Answered: Show that the vector-valued function…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.4: Multiple-angle Formulas

Problem 54E

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Question

Show that the vector-valued function shown below describes the function of a particle moving in a circle of radius 1 centered at a point (5,5,3) and lying in the plane 3x+3y-6z = 12

r(t) = (5i + 5j + 3k) + cos t
'
1
1
√.
+ sin t
1
13
1
ਤਾਂ
1

Expert Solution

Step 1

Part - 1 :

Given vector-valued function is

$r (t) = (5 i + 5 j + 3 k) + \cos t (\frac{1}{\sqrt{2}} i - \frac{1}{\sqrt{2}} j) + \sin t (\frac{1}{\sqrt{3}} i + \frac{1}{\sqrt{3}} j + \frac{1}{\sqrt{3}} k)$

$= (5 + \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}) i + (5 - \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}) j + (3 + \frac{\sin t}{\sqrt{3}}) k$

From this the parametric Equations of the given Curve are:

$x (t) = 5 + \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}$

$y (t) = 5 - \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}$

$z (t) = 3 + \frac{\sin t}{\sqrt{3}}$

Substitute these in the plane equation 3x+3y-6z = 12

$3 (5 + \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}) + 3 (5 - \frac{\cos t}{\sqrt{2}} + \frac{\sin t}{\sqrt{3}}) - 6 (3 + \frac{\sin t}{\sqrt{3}}) = 12$

$\Rightarrow$ $(15 + 15 - 18) + (\frac{3}{\sqrt{2}} - \frac{3}{\sqrt{2}}) \cos t + (\frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}} - \frac{6}{\sqrt{3}}) = 12$