Mixing
Show that the curve $(t^2, 2t-1, sqrt{t}), t > 0$ is a flow line of the velocity vector field $vec{F}(x,...
$begingroup$
Show that the curve $(t^2, 2t-1, sqrt{t}), t > 0$ is a flow line of the velocity vector field $vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
my attempt
A curve C described by $vec{r}(t)$ is flow line of vector field $vec{F}$ if $r^{-1}(t) = vec{F}(vec{r}(t))$
here $C(t) = (t^2, 2t - 1, sqrt{t})$ and $C'(t) = (2t, 2, frac{1}{2sqrt{t}})$
$vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
Now
$vec{F}(vec{c}(t)) = (2t-1+1, 2, frac{1}{2sqrt{t}}) = (2t, 2, frac{1}{2sqrt{t}})$
So $vec{F}(vec{c}(t)) = c^{-1}(t)$ so $c(t)$ is a flowline of F as wanted
Would this be correct?
multivariable-calculus
asked Jan 28 at 1:39
TinlerTinler
563315
$endgroup$
add a comment |
$begingroup$
Show that the curve $(t^2, 2t-1, sqrt{t}), t > 0$ is a flow line of the velocity vector field $vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
my attempt
A curve C described by $vec{r}(t)$ is flow line of vector field $vec{F}$ if $r^{-1}(t) = vec{F}(vec{r}(t))$
here $C(t) = (t^2, 2t - 1, sqrt{t})$ and $C'(t) = (2t, 2, frac{1}{2sqrt{t}})$
$vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
Now
$vec{F}(vec{c}(t)) = (2t-1+1, 2, frac{1}{2sqrt{t}}) = (2t, 2, frac{1}{2sqrt{t}})$
So $vec{F}(vec{c}(t)) = c^{-1}(t)$ so $c(t)$ is a flowline of F as wanted
Would this be correct?
multivariable-calculus
asked Jan 28 at 1:39
TinlerTinler
563315
$endgroup$
$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34
add a comment |
$begingroup$
Show that the curve $(t^2, 2t-1, sqrt{t}), t > 0$ is a flow line of the velocity vector field $vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
my attempt
A curve C described by $vec{r}(t)$ is flow line of vector field $vec{F}$ if $r^{-1}(t) = vec{F}(vec{r}(t))$
here $C(t) = (t^2, 2t - 1, sqrt{t})$ and $C'(t) = (2t, 2, frac{1}{2sqrt{t}})$
$vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
Now
$vec{F}(vec{c}(t)) = (2t-1+1, 2, frac{1}{2sqrt{t}}) = (2t, 2, frac{1}{2sqrt{t}})$
So $vec{F}(vec{c}(t)) = c^{-1}(t)$ so $c(t)$ is a flowline of F as wanted
Would this be correct?
multivariable-calculus
asked Jan 28 at 1:39
TinlerTinler
563315
$endgroup$
Show that the curve $(t^2, 2t-1, sqrt{t}), t > 0$ is a flow line of the velocity vector field $vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
my attempt
A curve C described by $vec{r}(t)$ is flow line of vector field $vec{F}$ if $r^{-1}(t) = vec{F}(vec{r}(t))$
here $C(t) = (t^2, 2t - 1, sqrt{t})$ and $C'(t) = (2t, 2, frac{1}{2sqrt{t}})$
$vec{F}(x, y, z) = (y+1, 2, frac{1}{2z})$
Now
$vec{F}(vec{c}(t)) = (2t-1+1, 2, frac{1}{2sqrt{t}}) = (2t, 2, frac{1}{2sqrt{t}})$
So $vec{F}(vec{c}(t)) = c^{-1}(t)$ so $c(t)$ is a flowline of F as wanted
Would this be correct?
multivariable-calculus
multivariable-calculus
asked Jan 28 at 1:39
TinlerTinler
563315
asked Jan 28 at 1:39
TinlerTinler
563315
asked Jan 28 at 1:39
TinlerTinler
563315
asked Jan 28 at 1:39
TinlerTinler
563315
asked Jan 28 at 1:39
TinlerTinler
563315
563315
$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34
add a comment |
$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34
$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34
$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34
add a comment |
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$begingroup$
I think it should be $r'(t)$, not $r^{-1}(t)$. Your reasoning seems correct to me.
$endgroup$
– GReyes
Jan 28 at 2:34