Chapter 14: Q. 14 (page 1119)

Generalize your answer to Exercise 12 to give a parametrization and a normal vector for the extension of any differentiable plane curve y = f(x) through a ≤ z ≤ b.

Short Answer

Expert verified

$Therefore, the required normal vector for the above cylinder is n = = \frac{\partial f}{\partial u} i - j .$

Step by step solution

Step 1. Given

The given plane curve is y = f(x) through a ≤ z ≤ b.

Step 2. Finding the equation

$The parametrization of the curve y = x^{2} . Let x = u then, y = x^{2} = u^{2} Hence, The parametrization of the curve y = x^{2} is, \{\begin{cases} x = u \\ y = u^{2} \end{cases} Let = v, then the parametrization of a " generalized cylinder " S, defined by extending the curve y = x ² upwards and downwards from = = - 2 to z = 3 is, \{\begin{cases} x = u \\ y = u^{2} \\ z = v \end{cases} Where, - 2 \leq v \leq 3 Therefore, the required smooth parametrization is r (u, v) = <u, u^{2}, v>$

Step 3. Find ru and rv.

$By above step, the surface S is parametrized by as follows : r (u, v) = <u, u^{2}, v> . T h e n, \begin{aligned} r_{u} = \frac{\partial}{\partial u} r (u, v) \\ = \frac{\partial}{\partial u} <u, u^{2}, v> \\ = ⟨ 1, 2 u, 0 ⟩ \end{aligned} a n d \begin{aligned} r_{v} = \frac{\partial}{\partial v} r (u, v) \\ = \frac{\partial}{\partial ν} <u, u^{2}, v> \\ = ⟨ 0, 0, 1 ⟩ \end{aligned}$

Step 4. Find n=rv×rv

$n = r_{v} \times r_{v} \begin{aligned} = <1, \frac{\partial f}{\partial u}, 0> \times ⟨ 0, 0, 1 ⟩ \\ = |\begin{array}{ccc} i & j & k \\ 1 & \partial f / \partial u & 0 \\ 0 & 0 & 1 \end{array}| \\ = [\frac{\partial f}{\partial u} \cdot 1 - 0 \cdot 0] i - [1 \cdot 1 - 0 \cdot 0] j + [1 \cdot 0 - \frac{\partial f}{\partial u} \cdot 0] k \\ = \frac{\partial f}{\partial u} i - 1 j + 0 k \\ = \frac{\partial f}{\partial u} i - j . \end{aligned} Therefore, the required normal vector for the above cylinder is n = = \frac{\partial f}{\partial u} i - j .$

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Generalize your answer to Exercise 12 to give a parametrization and a normal vector for the extension of any differentiable plane curve y = f(x) through a ≤ z ≤ b.

Short Answer

Step by step solution

Step 1. Given

Step 2. Finding the equation

Step 3. Find ru and rv.

Step 4. Find n=rv×rv

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