Explanation Given: y = ± 2 r x − p x 2 t ( y ) = R 2 − S 2 − R 2 − y 2 − r 2 − S 2 + r 2 − y 2 Formula used: Curvature is given as: k = | f ' ' ( y ) | ( 1 + ( f ' ( y ) ) 2 ) 3 2 Proof: We have been given the equation y = ± 2 r x − p x 2 . Let us solve this equation for x in terms of y . ⇒ y = ± 2 r x − p x 2 ⇒ y 2 = ( ± 2 r x − p x 2 ) 2 ⇒ y 2 = 2 r x − p x 2 ⇒ p x 2 − 2 r x + y 2 = 0 Let us solve this quadratic using quadratic formula to represent x in terms of y . ⇒ p x 2 − 2 r x + y 2 = 0 ⇒ x = − ( − 2 r ) ± 4 r 2 − 4 p y 2 2 p ⇒ x = 2 r ± 4 ( r 2 − p y 2 ) 2 p = 2 r ± 2 r 2 − p y 2 2 p ⇒ x = r ± r 2 − p y 2 p ⇒ x = p − 1 ( r − r 2 − p y 2 ) Since we have been given that after surgery the cross-section of cornea has the equation x = f ( y ) + t ( y ) . x ' = p − 1 ( r − r 2 − p y 2 ) + R 2 − S 2 − R 2 − y 2 − r 2 − S 2 + r 2 − y 2 We have to now find its curvature at point P ( y = 0 ) . Let us first find the derivatives of first and second order and then plug those expressions to get the curvature: ⇒ x = p − 1 ( r − r 2 − p y 2 ) + R 2 − S 2 − R 2 − y 2 − r 2 − S 2 + r 2 − y 2 ⇒ x ' = p − 1 ( 0 − − 2 p y 2 r 2 − p y 2 ) + 0 − − 2 y 2 R 2 − y 2 − 0 + − 2 y 2 r 2 − y 2 ⇒ x ' = p − 1 ( p y r 2 − p y 2 ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14.4, Problem 70E

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To show:

$x = f (y) = p^{- 1} (r - \sqrt{r^{2} - p y^{2}})$

Also, radius of curvature at point P $(y = 0)$ is $\frac{1}{R}$ .

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Chapter 14.4, Problem 69E

Chapter 14.4, Problem 71E

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Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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x = f ( y ) = p − 1 ( r − r 2 − p y 2 ) Also, radius of curvature at point P ( y = 0 ) is 1 R .

Solution Summary: The author explains how to solve the quadratic equation for x in terms of y.

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To show:

$x = f (y) = p^{- 1} (r - \sqrt{r^{2} - p y^{2}})$

Also, radius of curvature at point P $(y = 0)$ is $\frac{1}{R}$ .

Want to see the full answer?

Chapter 14 Solutions

x = f ( y ) = p − 1 ( r − r 2 − p y 2 ) Also, radius of curvature at point P ( y = 0 ) is 1 R .

Solution Summary: The author explains how to solve the quadratic equation for x in terms of y.

Concept explainers

Videos

To show: x=f(y)=p−1(r−r2−py2) Also, radius of curvature at point P(y=0) is 1R.

Want to see the full answer?

Chapter 14 Solutions

To show:

$x = f (y) = p^{- 1} (r - \sqrt{r^{2} - p y^{2}})$

Also, radius of curvature at point P $(y = 0)$ is $\frac{1}{R}$ .