Exercise 5.3 in Calculus Made Easy: are these answers equivalent?
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $mathrm dover mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)sqrta^2 - b^2$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$beginalign
mathrm dover mathrm dx(ay + bx) &= mathrm dover mathrm dxleft(by - ax + (x + y)sqrta^2 - b^2right)\
amathrm dyover mathrm dx + b &= bmathrm dyover mathrm dx - a + left(1 + dyover dxright)sqrta^2 - b^2\
left(a - b - sqrta^2 - b^2right)mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b\
mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2
endalign$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$beginalign
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\
left[(a - b)^2 - left(a^2 - b^2right)right]y^2 &= left[left(a^2 - b^2right) - (a + b)^2right]x^2\
2b(b - a)y^2 &= -2b(b + a)x^2\
y &= sqrta + b over a - bx\
mathrm dyover mathrm dx &= sqrta + b over a - b
endalign$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
calculus
asked May 9 '17 at 13:34
Alex D
1875
add a comment |Â
up vote
9
down vote
favorite
Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $mathrm dover mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)sqrta^2 - b^2$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$beginalign
mathrm dover mathrm dx(ay + bx) &= mathrm dover mathrm dxleft(by - ax + (x + y)sqrta^2 - b^2right)\
amathrm dyover mathrm dx + b &= bmathrm dyover mathrm dx - a + left(1 + dyover dxright)sqrta^2 - b^2\
left(a - b - sqrta^2 - b^2right)mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b\
mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2
endalign$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$beginalign
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\
left[(a - b)^2 - left(a^2 - b^2right)right]y^2 &= left[left(a^2 - b^2right) - (a + b)^2right]x^2\
2b(b - a)y^2 &= -2b(b + a)x^2\
y &= sqrta + b over a - bx\
mathrm dyover mathrm dx &= sqrta + b over a - b
endalign$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
calculus
asked May 9 '17 at 13:34
Alex D
1875
2
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $mathrm dover mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)sqrta^2 - b^2$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$beginalign
mathrm dover mathrm dx(ay + bx) &= mathrm dover mathrm dxleft(by - ax + (x + y)sqrta^2 - b^2right)\
amathrm dyover mathrm dx + b &= bmathrm dyover mathrm dx - a + left(1 + dyover dxright)sqrta^2 - b^2\
left(a - b - sqrta^2 - b^2right)mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b\
mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2
endalign$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$beginalign
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\
left[(a - b)^2 - left(a^2 - b^2right)right]y^2 &= left[left(a^2 - b^2right) - (a + b)^2right]x^2\
2b(b - a)y^2 &= -2b(b + a)x^2\
y &= sqrta + b over a - bx\
mathrm dyover mathrm dx &= sqrta + b over a - b
endalign$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
calculus
asked May 9 '17 at 13:34
Alex D
1875
Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $mathrm dover mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)sqrta^2 - b^2$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$beginalign
mathrm dover mathrm dx(ay + bx) &= mathrm dover mathrm dxleft(by - ax + (x + y)sqrta^2 - b^2right)\
amathrm dyover mathrm dx + b &= bmathrm dyover mathrm dx - a + left(1 + dyover dxright)sqrta^2 - b^2\
left(a - b - sqrta^2 - b^2right)mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b\
mathrm dyover mathrm dx &= sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2
endalign$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$beginalign
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\
left[(a - b)^2 - left(a^2 - b^2right)right]y^2 &= left[left(a^2 - b^2right) - (a + b)^2right]x^2\
2b(b - a)y^2 &= -2b(b + a)x^2\
y &= sqrta + b over a - bx\
mathrm dyover mathrm dx &= sqrta + b over a - b
endalign$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
calculus
calculus
asked May 9 '17 at 13:34
Alex D
1875
asked May 9 '17 at 13:34
Alex D
1875
edited Sep 9 at 9:29
asked May 9 '17 at 13:34
Alex D
1875
asked May 9 '17 at 13:34
Alex D
1875
asked May 9 '17 at 13:34
Alex D
1875
1875
2
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10
add a comment |Â
2
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10
2
2
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
14
down vote
accepted
$$sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2=fracsqrta+bleft(sqrta-b-sqrta+bright)sqrta-bleft(sqrta-b-sqrta+bright)=fracsqrta+bsqrta-b$$
answered May 9 '17 at 13:41
CY Aries
16.6k11741
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
$$sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2=fracsqrta+bleft(sqrta-b-sqrta+bright)sqrta-bleft(sqrta-b-sqrta+bright)=fracsqrta+bsqrta-b$$
answered May 9 '17 at 13:41
CY Aries
16.6k11741
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
add a comment |Â
up vote
14
down vote
accepted
$$sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2=fracsqrta+bleft(sqrta-b-sqrta+bright)sqrta-bleft(sqrta-b-sqrta+bright)=fracsqrta+bsqrta-b$$
answered May 9 '17 at 13:41
CY Aries
16.6k11741
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
add a comment |Â
up vote
14
down vote
accepted
up vote
14
down vote
accepted
$$sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2=fracsqrta+bleft(sqrta-b-sqrta+bright)sqrta-bleft(sqrta-b-sqrta+bright)=fracsqrta+bsqrta-b$$
answered May 9 '17 at 13:41
CY Aries
16.6k11741
$$sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2=fracsqrta+bleft(sqrta-b-sqrta+bright)sqrta-bleft(sqrta-b-sqrta+bright)=fracsqrta+bsqrta-b$$
answered May 9 '17 at 13:41
CY Aries
16.6k11741
answered May 9 '17 at 13:41
CY Aries
16.6k11741
answered May 9 '17 at 13:41
CY Aries
16.6k11741
answered May 9 '17 at 13:41
CY Aries
16.6k11741
16.6k11741
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
add a comment |Â
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
Wow! How did you "see" that?
– Alex D
May 9 '17 at 13:46
2
2
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
Given the desired endpoint, it was probably just a matter of seeing whether you can factor the desired numerator out of the top, and the desired denominator out of the bottom. Turns out, that worked in this case.
– G Tony Jacobs
May 9 '17 at 14:01
2
2
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
@AlexD Also, you have a Difference of Two Squares ($a^2-b^2$) which is one of the "I need to factor this" situations.
– AlgorithmsX
May 9 '17 at 15:50
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2273072%2fexercise-5-3-in-calculus-made-easy-are-these-answers-equivalent%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Note that if you had done the same work before differentiating to isolate $y$ that you did afterwards to isolate $fracdydx$, you would have found $$y = sqrta^2 - b^2 - a - b over a - b - sqrta^2 - b^2x$$which at least tells you the issue was strictly algebraic. Note also that Dr. Thompson can only get away with his manipulations because he assumes that $x$ and $y$ are positive. Otherwise his method would require additional care that he ignores.
– Paul Sinclair
May 9 '17 at 17:10