Find $frac{partial y}{partial z}$ of the surface $g(s,t)=(s^2+2t,s+t,e^{st})$ near $g(1, 1) = (3, 2, e)$.
$begingroup$
Consider the surface given by $g(s, t) = (s^2 + 2t, s + t, e^{st})$.
Think of $y$ as a function of $x$ and $z$. Find $dfrac{partial y}{partial z}(3,e)$ near $g(1, 1) = (3, 2, e)$.
really tough question because i cant understand how it is related to implicit functions like what should be my $F(x,y,z)$.
thanks a lot for your help.
real-analysis multivariable-calculus partial-derivative parametrization implicit-function
edited Dec 8 '18 at 0:09
Batominovski
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
$endgroup$
|
show 1 more comment
$begingroup$
Consider the surface given by $g(s, t) = (s^2 + 2t, s + t, e^{st})$.
Think of $y$ as a function of $x$ and $z$. Find $dfrac{partial y}{partial z}(3,e)$ near $g(1, 1) = (3, 2, e)$.
really tough question because i cant understand how it is related to implicit functions like what should be my $F(x,y,z)$.
thanks a lot for your help.
real-analysis multivariable-calculus partial-derivative parametrization implicit-function
edited Dec 8 '18 at 0:09
Batominovski
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
$endgroup$
$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
$endgroup$
– user10354138
Dec 7 '18 at 17:36
$begingroup$
my bad its s^2 i dont know how to write that in language really sorry
$endgroup$
– Mather
Dec 7 '18 at 17:37
$begingroup$
i have editied it
$endgroup$
– Mather
Dec 7 '18 at 17:39
$begingroup$
i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
$endgroup$
– Mather
Dec 7 '18 at 18:09
$begingroup$
anyone ? i really need help i am staring at the question for the 10th time i really need a hint
$endgroup$
– Mather
Dec 7 '18 at 18:33
|
show 1 more comment
$begingroup$
Consider the surface given by $g(s, t) = (s^2 + 2t, s + t, e^{st})$.
Think of $y$ as a function of $x$ and $z$. Find $dfrac{partial y}{partial z}(3,e)$ near $g(1, 1) = (3, 2, e)$.
really tough question because i cant understand how it is related to implicit functions like what should be my $F(x,y,z)$.
thanks a lot for your help.
real-analysis multivariable-calculus partial-derivative parametrization implicit-function
edited Dec 8 '18 at 0:09
Batominovski
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
$endgroup$
Consider the surface given by $g(s, t) = (s^2 + 2t, s + t, e^{st})$.
Think of $y$ as a function of $x$ and $z$. Find $dfrac{partial y}{partial z}(3,e)$ near $g(1, 1) = (3, 2, e)$.
really tough question because i cant understand how it is related to implicit functions like what should be my $F(x,y,z)$.
thanks a lot for your help.
real-analysis multivariable-calculus partial-derivative parametrization implicit-function
real-analysis multivariable-calculus partial-derivative parametrization implicit-function
edited Dec 8 '18 at 0:09
Batominovski
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
edited Dec 8 '18 at 0:09
Batominovski
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
edited Dec 8 '18 at 0:09
Batominovski
1
edited Dec 8 '18 at 0:09
Batominovski
1
edited Dec 8 '18 at 0:09
Batominovski
1
1
asked Dec 7 '18 at 17:30
Mather Mather
1947
asked Dec 7 '18 at 17:30
Mather Mather
1947
asked Dec 7 '18 at 17:30
Mather Mather
1947
1947
$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
$endgroup$
– user10354138
Dec 7 '18 at 17:36
$begingroup$
my bad its s^2 i dont know how to write that in language really sorry
$endgroup$
– Mather
Dec 7 '18 at 17:37
$begingroup$
i have editied it
$endgroup$
– Mather
Dec 7 '18 at 17:39
$begingroup$
i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
$endgroup$
– Mather
Dec 7 '18 at 18:09
$begingroup$
anyone ? i really need help i am staring at the question for the 10th time i really need a hint
$endgroup$
– Mather
Dec 7 '18 at 18:33
|
show 1 more comment
$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
$endgroup$
– user10354138
Dec 7 '18 at 17:36
$begingroup$
my bad its s^2 i dont know how to write that in language really sorry
$endgroup$
– Mather
Dec 7 '18 at 17:37
$begingroup$
i have editied it
$endgroup$
– Mather
Dec 7 '18 at 17:39
$begingroup$
i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
$endgroup$
– Mather
Dec 7 '18 at 18:09
$begingroup$
anyone ? i really need help i am staring at the question for the 10th time i really need a hint
$endgroup$
– Mather
Dec 7 '18 at 18:33
$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
$endgroup$
– user10354138
Dec 7 '18 at 17:36
$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
$endgroup$
– user10354138
Dec 7 '18 at 17:36
$begingroup$
my bad its s^2 i dont know how to write that in language really sorry
$endgroup$
– Mather
Dec 7 '18 at 17:37
$begingroup$
my bad its s^2 i dont know how to write that in language really sorry
$endgroup$
– Mather
Dec 7 '18 at 17:37
$begingroup$
i have editied it
$endgroup$
– Mather
Dec 7 '18 at 17:39
$begingroup$
i have editied it
$endgroup$
– Mather
Dec 7 '18 at 17:39
$begingroup$
i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
$endgroup$
– Mather
Dec 7 '18 at 18:09
$begingroup$
i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
$endgroup$
– Mather
Dec 7 '18 at 18:09
$begingroup$
anyone ? i really need help i am staring at the question for the 10th time i really need a hint
$endgroup$
– Mather
Dec 7 '18 at 18:33
$begingroup$
anyone ? i really need help i am staring at the question for the 10th time i really need a hint
$endgroup$
– Mather
Dec 7 '18 at 18:33
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Note that $(x,y,z)=big(s^2+2t,s+t,exp(st)big)$ implies
$$text{d}x=2s,text{d}s+2,text{d}t,,tag{1}$$
$$text{d}y=text{d}s+text{d}t,,tag{2}$$
and
$$text{d}z=texp(st),text{d}s+sexp(st),text{d}t,.tag{3}$$
Thus, the differentials at $(s,t)=(1,1)$ are
$$text{d}x=2,text{d}s+2,text{d}t,,$$
$$text{d}y=text{d}s+text{d}t,,$$
and
$$text{d}z=text{e},text{d}s+text{e},text{d}t,.$$
To compute $dfrac{text{d}y}{text{d}z}(s=1,t=1)$, we require that $text{d}x$ and $text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $dfrac{text{d}y}{text{d}z}(s=1,t=1)$ may not exist (see a comment below).
Here is a way to compute $dfrac{text{d}y}{text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2neq t$). Using (1) and (3), we obtain
$$text{d}s=frac{sexp(st),text{d}x-2,text{d}z}{2,(s^2-t),exp(st)}$$
and
$$text{d}t=frac{2s,text{d}z-texp(st),text{d}x}{2,(s^2-t),exp(st)},.$$
By (2), we get
$$text{d}y=text{d}s+text{d}t=frac{s-t}{2,(s^2-t)},text{d}x+frac{s-1}{(s^2-t),exp(st)},text{d}z,.$$
Thus,
$$frac{partial y}{partial x}(s,t)=frac{s-t}{2,(s^2-t)}text{ and }frac{partial y}{partial z}(s,t)=frac{s-1}{(s^2-t),exp(st)},.$$
When $sigma^2=tau$, we see that
$$lim_{(s,t)to(sigma,tau)},frac{partial y}{partial x}(s,t)text{ and }lim_{(s,t)to(sigma,tau)},frac{partial y}{partial z}(s,t)$$
do not exist. Hence, $dfrac{partial y}{partial x}(s,t)$ and $dfrac{partial y}{partial z}(s,t)$ do not exist along the curve $s^2=t$.
As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
$endgroup$
1
$begingroup$
I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
$endgroup$
– user10354138
Dec 8 '18 at 6:33
add a comment |
$begingroup$
Hint: the cross product $frac{partial g}{partial s} times frac{partial g}{partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, frac{partial y}{partial z}, 1)$ which must therefore be perpendicular to this normal vector.
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
$endgroup$
1
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
1
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
add a comment |
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2 Answers
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$begingroup$
Note that $(x,y,z)=big(s^2+2t,s+t,exp(st)big)$ implies
$$text{d}x=2s,text{d}s+2,text{d}t,,tag{1}$$
$$text{d}y=text{d}s+text{d}t,,tag{2}$$
and
$$text{d}z=texp(st),text{d}s+sexp(st),text{d}t,.tag{3}$$
Thus, the differentials at $(s,t)=(1,1)$ are
$$text{d}x=2,text{d}s+2,text{d}t,,$$
$$text{d}y=text{d}s+text{d}t,,$$
and
$$text{d}z=text{e},text{d}s+text{e},text{d}t,.$$
To compute $dfrac{text{d}y}{text{d}z}(s=1,t=1)$, we require that $text{d}x$ and $text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $dfrac{text{d}y}{text{d}z}(s=1,t=1)$ may not exist (see a comment below).
Here is a way to compute $dfrac{text{d}y}{text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2neq t$). Using (1) and (3), we obtain
$$text{d}s=frac{sexp(st),text{d}x-2,text{d}z}{2,(s^2-t),exp(st)}$$
and
$$text{d}t=frac{2s,text{d}z-texp(st),text{d}x}{2,(s^2-t),exp(st)},.$$
By (2), we get
$$text{d}y=text{d}s+text{d}t=frac{s-t}{2,(s^2-t)},text{d}x+frac{s-1}{(s^2-t),exp(st)},text{d}z,.$$
Thus,
$$frac{partial y}{partial x}(s,t)=frac{s-t}{2,(s^2-t)}text{ and }frac{partial y}{partial z}(s,t)=frac{s-1}{(s^2-t),exp(st)},.$$
When $sigma^2=tau$, we see that
$$lim_{(s,t)to(sigma,tau)},frac{partial y}{partial x}(s,t)text{ and }lim_{(s,t)to(sigma,tau)},frac{partial y}{partial z}(s,t)$$
do not exist. Hence, $dfrac{partial y}{partial x}(s,t)$ and $dfrac{partial y}{partial z}(s,t)$ do not exist along the curve $s^2=t$.
As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
$endgroup$
1
$begingroup$
I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
$endgroup$
– user10354138
Dec 8 '18 at 6:33
add a comment |
$begingroup$
Note that $(x,y,z)=big(s^2+2t,s+t,exp(st)big)$ implies
$$text{d}x=2s,text{d}s+2,text{d}t,,tag{1}$$
$$text{d}y=text{d}s+text{d}t,,tag{2}$$
and
$$text{d}z=texp(st),text{d}s+sexp(st),text{d}t,.tag{3}$$
Thus, the differentials at $(s,t)=(1,1)$ are
$$text{d}x=2,text{d}s+2,text{d}t,,$$
$$text{d}y=text{d}s+text{d}t,,$$
and
$$text{d}z=text{e},text{d}s+text{e},text{d}t,.$$
To compute $dfrac{text{d}y}{text{d}z}(s=1,t=1)$, we require that $text{d}x$ and $text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $dfrac{text{d}y}{text{d}z}(s=1,t=1)$ may not exist (see a comment below).
Here is a way to compute $dfrac{text{d}y}{text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2neq t$). Using (1) and (3), we obtain
$$text{d}s=frac{sexp(st),text{d}x-2,text{d}z}{2,(s^2-t),exp(st)}$$
and
$$text{d}t=frac{2s,text{d}z-texp(st),text{d}x}{2,(s^2-t),exp(st)},.$$
By (2), we get
$$text{d}y=text{d}s+text{d}t=frac{s-t}{2,(s^2-t)},text{d}x+frac{s-1}{(s^2-t),exp(st)},text{d}z,.$$
Thus,
$$frac{partial y}{partial x}(s,t)=frac{s-t}{2,(s^2-t)}text{ and }frac{partial y}{partial z}(s,t)=frac{s-1}{(s^2-t),exp(st)},.$$
When $sigma^2=tau$, we see that
$$lim_{(s,t)to(sigma,tau)},frac{partial y}{partial x}(s,t)text{ and }lim_{(s,t)to(sigma,tau)},frac{partial y}{partial z}(s,t)$$
do not exist. Hence, $dfrac{partial y}{partial x}(s,t)$ and $dfrac{partial y}{partial z}(s,t)$ do not exist along the curve $s^2=t$.
As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
$endgroup$
1
$begingroup$
I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
$endgroup$
– user10354138
Dec 8 '18 at 6:33
add a comment |
$begingroup$
Note that $(x,y,z)=big(s^2+2t,s+t,exp(st)big)$ implies
$$text{d}x=2s,text{d}s+2,text{d}t,,tag{1}$$
$$text{d}y=text{d}s+text{d}t,,tag{2}$$
and
$$text{d}z=texp(st),text{d}s+sexp(st),text{d}t,.tag{3}$$
Thus, the differentials at $(s,t)=(1,1)$ are
$$text{d}x=2,text{d}s+2,text{d}t,,$$
$$text{d}y=text{d}s+text{d}t,,$$
and
$$text{d}z=text{e},text{d}s+text{e},text{d}t,.$$
To compute $dfrac{text{d}y}{text{d}z}(s=1,t=1)$, we require that $text{d}x$ and $text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $dfrac{text{d}y}{text{d}z}(s=1,t=1)$ may not exist (see a comment below).
Here is a way to compute $dfrac{text{d}y}{text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2neq t$). Using (1) and (3), we obtain
$$text{d}s=frac{sexp(st),text{d}x-2,text{d}z}{2,(s^2-t),exp(st)}$$
and
$$text{d}t=frac{2s,text{d}z-texp(st),text{d}x}{2,(s^2-t),exp(st)},.$$
By (2), we get
$$text{d}y=text{d}s+text{d}t=frac{s-t}{2,(s^2-t)},text{d}x+frac{s-1}{(s^2-t),exp(st)},text{d}z,.$$
Thus,
$$frac{partial y}{partial x}(s,t)=frac{s-t}{2,(s^2-t)}text{ and }frac{partial y}{partial z}(s,t)=frac{s-1}{(s^2-t),exp(st)},.$$
When $sigma^2=tau$, we see that
$$lim_{(s,t)to(sigma,tau)},frac{partial y}{partial x}(s,t)text{ and }lim_{(s,t)to(sigma,tau)},frac{partial y}{partial z}(s,t)$$
do not exist. Hence, $dfrac{partial y}{partial x}(s,t)$ and $dfrac{partial y}{partial z}(s,t)$ do not exist along the curve $s^2=t$.
As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
$endgroup$
Note that $(x,y,z)=big(s^2+2t,s+t,exp(st)big)$ implies
$$text{d}x=2s,text{d}s+2,text{d}t,,tag{1}$$
$$text{d}y=text{d}s+text{d}t,,tag{2}$$
and
$$text{d}z=texp(st),text{d}s+sexp(st),text{d}t,.tag{3}$$
Thus, the differentials at $(s,t)=(1,1)$ are
$$text{d}x=2,text{d}s+2,text{d}t,,$$
$$text{d}y=text{d}s+text{d}t,,$$
and
$$text{d}z=text{e},text{d}s+text{e},text{d}t,.$$
To compute $dfrac{text{d}y}{text{d}z}(s=1,t=1)$, we require that $text{d}x$ and $text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $dfrac{text{d}y}{text{d}z}(s=1,t=1)$ may not exist (see a comment below).
Here is a way to compute $dfrac{text{d}y}{text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2neq t$). Using (1) and (3), we obtain
$$text{d}s=frac{sexp(st),text{d}x-2,text{d}z}{2,(s^2-t),exp(st)}$$
and
$$text{d}t=frac{2s,text{d}z-texp(st),text{d}x}{2,(s^2-t),exp(st)},.$$
By (2), we get
$$text{d}y=text{d}s+text{d}t=frac{s-t}{2,(s^2-t)},text{d}x+frac{s-1}{(s^2-t),exp(st)},text{d}z,.$$
Thus,
$$frac{partial y}{partial x}(s,t)=frac{s-t}{2,(s^2-t)}text{ and }frac{partial y}{partial z}(s,t)=frac{s-1}{(s^2-t),exp(st)},.$$
When $sigma^2=tau$, we see that
$$lim_{(s,t)to(sigma,tau)},frac{partial y}{partial x}(s,t)text{ and }lim_{(s,t)to(sigma,tau)},frac{partial y}{partial z}(s,t)$$
do not exist. Hence, $dfrac{partial y}{partial x}(s,t)$ and $dfrac{partial y}{partial z}(s,t)$ do not exist along the curve $s^2=t$.
As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
edited Dec 8 '18 at 9:13
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
answered Dec 7 '18 at 23:56
BatominovskiBatominovski
1
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1
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I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
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– user10354138
Dec 8 '18 at 6:33
add a comment |
1
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I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
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– user10354138
Dec 8 '18 at 6:33
1
1
$begingroup$
I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
$endgroup$
– user10354138
Dec 8 '18 at 6:33
$begingroup$
I believe you have it backwards --- if the Jacobian matrix is nonsingular then $frac{partial y}{partial z}$ can be computed this way and equal to blah, but even if the Jacobian matrix is singular it may still be possible to have $frac{partial y}{partial z}$, trivial example: $x=s^3,z=t^3$ and $y=(st)^N$ for $Ngeq 3$, at $s=t=0$.
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– user10354138
Dec 8 '18 at 6:33
add a comment |
$begingroup$
Hint: the cross product $frac{partial g}{partial s} times frac{partial g}{partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, frac{partial y}{partial z}, 1)$ which must therefore be perpendicular to this normal vector.
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
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1
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Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
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– Daniel Schepler
Dec 8 '18 at 0:37
1
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I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
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– Batominovski
Dec 8 '18 at 0:43
add a comment |
$begingroup$
Hint: the cross product $frac{partial g}{partial s} times frac{partial g}{partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, frac{partial y}{partial z}, 1)$ which must therefore be perpendicular to this normal vector.
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
$endgroup$
1
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
1
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
add a comment |
$begingroup$
Hint: the cross product $frac{partial g}{partial s} times frac{partial g}{partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, frac{partial y}{partial z}, 1)$ which must therefore be perpendicular to this normal vector.
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
$endgroup$
Hint: the cross product $frac{partial g}{partial s} times frac{partial g}{partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, frac{partial y}{partial z}, 1)$ which must therefore be perpendicular to this normal vector.
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
answered Dec 8 '18 at 0:08
Daniel ScheplerDaniel Schepler
8,4041618
8,4041618
1
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
1
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
add a comment |
1
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
1
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
1
1
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
$begingroup$
Except in this case, the cross product turns out to be 0 at that particular point - indicating that the surface parametrization is singular at $(s,t)=(1,1)$... In particular, $g_s = g_t = (2,1,e)$, so no vector $(0, ?, 1)$ can be a linear combination - implying that even if $y$ can be expressed as a function of $x$ and $z$ in a neighborhood of $(3,2,e)$ (which is doubtful), that function can't possibly be differentiable at $(3,e)$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:37
1
1
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
$begingroup$
I don't think $y$ can be expressed as a function of $x$ and $z$ near $(s,t)=(1,1)$ or $(x,z)=(3,text{e})$. After closer inspection, I found out that this is a branch point (or whichever technical phrase for this situation). To solve for $y$, given $x$ and $z$, you need to solve a cubic polynomial equation. The point $(x,z)=(3,text{e})$ is where two roots of the cubic coincide. To be precise, the values of $s$ there are $-2$ and $1$, where $1$ occurs with multiplicity $2$.
$endgroup$
– Batominovski
Dec 8 '18 at 0:43
add a comment |
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$begingroup$
If $(x,y,z)=(2s+2t,s+t,e^{st})$ then $y=frac12x$ is independent of $z$,
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– user10354138
Dec 7 '18 at 17:36
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my bad its s^2 i dont know how to write that in language really sorry
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– Mather
Dec 7 '18 at 17:37
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i have editied it
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– Mather
Dec 7 '18 at 17:39
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i tried but i always get one equation 2Fx + Fy + eFz = 0 where Fx means partial derivitive
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– Mather
Dec 7 '18 at 18:09
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anyone ? i really need help i am staring at the question for the 10th time i really need a hint
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– Mather
Dec 7 '18 at 18:33