exercise:39ba8c90c5: Difference between revisions
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Let <math>P:[a,b] \goesto \R^2</math> and <math>Q:[c,d]\goesto R^2</math> | Let <math>P:[a,b] \goesto \R^2</math> and <math>Q:[c,d]\goesto R^2</math> be two parametrizations of the same curve <math>C</math> such that all four coordinate functions are continuously differentiable. | ||
be two parametrizations of the same curve | (A function is '''continuously differentiable '''if its derivative exists and is continuous at every number in its domain.) | ||
<math>C</math> such that all four coordinate functions | Then <math>P</math> and <math>Q</math> are called '''equivalent parametrizations''' of <math>C</math> if there exists a continuously differentiable function <math>f</math> with domain <math>[a,b]</math> and range <math>[c,d]</math> which has a continuously differentiable inverse function, and in addition satisfies | ||
are continuously differentiable. | |||
(A function is '''continuously differentiable''' | |||
if its derivative exists and is continuous at | |||
every number in its domain.) | |||
Then <math>P</math> and <math>Q</math> are called | |||
'''equivalent parametrizations''' of <math>C</math> | |||
if there exists a continuously differentiable | |||
function <math>f</math> with domain <math>[a,b]</math> and range <math>[c,d]</math> | |||
which has a continuously differentiable inverse | |||
function, and in addition satisfies | |||
(i) <math>f(a) = c</math> and <math>f(b) = d</math>, | (i) <math>f(a) = c</math> and <math>f(b) = d</math>, | ||
(ii) <math>P(t) = Q(f(t)),</math> for every <math>t</math> in <math>[a,b]</math>. | (ii) <math>P(t) = Q(f(t)),</math> for every <math>t</math> in <math>[a,b]</math>. | ||
<ul style{{=}}"list-style-type:lower-alpha"><li>Using the Chain Rule and the Change of Variable | <ul style{{=}}"list-style-type:lower-alpha"><li>Using the Chain Rule and the Change of Variable | ||
Theorem for Definite Integrals | Theorem for Definite Integrals (for the latter, see ([[guide:Eb0a63a097#theorem-6|Theorem]]), prove that equivalent parametrizations assign the same arc length to <math>C</math>.</li> | ||
(for the latter, see Theorem | |||
prove that equivalent parametrizations assign | |||
the same arc length to <math>C</math>.</li> | |||
<li>Show that | <li>Show that | ||
Latest revision as of 22:38, 25 November 2024
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[/math]
Let [math]P:[a,b] \goesto \R^2[/math] and [math]Q:[c,d]\goesto R^2[/math] be two parametrizations of the same curve [math]C[/math] such that all four coordinate functions are continuously differentiable. (A function is continuously differentiable if its derivative exists and is continuous at every number in its domain.) Then [math]P[/math] and [math]Q[/math] are called equivalent parametrizations of [math]C[/math] if there exists a continuously differentiable function [math]f[/math] with domain [math][a,b][/math] and range [math][c,d][/math] which has a continuously differentiable inverse function, and in addition satisfies (i) [math]f(a) = c[/math] and [math]f(b) = d[/math], (ii) [math]P(t) = Q(f(t)),[/math] for every [math]t[/math] in [math][a,b][/math].
- Using the Chain Rule and the Change of Variable Theorem for Definite Integrals (for the latter, see (Theorem), prove that equivalent parametrizations assign the same arc length to [math]C[/math].
- Show that
[[math]] P(t) = (\cos t, \sin t), \quad 0 \leq t \leq \frac{\pi}2 , [[/math]][[math]] Q(s) = \left( \frac{1-s^2}{1+s^2}, \frac{2s}{1+s^2}\right), \quad 0 \leq s \leq 1 , [[/math]]are equivalent parametrizations of the same curve [math]C[/math], and identify the curve.
- Show that
[[math]] P(t) = (\cos t, \sin t), \quad 0 \leq t \leq 2\pi , [[/math]]and[[math]] Q(s) = (\cos 5t, \sin 5t), \quad 0 \leq t \leq 2\pi , [[/math]]are nonequivalent parametrizations of the circle.