Explanation Let f ( x , y ) be a continuous function on a closed, bounded domain D in R 2 . Then: (i) f ( x , y ) takes both the minimum and maximum values on D . (ii) The extreme values occur either at critical points in the interior of D or at points on the boundary of D . Given: The equation of second degree curve a x 2 + b x y + c y 2 + d x + e y + f = 0 The following figure is given: Proof: The equation of second degree curve is a x 2 + b x y + c y 2 + d x + e y + f = 0 (1) If the polar coordinates of P with respect to the rotated axes are ( r , α ) , then the line from the origin to P has length r and makes an angle of α with rotated x -axis( x ˜ axis). Since the x ˜ -axis makes an angle of θ with the x axis. It follows that the line from the origin to P makes an angle of α + θ with the x axis. So, the polar coordinates of P with respect to the standard axes are ( r , α + θ ) Consider ( x ˜ , y ˜ ) rectangular coordinates of P with respect to the rotated axes and ( x , y ) for the standard axes. x = r cos ( α + θ ) ⇒ x = ( r cos α ) cos θ − ( r sin α ) sin θ ⇒ x = x ˜ cos θ − y ˜ sin θ ( 2 ) y = r sin ( α + θ ) ⇒ y = ( r sin α ) cos θ + ( r cos α ) sin θ ⇒ y = y ˜ cos θ + x ˜ sin θ ⇒ y = x ˜ sin θ + y ˜ cos θ ... To determine (b) To Show: That if b ≠ 0 , then we can obtain b ′ = 0 for θ = 1 2 cot − 1 ( a − c ) b

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 12.5, Problem 78E

To determine

(a)

To Show:

$b^{'} = b \cos 2 θ + (c - a) \sin 2 θ$

To determine

(b)

To Show:

That if $b \neq 0$ , then we can obtain $b^{'} = 0$ for $θ = \frac{1}{2} \cot^{- 1} \frac{(a - c)}{b}$

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Chapter 12.5, Problem 77E

Chapter 12, Problem 1CRE

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Chapter 12 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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(a) To Show: b ′ = b cos 2 θ + ( c − a ) sin 2 θ

Solution Summary: The following figure illustrates the equation of second degree curve: f (x, y) takes both the minimum and maximum values on D, and the line from the origin to P makes an angle of with the x

Concept explainers

Videos

(a)

To Show:

$b^{'} = b \cos 2 θ + (c - a) \sin 2 θ$

(b)

To Show:

That if $b \neq 0$ , then we can obtain $b^{'} = 0$ for $θ = \frac{1}{2} \cot^{- 1} \frac{(a - c)}{b}$

Want to see the full answer?

Chapter 12 Solutions

(a) To Show: b ′ = b cos 2 θ + ( c − a ) sin 2 θ

Solution Summary: The following figure illustrates the equation of second degree curve: f (x, y) takes both the minimum and maximum values on D, and the line from the origin to P makes an angle of with the x

Concept explainers

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(a) To Show: b′=bcos2θ+(c−a)sin2θ

(b) To Show: That if b≠0, then we can obtain b′=0 for θ=12cot−1(a−c)b

Want to see the full answer?

Chapter 12 Solutions

(a)

To Show:

$b^{'} = b \cos 2 θ + (c - a) \sin 2 θ$

(b)

To Show:

That if $b \neq 0$ , then we can obtain $b^{'} = 0$ for $θ = \frac{1}{2} \cot^{- 1} \frac{(a - c)}{b}$