Answered: find an equation of the normal plane…

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Calculus

find an equation of the normal plane through the point t=pi/3. r(t) = 2sin(t)i+. 4cos(t)j +tk

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section: Chapter Questions

Problem 18T

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Concept explainers

Vector Arithmetic

Vectors are those objects which have a magnitude along with the direction. In vector arithmetic, we will see how arithmetic operators like addition and multiplication are used on any two vectors. Arithmetic in basic means dealing with numbers. Here, magnitude means the length or the size of an object. The notation used is the arrow over the head of the vector indicating its direction.

Vector Calculus

Vector calculus is an important branch of mathematics and it relates two important branches of mathematics namely vector and calculus.

Question

find an equation of the normal plane through the point t=pi/3.

r(t) = 2sin(t)i+. 4cos(t)j +tk

Expert Solution

Step 1

Given curve is

$r (t) = 2 \sin (t) i + 4 \cos (t) j + t k r' (t) = 2 \cos (t) i - 4 \sin (t) j + k$

At t=pi/3

$r (\frac{π}{3}) = 2 \sin (\frac{π}{3}) i + 4 \cos (\frac{π}{3}) j + t k = \sqrt{3} i + 2 j + \frac{π}{3} k r' (\frac{π}{3}) = 2 \cos (\frac{π}{3}) i - 4 \sin (\frac{π}{3}) j + k = i - 2 \sqrt{3} j + k$

Now, unit tangent vector at point $(\sqrt{3}, 2, \frac{π}{3})$ is

$T (\frac{π}{3}) = \frac{r' (\frac{π}{3})}{|r' (\frac{π}{3})|} = \frac{i - 2 \sqrt{3} j + k}{\sqrt{1 + 12 + 1}} = \frac{i - 2 \sqrt{3} j + k}{\sqrt{14}}$

Step 2

Now, equation of normal plane at point $(\sqrt{3}, 2, \frac{π}{3})$ is

$\frac{1}{\sqrt{14}} (x - \sqrt{3}) - \frac{2 \sqrt{3}}{\sqrt{14}} (y - 2) + \frac{1}{\sqrt{14}} (z - \frac{π}{3}) = 0 (x - \sqrt{3}) - 2 \sqrt{3} (y - 2) + (z - \frac{π}{3}) = 0 x - 2 \sqrt{3} y + z = \frac{π}{3} - 3 \sqrt{3}$

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