edge"><title>rezvan | Mathematical analysis in several variables: Part 14 - Curves</title><link rel="icon" type="image/png" href=images/icon.png /><meta name="viewport" content="width=device-width, initial-scale=1"> | ||
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| content="Introduction In this part we’ll cover curves and how we can parameterize curve. | ||
| Definition Let, $f(t)$, $g(t)$ and $h(t)$ be continuous functions on some interval, $I$. The connection of points $(x, y, z)$, where $x = f(t)$, $y = g(t)$ and $z = h(t)$, is a curve. | ||
| Equivalently, we can say that: $$ \vec{r}(t) = \langle f(t), g(t), h(t) \rangle $$ | ||
| We call this the parameterization of the curve. Where $t$ is the parameter." /> | ||
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| <meta property="og:description" content="Introduction In this part we’ll cover curves and how we can parameterize curve. | ||
| Definition Let, $f(t)$, $g(t)$ and $h(t)$ be continuous functions on some interval, $I$. The connection of points $(x, y, z)$, where $x = f(t)$, $y = g(t)$ and $z = h(t)$, is a curve. | ||
| Equivalently, we can say that: $$ \vec{r}(t) = \langle f(t), g(t), h(t) \rangle $$ | ||
| We call this the parameterization of the curve. Where $t$ is the parameter." /> | ||
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| <meta name="twitter:title" content="Mathematical analysis in several variables: Part 14 - Curves"/> | ||
| <meta name="twitter:description" content="Introduction In this part we’ll cover curves and how we can parameterize curve. | ||
| Definition Let, $f(t)$, $g(t)$ and $h(t)$ be continuous functions on some interval, $I$. The connection of points $(x, y, z)$, where $x = f(t)$, $y = g(t)$ and $z = h(t)$, is a curve. | ||
| Equivalently, we can say that: $$ \vec{r}(t) = \langle f(t), g(t), h(t) \rangle $$ | ||
| We call this the parameterization of the curve. Where $t$ is the parameter."/> | ||
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| <div class="title"> | ||
| <h1 class="title">Mathematical analysis in several variables: Part 14 - Curves</h1> | ||
| <div class="meta">Posted on Sep 27, 2023</div> | ||
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| <section class="body"> | ||
| <h3 id="introduction">Introduction</h3> | ||
| <p>In this part we’ll cover curves and how we can parameterize curve.</p> | ||
| <h3 id="definition">Definition</h3> | ||
| <p>Let, $f(t)$, $g(t)$ and $h(t)$ be continuous functions on some interval, $I$. The connection of points $(x, y, z)$, where $x = f(t)$, $y = g(t)$ and $z = h(t)$, is a curve.</p> | ||
| <p>Equivalently, we can say that: | ||
| $$ | ||
| \vec{r}(t) = \langle f(t), g(t), h(t) \rangle | ||
| $$</p> | ||
| <p>We call this the <strong>parameterization</strong> of the curve. Where $t$ is the parameter.</p> | ||
| <h3 id="example">Example</h3> | ||
| <p>Given the function $y = x$. In the interval, $0 \leq t \leq 1$.</p> | ||
| <p>We can parameterize and say: | ||
| $$ | ||
| \vec{r}(t) = \langle t, t \rangle \ | \ 0 \leq t \leq 1 | ||
| $$</p> | ||
| <p>But we could also say that: | ||
| $$ | ||
| \vec{r}(t) = \langle t^2, t^2 \rangle \ | \ 0 \leq t \leq 1 | ||
| $$</p> | ||
| <p>So, there isn’t necessarily a unique parameterization for a function.</p> | ||
| <p>We can also change the “direction”: | ||
| $$ | ||
| \vec{r}(t) = \langle 1 - t, 1 - t \rangle \ | \ 0 \leq t \leq 1 | ||
| $$</p> | ||
| <p>Let’s try for a more complex function. Given the unit circle, $x^2 + y^2 = 1$. | ||
| $$ | ||
| \vec{r}(t) = \langle cos(t), sin(t) \rangle \ | \ 0 \leq t \leq 2\pi | ||
| $$</p> | ||
| <p>In general: | ||
| $$ | ||
| \vec{r}(t) = \langle t, f(t) \rangle \ | \ a \leq t \leq b | ||
| $$</p> | ||
| <h3 id="definition-1">Definition</h3> | ||
| <p>If a curve, $C$, is given by a parameterization: | ||
| $$ | ||
| \vec{r}(t) = \langle f(t), g(t), h(t) | ||
| $$</p> | ||
| <p>Then: | ||
| $$ | ||
| \vec{r}\ ^\prime(t) = \langle f^\prime(t), g^\prime(t), h^\prime(t) | ||
| $$</p> | ||
| <p>is <strong>the tangent vector</strong>.</p> | ||
| <h3 id="definition-2">Definition</h3> | ||
| <p>$$ | ||
| \vec{T}(t) = \dfrac{\vec{r}\ ^\prime(t)}{|\vec{r}\ ^\prime(t)|} | ||
| $$</p> | ||
| <p>is <strong>the unit tangent vector</strong>.</p> | ||
| <h3 id="examples">Examples</h3> | ||
| <p>For the curve $\vec{r}(t) = \sqrt(t) \vec{i} + (2 - t)\vec{j}$.</p> | ||
| <p>Find $\vec{T}(t)$ at $t = 1$.</p> | ||
| <p>Let’s rewrite $\vec{r}(t)$ in usual notation: | ||
| $$ | ||
| \vec{r}(t) = \langle \sqrt{t}, 2 - t \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{r}\ ^\prime (t) = \langle \dfrac{1}{2\sqrt{t}}, -1 \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{r}\ ^\prime (1) = \langle \dfrac{1}{2}, -1 \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| |\vec{r}\ ^\prime (1)| = \ldots = \dfrac{\sqrt{5}}{2} | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{T}(1) = \dfrac{\vec{r}\ ^\prime(1)}{|\vec{r}\ ^\prime(1)|} | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{T}(1) = \dfrac{\langle \dfrac{1}{2}, -1 \rangle}{\dfrac{\sqrt{5}}{2}} | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{T}(1) = \langle \dfrac{1}{\sqrt{5}}, - \dfrac{2}{\sqrt{5}} \rangle | ||
| $$</p> | ||
| <h3 id="parameterization-over-line">Parameterization over line</h3> | ||
| <p>Given a line that passes through the point $(x_0, y_0, z_0)$, with a direction of the vector, $\vec{v} = \langle a, b, c \rangle$.</p> | ||
| <p>If we want to parameterize this line, we can choose another point that this vector passes through as: | ||
| $$ | ||
| \vec{r}(t) = \vec{r}(t_0) + t\vec{v} | ||
| $$</p> | ||
| <p>$$ | ||
| (x(t), y(t), z(t)) = (x_0, y_0, z_0) + t(a, b, c) | ||
| $$</p> | ||
| <p>This means that: | ||
| $$ | ||
| x(t) = x_0 + ta | ||
| $$</p> | ||
| <p>$$ | ||
| y(t) = y_0 + tb | ||
| $$</p> | ||
| <p>$$ | ||
| z(t) = z_0 + tc | ||
| $$</p> | ||
| <h3 id="example-1">Example</h3> | ||
| <p>Find the parameterization equation to the tangent line to the helix, $x = 2cos(t), y = sin(t), z = t$. At point $(0, 1, \dfrac{\pi}{2})$.</p> | ||
| <p>From this we easily see that $t = \dfrac{\pi}{2}$.</p> | ||
| <p>The tangent line passes through $\vec{T}(\dfrac{\pi}{2})$.</p> | ||
| <p>Let’s find this.</p> | ||
| <p>$$ | ||
| \vec{r}(t) = \langle 2 cos(t), sin(t), t \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{r}\ ^\prime(t) = \langle -2 sin(t), cos(t), 1 \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{r}\ ^\prime(\dfrac{\pi}{2}) = \langle -2, 0, 1 \rangle | ||
| $$</p> | ||
| <p>$$ | ||
| |\vec{r}\ ^\prime(\dfrac{\pi}{2})| = \sqrt{(-2)^2 + 0^2 + 1^2} = \sqrt{5} | ||
| $$</p> | ||
| <p>$$ | ||
| \vec{T}(\dfrac{\pi}{2}) = \langle - \dfrac{2}{\sqrt{5}}, 0, \dfrac{1}{\sqrt{5}} \rangle | ||
| $$</p> | ||
| <p>Now let’s set this into our equation: | ||
| $$ | ||
| x(t) = 0 + t \cdot -\dfrac{2}{\sqrt{5}} = \boxed{-\dfrac{2t}{\sqrt{5}}} | ||
| $$</p> | ||
| <p>$$ | ||
| y(t) = 1 + t \cdot 0 = \boxed{1} | ||
| $$</p> | ||
| <p>$$ | ||
| z(t) = \dfrac{\pi}{2} + t \dfrac{1}{\sqrt{5}} = \boxed{\dfrac{\pi}{2} + \dfrac{t}{\sqrt{5}}} | ||
| $$</p> | ||
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