Mixing
Eliminate the parameter to find a Cartesian equation of the curve with $x = t^2$
$begingroup$
The problem requests I "eliminate the parameter to find a Cartesian equation of the curve."
The given equations are:
$x = t^2 $
and $y = t^5 $
I wasn't having trouble with these problems until this one, where when attempting to isolate $t$ in the $x$ equation, you get two different equations:
$ t = -(x)^{frac 12} $ and $ t = (x)^{frac 12} $
How do I go about accounting for both eqations when substituting my $t$ in the y equation $y = t^5 $?
calculus parametric
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
$endgroup$
add a comment |
$begingroup$
The problem requests I "eliminate the parameter to find a Cartesian equation of the curve."
The given equations are:
$x = t^2 $
and $y = t^5 $
I wasn't having trouble with these problems until this one, where when attempting to isolate $t$ in the $x$ equation, you get two different equations:
$ t = -(x)^{frac 12} $ and $ t = (x)^{frac 12} $
How do I go about accounting for both eqations when substituting my $t$ in the y equation $y = t^5 $?
calculus parametric
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
$endgroup$
2
$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27
add a comment |
$begingroup$
The problem requests I "eliminate the parameter to find a Cartesian equation of the curve."
The given equations are:
$x = t^2 $
and $y = t^5 $
I wasn't having trouble with these problems until this one, where when attempting to isolate $t$ in the $x$ equation, you get two different equations:
$ t = -(x)^{frac 12} $ and $ t = (x)^{frac 12} $
How do I go about accounting for both eqations when substituting my $t$ in the y equation $y = t^5 $?
calculus parametric
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
$endgroup$
The problem requests I "eliminate the parameter to find a Cartesian equation of the curve."
The given equations are:
$x = t^2 $
and $y = t^5 $
I wasn't having trouble with these problems until this one, where when attempting to isolate $t$ in the $x$ equation, you get two different equations:
$ t = -(x)^{frac 12} $ and $ t = (x)^{frac 12} $
How do I go about accounting for both eqations when substituting my $t$ in the y equation $y = t^5 $?
calculus parametric
calculus parametric
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
edited Aug 14 '18 at 18:07
Mohammad Zuhair Khan
1,5122525
1,5122525
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
asked Aug 14 '18 at 16:56
ElaredoElaredo
272
272
2
$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27
add a comment |
2
$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27
2
2
$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Scrutiny of the two parametric equations
$x = t^2, tag 1$
$y = t^5, tag 2$
indicates that, if we assume the parameter $t in Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $Bbb R$, $y$ will also take any value in $Bbb R = (-infty, infty)$, whereas $x$ is restricted to lie in the range $[0, infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, tag 3$
and
$y^2 = (t^5)^2 = t^{10}, tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, tag 5$
or
$y^2 - x^5 = 0; tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t in Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x in [0, infty)$, but allow $y$ to take on any value in $Bbb R$; therefore we may always find $t in Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t mapsto t^5 tag 7$
is a differentiable bijection from $Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} ge 0; tag 8$
we may then apply the function
$sqrt[5]{cdot}: Bbb R_{ge 0} to Bbb R_{ge 0}, tag 9$
to each side of (8) to find
$x = sqrt[5]{x^5} = sqrt[5]{t^{10}} = t^2, tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
How about letting x be the square root of t and y be the fifth root of t. Then, you can set them equal as in $ y^{1/5}=pmsqrt{x}$. Then, when you quintuple the power of both sides you would have $y=pmsqrt{x^5}$.
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
$endgroup$
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Scrutiny of the two parametric equations
$x = t^2, tag 1$
$y = t^5, tag 2$
indicates that, if we assume the parameter $t in Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $Bbb R$, $y$ will also take any value in $Bbb R = (-infty, infty)$, whereas $x$ is restricted to lie in the range $[0, infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, tag 3$
and
$y^2 = (t^5)^2 = t^{10}, tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, tag 5$
or
$y^2 - x^5 = 0; tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t in Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x in [0, infty)$, but allow $y$ to take on any value in $Bbb R$; therefore we may always find $t in Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t mapsto t^5 tag 7$
is a differentiable bijection from $Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} ge 0; tag 8$
we may then apply the function
$sqrt[5]{cdot}: Bbb R_{ge 0} to Bbb R_{ge 0}, tag 9$
to each side of (8) to find
$x = sqrt[5]{x^5} = sqrt[5]{t^{10}} = t^2, tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
Scrutiny of the two parametric equations
$x = t^2, tag 1$
$y = t^5, tag 2$
indicates that, if we assume the parameter $t in Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $Bbb R$, $y$ will also take any value in $Bbb R = (-infty, infty)$, whereas $x$ is restricted to lie in the range $[0, infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, tag 3$
and
$y^2 = (t^5)^2 = t^{10}, tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, tag 5$
or
$y^2 - x^5 = 0; tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t in Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x in [0, infty)$, but allow $y$ to take on any value in $Bbb R$; therefore we may always find $t in Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t mapsto t^5 tag 7$
is a differentiable bijection from $Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} ge 0; tag 8$
we may then apply the function
$sqrt[5]{cdot}: Bbb R_{ge 0} to Bbb R_{ge 0}, tag 9$
to each side of (8) to find
$x = sqrt[5]{x^5} = sqrt[5]{t^{10}} = t^2, tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
$endgroup$
add a comment |
$begingroup$
Scrutiny of the two parametric equations
$x = t^2, tag 1$
$y = t^5, tag 2$
indicates that, if we assume the parameter $t in Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $Bbb R$, $y$ will also take any value in $Bbb R = (-infty, infty)$, whereas $x$ is restricted to lie in the range $[0, infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, tag 3$
and
$y^2 = (t^5)^2 = t^{10}, tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, tag 5$
or
$y^2 - x^5 = 0; tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t in Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x in [0, infty)$, but allow $y$ to take on any value in $Bbb R$; therefore we may always find $t in Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t mapsto t^5 tag 7$
is a differentiable bijection from $Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} ge 0; tag 8$
we may then apply the function
$sqrt[5]{cdot}: Bbb R_{ge 0} to Bbb R_{ge 0}, tag 9$
to each side of (8) to find
$x = sqrt[5]{x^5} = sqrt[5]{t^{10}} = t^2, tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
$endgroup$
Scrutiny of the two parametric equations
$x = t^2, tag 1$
$y = t^5, tag 2$
indicates that, if we assume the parameter $t in Bbb R$ with no other restrictions, so that $t$ is allowed to vary over all of $Bbb R$, $y$ will also take any value in $Bbb R = (-infty, infty)$, whereas $x$ is restricted to lie in the range $[0, infty)$; this is of course intuitively seen to be the case since odd powers of $t$ preserve its sign, whereas even powers "fold" the $t$-axis back on itself at $t = 0$, so that both negative and positive values of $t$ map to the positives in $Bbb R$; $0$ of course maps to itself for both even and odd exponents.
We note that, algebraically, we have
$x^5 = (t^2)^5 = t^{10}, tag 3$
and
$y^2 = (t^5)^2 = t^{10}, tag 4$
so that a necessary condition for (1), (2) is
$y^2 = x^5, tag 5$
or
$y^2 - x^5 = 0; tag 6$
every point $(x, y)$ satisfying (1), (2) also obeys (6).
Going the other way, that is, trying to work backwards from (5)-(6) towards (1)-(2), we would like to find, for any pair $(x, y)$ obeying (5), (6), some $t in Bbb R$ such that (1)-(2) bind. We observe that (5)-(6) restrict $x in [0, infty)$, but allow $y$ to take on any value in $Bbb R$; therefore we may always find $t in Bbb R$ such that (2) holds; furthermore, such a $t$ is uniquely determined by $y$, since the map
$t mapsto t^5 tag 7$
is a differentiable bijection from $Bbb R$ to itself; accepting then that $y = t^5$, (5) yields
$x^5 = y^2 = t^{10} ge 0; tag 8$
we may then apply the function
$sqrt[5]{cdot}: Bbb R_{ge 0} to Bbb R_{ge 0}, tag 9$
to each side of (8) to find
$x = sqrt[5]{x^5} = sqrt[5]{t^{10}} = t^2, tag{10}$
re-creating the original parametric form (1)-(2). We thus see that (5)-(6) are also sufficient for (1)-(2); thus the Cartesian forms (5)-(6) are equivalent to the parametric presentation (1)-(2).
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
edited Aug 15 '18 at 19:34
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
answered Aug 15 '18 at 19:23
Robert LewisRobert Lewis
45.1k23065
45.1k23065
add a comment |
add a comment |
$begingroup$
How about letting x be the square root of t and y be the fifth root of t. Then, you can set them equal as in $ y^{1/5}=pmsqrt{x}$. Then, when you quintuple the power of both sides you would have $y=pmsqrt{x^5}$.
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
$endgroup$
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
add a comment |
$begingroup$
How about letting x be the square root of t and y be the fifth root of t. Then, you can set them equal as in $ y^{1/5}=pmsqrt{x}$. Then, when you quintuple the power of both sides you would have $y=pmsqrt{x^5}$.
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
$endgroup$
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
add a comment |
$begingroup$
How about letting x be the square root of t and y be the fifth root of t. Then, you can set them equal as in $ y^{1/5}=pmsqrt{x}$. Then, when you quintuple the power of both sides you would have $y=pmsqrt{x^5}$.
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
$endgroup$
How about letting x be the square root of t and y be the fifth root of t. Then, you can set them equal as in $ y^{1/5}=pmsqrt{x}$. Then, when you quintuple the power of both sides you would have $y=pmsqrt{x^5}$.
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
edited Jan 6 at 18:23
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
answered Aug 14 '18 at 17:04
poetasispoetasis
412117
412117
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
add a comment |
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
$begingroup$
This misses the branch of the curve with $y < 0$, unfortunately.
$endgroup$
– Matthew Leingang
Aug 14 '18 at 17:17
1
1
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
@MatthewLeingang I think the problem can be avoided if we use $pm$.
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:20
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
Then you’re not describing a single equation.
$endgroup$
– Matthew Leingang
Aug 15 '18 at 1:03
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
The square root of x implies plus and minus. If the square root of x is negative, y will be negative because raising a negative to the fifth power yields a negative.
$endgroup$
– poetasis
Aug 15 '18 at 16:23
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
$begingroup$
@Matthew Leingang Please read my previous comment and let me know if it satisfies or supplies the negative you found missing in my single equation.
$endgroup$
– poetasis
Aug 16 '18 at 16:37
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$begingroup$
How about $y^2 = x^5$? Or you may be wanting $y = pm x^2sqrt x$
$endgroup$
– Mohammad Zuhair Khan
Aug 14 '18 at 17:02
$begingroup$
Graphing y^2 = x^5 shows that its correct. Thank you for the suggestion, I never even considered that as an option.
$endgroup$
– Elaredo
Aug 14 '18 at 17:59
$begingroup$
@MohammadZuhairKhan yep
$endgroup$
– Cloud JR
Aug 14 '18 at 18:10
$begingroup$
You can also consider a rotated base, after all $x=|y|^{frac 25}=sqrt[5]{y^2}$ is perfectly valid too.
$endgroup$
– zwim
Jan 6 at 18:27