Mixing
Query on the Problem: Find all points on the curve $x^2y^2+xy=2$ where the slope of the tangent line is $-1$.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
A problem on James Stewart's book Essential Calculus with Differential Equations Section 2.6 Number 41 states that:
"Find all points on the curve
beginequation
x^2y^2+xy=2
endequation
where the slope of the tangent line is $-1$."
By reading the problem, I thought that the problem is easy and actually it is. Until I stumbled into something. Below is my solution to the problem and my questions will be written after I wrote my solution.
$textbfSolution:$
To find all points $(x,y)$ such that the above equation has tangent line whose slope is $-1$, we will just simply use implicit differentiation to find $y'$ and equate it to $-1$; since $left.y'right|_(x,y)$ is the slope of the tangent line to the curve of the given equation at point $(x,y)$.
Doing that gives the equation
beginequation
y'=-frac2xy^2+y2x^2y+x=-1.
endequation
The equation above implies $2xy^2+y=2x^2y+x$.
After some simplifications and by factoring by grouping we will arrive at:
beginequation
(y-x)(2xy+1)=0.
endequation
This means that $x-y=0$ that is $x=y$ or $2xy+1=0$. Knowing that $x=y$, we arrived at $2x^2+1=0$ whose roots are both imaginary.
At this point, I concluded that there is no such point $(x,y)$ such that $x^2y^2+xy=2$ has a tangent line with slope $-1$. But I stumbled for a while and think that it is irrational for the question to be included as an exercise in the book if it has no solution.
Where did I go wrong? Why am I wrong if i stopped at this part? Did I missed something?
Did I missed the other implication of $x=y$? That is; If $x=y$ then the original equation becomes:
beginequation
x^2x^2+xx=2
endequation
From here we get $x=pm1$ and from here it follows that $y=1$ if $x=1$ and $y=-1$ if $x=-1$. Which means, the points at which the graph of the equation $x^2y^2+xy=2$ has a tangent lines with slope $-1$ are $(-1,-1)$ and $(1,1)$.
Is this answer correct?
multivariable-calculus implicit-differentiation tangent-line
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
asked Jul 17 at 8:12
Jr Antalan
1,2221722
add a comment |Â
up vote
0
down vote
favorite
A problem on James Stewart's book Essential Calculus with Differential Equations Section 2.6 Number 41 states that:
"Find all points on the curve
beginequation
x^2y^2+xy=2
endequation
where the slope of the tangent line is $-1$."
By reading the problem, I thought that the problem is easy and actually it is. Until I stumbled into something. Below is my solution to the problem and my questions will be written after I wrote my solution.
$textbfSolution:$
To find all points $(x,y)$ such that the above equation has tangent line whose slope is $-1$, we will just simply use implicit differentiation to find $y'$ and equate it to $-1$; since $left.y'right|_(x,y)$ is the slope of the tangent line to the curve of the given equation at point $(x,y)$.
Doing that gives the equation
beginequation
y'=-frac2xy^2+y2x^2y+x=-1.
endequation
The equation above implies $2xy^2+y=2x^2y+x$.
After some simplifications and by factoring by grouping we will arrive at:
beginequation
(y-x)(2xy+1)=0.
endequation
This means that $x-y=0$ that is $x=y$ or $2xy+1=0$. Knowing that $x=y$, we arrived at $2x^2+1=0$ whose roots are both imaginary.
At this point, I concluded that there is no such point $(x,y)$ such that $x^2y^2+xy=2$ has a tangent line with slope $-1$. But I stumbled for a while and think that it is irrational for the question to be included as an exercise in the book if it has no solution.
Where did I go wrong? Why am I wrong if i stopped at this part? Did I missed something?
Did I missed the other implication of $x=y$? That is; If $x=y$ then the original equation becomes:
beginequation
x^2x^2+xx=2
endequation
From here we get $x=pm1$ and from here it follows that $y=1$ if $x=1$ and $y=-1$ if $x=-1$. Which means, the points at which the graph of the equation $x^2y^2+xy=2$ has a tangent lines with slope $-1$ are $(-1,-1)$ and $(1,1)$.
Is this answer correct?
multivariable-calculus implicit-differentiation tangent-line
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
asked Jul 17 at 8:12
Jr Antalan
1,2221722
6
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A problem on James Stewart's book Essential Calculus with Differential Equations Section 2.6 Number 41 states that:
"Find all points on the curve
beginequation
x^2y^2+xy=2
endequation
where the slope of the tangent line is $-1$."
By reading the problem, I thought that the problem is easy and actually it is. Until I stumbled into something. Below is my solution to the problem and my questions will be written after I wrote my solution.
$textbfSolution:$
To find all points $(x,y)$ such that the above equation has tangent line whose slope is $-1$, we will just simply use implicit differentiation to find $y'$ and equate it to $-1$; since $left.y'right|_(x,y)$ is the slope of the tangent line to the curve of the given equation at point $(x,y)$.
Doing that gives the equation
beginequation
y'=-frac2xy^2+y2x^2y+x=-1.
endequation
The equation above implies $2xy^2+y=2x^2y+x$.
After some simplifications and by factoring by grouping we will arrive at:
beginequation
(y-x)(2xy+1)=0.
endequation
This means that $x-y=0$ that is $x=y$ or $2xy+1=0$. Knowing that $x=y$, we arrived at $2x^2+1=0$ whose roots are both imaginary.
At this point, I concluded that there is no such point $(x,y)$ such that $x^2y^2+xy=2$ has a tangent line with slope $-1$. But I stumbled for a while and think that it is irrational for the question to be included as an exercise in the book if it has no solution.
Where did I go wrong? Why am I wrong if i stopped at this part? Did I missed something?
Did I missed the other implication of $x=y$? That is; If $x=y$ then the original equation becomes:
beginequation
x^2x^2+xx=2
endequation
From here we get $x=pm1$ and from here it follows that $y=1$ if $x=1$ and $y=-1$ if $x=-1$. Which means, the points at which the graph of the equation $x^2y^2+xy=2$ has a tangent lines with slope $-1$ are $(-1,-1)$ and $(1,1)$.
Is this answer correct?
multivariable-calculus implicit-differentiation tangent-line
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
asked Jul 17 at 8:12
Jr Antalan
1,2221722
A problem on James Stewart's book Essential Calculus with Differential Equations Section 2.6 Number 41 states that:
"Find all points on the curve
beginequation
x^2y^2+xy=2
endequation
where the slope of the tangent line is $-1$."
By reading the problem, I thought that the problem is easy and actually it is. Until I stumbled into something. Below is my solution to the problem and my questions will be written after I wrote my solution.
$textbfSolution:$
To find all points $(x,y)$ such that the above equation has tangent line whose slope is $-1$, we will just simply use implicit differentiation to find $y'$ and equate it to $-1$; since $left.y'right|_(x,y)$ is the slope of the tangent line to the curve of the given equation at point $(x,y)$.
Doing that gives the equation
beginequation
y'=-frac2xy^2+y2x^2y+x=-1.
endequation
The equation above implies $2xy^2+y=2x^2y+x$.
After some simplifications and by factoring by grouping we will arrive at:
beginequation
(y-x)(2xy+1)=0.
endequation
This means that $x-y=0$ that is $x=y$ or $2xy+1=0$. Knowing that $x=y$, we arrived at $2x^2+1=0$ whose roots are both imaginary.
At this point, I concluded that there is no such point $(x,y)$ such that $x^2y^2+xy=2$ has a tangent line with slope $-1$. But I stumbled for a while and think that it is irrational for the question to be included as an exercise in the book if it has no solution.
Where did I go wrong? Why am I wrong if i stopped at this part? Did I missed something?
Did I missed the other implication of $x=y$? That is; If $x=y$ then the original equation becomes:
beginequation
x^2x^2+xx=2
endequation
From here we get $x=pm1$ and from here it follows that $y=1$ if $x=1$ and $y=-1$ if $x=-1$. Which means, the points at which the graph of the equation $x^2y^2+xy=2$ has a tangent lines with slope $-1$ are $(-1,-1)$ and $(1,1)$.
Is this answer correct?
multivariable-calculus implicit-differentiation tangent-line
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
asked Jul 17 at 8:12
Jr Antalan
1,2221722
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
edited Jul 17 at 10:35
José Carlos Santos
114k1698177
114k1698177
asked Jul 17 at 8:12
Jr Antalan
1,2221722
asked Jul 17 at 8:12
Jr Antalan
1,2221722
asked Jul 17 at 8:12
Jr Antalan
1,2221722
1,2221722
6
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59
add a comment |Â
6
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59
6
6
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
3
down vote
accepted
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $pm1$. So $pm(1,1)$ are solutions of your problem. See the image below.
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation
$$
fracpartial fpartial x(a,b)(x-a)+
fracpartial fpartial y(a,b)(y-b)=0
$$
The slope is $-1$ if and only if
$$
fracpartial fpartial x(a,b)=fracpartial fpartial y(a,b)
$$
This leads to
$$
begincases
a^2b^2+ab=2 \[4px]
2ab^2+b=2a^2b+a
endcases
$$
The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.
The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=pm1$.
The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.
The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.
Actually the curve has singular points at infinity. Indeed, the homogeneous equation is
$$
F(x,y,z)=x^2y^2+xyz^2-2z^4=0
$$
and
$$
fracpartial Fpartial x=2xy^2+yz^2
qquad
fracpartial Fpartial y=2x^2y+xz^2
qquad
fracpartial Fpartial z=2xyz-8z^3
$$
The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.
answered Jul 17 at 9:04
egreg
164k1180187
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
add a comment |Â
up vote
0
down vote
Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.
The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = pm 1$, while the second has no solution for $y' = -1.$
answered Jul 17 at 8:56
md2perpe
5,93011022
add a comment |Â
up vote
0
down vote
I get it now, the problem in what I did which results into an imaginary value of $x$ is that the two implications of the equation
beginequation
(y-x)(2xy+1)=0
endequation
which is $x=y$ or $y=-frac12x$ must be used in the original equation and not to be used into the equation $(y-x)(2xy+1)=0$ in which the two implications were derived.
Doing that leads to the correct answer $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 8:31
Jr Antalan
1,2221722
add a comment |Â
up vote
0
down vote
We may find the derivative $dfracrm dyrm dx$ as below
$$2xy^2+ x^2 cdot 2y cdot fracrm dyrm dx+y+x fracrm dyrm dx=0,$$ thus $$fracrm dyrm dx=-frac2xy^2+y2x^2y+x=-fracy(2xy+1)x(2xy+1)=-fracyx.$$
Let $$fracrm dyrm dx=-fracyx=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$
As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 10:16
mengdie1982
2,972216
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $pm1$. So $pm(1,1)$ are solutions of your problem. See the image below.
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
accepted
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $pm1$. So $pm(1,1)$ are solutions of your problem. See the image below.
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $pm1$. So $pm(1,1)$ are solutions of your problem. See the image below.
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $pm1$. So $pm(1,1)$ are solutions of your problem. See the image below.
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
edited Jul 17 at 8:36
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
answered Jul 17 at 8:28
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
1
down vote
The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation
$$
fracpartial fpartial x(a,b)(x-a)+
fracpartial fpartial y(a,b)(y-b)=0
$$
The slope is $-1$ if and only if
$$
fracpartial fpartial x(a,b)=fracpartial fpartial y(a,b)
$$
This leads to
$$
begincases
a^2b^2+ab=2 \[4px]
2ab^2+b=2a^2b+a
endcases
$$
The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.
The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=pm1$.
The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.
The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.
Actually the curve has singular points at infinity. Indeed, the homogeneous equation is
$$
F(x,y,z)=x^2y^2+xyz^2-2z^4=0
$$
and
$$
fracpartial Fpartial x=2xy^2+yz^2
qquad
fracpartial Fpartial y=2x^2y+xz^2
qquad
fracpartial Fpartial z=2xyz-8z^3
$$
The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.
answered Jul 17 at 9:04
egreg
164k1180187
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
add a comment |Â
up vote
1
down vote
The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation
$$
fracpartial fpartial x(a,b)(x-a)+
fracpartial fpartial y(a,b)(y-b)=0
$$
The slope is $-1$ if and only if
$$
fracpartial fpartial x(a,b)=fracpartial fpartial y(a,b)
$$
This leads to
$$
begincases
a^2b^2+ab=2 \[4px]
2ab^2+b=2a^2b+a
endcases
$$
The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.
The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=pm1$.
The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.
The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.
Actually the curve has singular points at infinity. Indeed, the homogeneous equation is
$$
F(x,y,z)=x^2y^2+xyz^2-2z^4=0
$$
and
$$
fracpartial Fpartial x=2xy^2+yz^2
qquad
fracpartial Fpartial y=2x^2y+xz^2
qquad
fracpartial Fpartial z=2xyz-8z^3
$$
The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.
answered Jul 17 at 9:04
egreg
164k1180187
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation
$$
fracpartial fpartial x(a,b)(x-a)+
fracpartial fpartial y(a,b)(y-b)=0
$$
The slope is $-1$ if and only if
$$
fracpartial fpartial x(a,b)=fracpartial fpartial y(a,b)
$$
This leads to
$$
begincases
a^2b^2+ab=2 \[4px]
2ab^2+b=2a^2b+a
endcases
$$
The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.
The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=pm1$.
The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.
The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.
Actually the curve has singular points at infinity. Indeed, the homogeneous equation is
$$
F(x,y,z)=x^2y^2+xyz^2-2z^4=0
$$
and
$$
fracpartial Fpartial x=2xy^2+yz^2
qquad
fracpartial Fpartial y=2x^2y+xz^2
qquad
fracpartial Fpartial z=2xyz-8z^3
$$
The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.
answered Jul 17 at 9:04
egreg
164k1180187
The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation
$$
fracpartial fpartial x(a,b)(x-a)+
fracpartial fpartial y(a,b)(y-b)=0
$$
The slope is $-1$ if and only if
$$
fracpartial fpartial x(a,b)=fracpartial fpartial y(a,b)
$$
This leads to
$$
begincases
a^2b^2+ab=2 \[4px]
2ab^2+b=2a^2b+a
endcases
$$
The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.
The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=pm1$.
The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.
The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.
Actually the curve has singular points at infinity. Indeed, the homogeneous equation is
$$
F(x,y,z)=x^2y^2+xyz^2-2z^4=0
$$
and
$$
fracpartial Fpartial x=2xy^2+yz^2
qquad
fracpartial Fpartial y=2x^2y+xz^2
qquad
fracpartial Fpartial z=2xyz-8z^3
$$
The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.
answered Jul 17 at 9:04
egreg
164k1180187
answered Jul 17 at 9:04
egreg
164k1180187
answered Jul 17 at 9:04
egreg
164k1180187
answered Jul 17 at 9:04
egreg
164k1180187
164k1180187
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
add a comment |Â
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.
– Jr Antalan
Jul 17 at 9:20
add a comment |Â
up vote
0
down vote
Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.
The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = pm 1$, while the second has no solution for $y' = -1.$
answered Jul 17 at 8:56
md2perpe
5,93011022
add a comment |Â
up vote
0
down vote
Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.
The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = pm 1$, while the second has no solution for $y' = -1.$
answered Jul 17 at 8:56
md2perpe
5,93011022
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.
The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = pm 1$, while the second has no solution for $y' = -1.$
answered Jul 17 at 8:56
md2perpe
5,93011022
Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.
The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = pm 1$, while the second has no solution for $y' = -1.$
answered Jul 17 at 8:56
md2perpe
5,93011022
answered Jul 17 at 8:56
md2perpe
5,93011022
answered Jul 17 at 8:56
md2perpe
5,93011022
answered Jul 17 at 8:56
md2perpe
5,93011022
5,93011022
add a comment |Â
add a comment |Â
up vote
0
down vote
I get it now, the problem in what I did which results into an imaginary value of $x$ is that the two implications of the equation
beginequation
(y-x)(2xy+1)=0
endequation
which is $x=y$ or $y=-frac12x$ must be used in the original equation and not to be used into the equation $(y-x)(2xy+1)=0$ in which the two implications were derived.
Doing that leads to the correct answer $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 8:31
Jr Antalan
1,2221722
add a comment |Â
up vote
0
down vote
I get it now, the problem in what I did which results into an imaginary value of $x$ is that the two implications of the equation
beginequation
(y-x)(2xy+1)=0
endequation
which is $x=y$ or $y=-frac12x$ must be used in the original equation and not to be used into the equation $(y-x)(2xy+1)=0$ in which the two implications were derived.
Doing that leads to the correct answer $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 8:31
Jr Antalan
1,2221722
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I get it now, the problem in what I did which results into an imaginary value of $x$ is that the two implications of the equation
beginequation
(y-x)(2xy+1)=0
endequation
which is $x=y$ or $y=-frac12x$ must be used in the original equation and not to be used into the equation $(y-x)(2xy+1)=0$ in which the two implications were derived.
Doing that leads to the correct answer $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 8:31
Jr Antalan
1,2221722
I get it now, the problem in what I did which results into an imaginary value of $x$ is that the two implications of the equation
beginequation
(y-x)(2xy+1)=0
endequation
which is $x=y$ or $y=-frac12x$ must be used in the original equation and not to be used into the equation $(y-x)(2xy+1)=0$ in which the two implications were derived.
Doing that leads to the correct answer $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 8:31
Jr Antalan
1,2221722
edited Jul 17 at 8:58
answered Jul 17 at 8:31
Jr Antalan
1,2221722
answered Jul 17 at 8:31
Jr Antalan
1,2221722
answered Jul 17 at 8:31
Jr Antalan
1,2221722
1,2221722
add a comment |Â
add a comment |Â
up vote
0
down vote
We may find the derivative $dfracrm dyrm dx$ as below
$$2xy^2+ x^2 cdot 2y cdot fracrm dyrm dx+y+x fracrm dyrm dx=0,$$ thus $$fracrm dyrm dx=-frac2xy^2+y2x^2y+x=-fracy(2xy+1)x(2xy+1)=-fracyx.$$
Let $$fracrm dyrm dx=-fracyx=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$
As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 10:16
mengdie1982
2,972216
add a comment |Â
up vote
0
down vote
We may find the derivative $dfracrm dyrm dx$ as below
$$2xy^2+ x^2 cdot 2y cdot fracrm dyrm dx+y+x fracrm dyrm dx=0,$$ thus $$fracrm dyrm dx=-frac2xy^2+y2x^2y+x=-fracy(2xy+1)x(2xy+1)=-fracyx.$$
Let $$fracrm dyrm dx=-fracyx=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$
As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 10:16
mengdie1982
2,972216
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We may find the derivative $dfracrm dyrm dx$ as below
$$2xy^2+ x^2 cdot 2y cdot fracrm dyrm dx+y+x fracrm dyrm dx=0,$$ thus $$fracrm dyrm dx=-frac2xy^2+y2x^2y+x=-fracy(2xy+1)x(2xy+1)=-fracyx.$$
Let $$fracrm dyrm dx=-fracyx=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$
As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 10:16
mengdie1982
2,972216
We may find the derivative $dfracrm dyrm dx$ as below
$$2xy^2+ x^2 cdot 2y cdot fracrm dyrm dx+y+x fracrm dyrm dx=0,$$ thus $$fracrm dyrm dx=-frac2xy^2+y2x^2y+x=-fracy(2xy+1)x(2xy+1)=-fracyx.$$
Let $$fracrm dyrm dx=-fracyx=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$
As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.
answered Jul 17 at 10:16
mengdie1982
2,972216
answered Jul 17 at 10:16
mengdie1982
2,972216
answered Jul 17 at 10:16
mengdie1982
2,972216
answered Jul 17 at 10:16
mengdie1982
2,972216
2,972216
add a comment |Â
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%2f2854283%2fquery-on-the-problem-find-all-points-on-the-curve-x2y2xy-2-where-the-slope%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
6
$x=y$ OR $2xy+1=0$, as you stated. But then, you suddenly proceed to reason as if it was an "AND" (you substitute the first equation inside the second). I think this is where the problem lies.
– Suzet
Jul 17 at 8:15
This problem is much easier if you realize this equation is $(xy-1)(xy+2)=0.$
– Thomas Andrews
Jul 17 at 8:59