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Given a specific curve $alpha(s)$, Prove that $kappa(s) geq 1/R$, where $kappa(s)$ is the curvature of...
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Suppose $alpha$ is an arclength-parameterized curve with the property that $|alpha(s)| leq |alpha(s_o)| = R,forall s$ sufficiently close to $s_o$. Prove that $kappa(s) geq 1/R$, where $kappa(s)$ is the curvature of $alpha$ at $s$.
A hint is given to define a function $f(s)=|alpha(s)|^2$ and to consider it's second derivative at $s_o$
$frac{d}{ds} f(s)=frac{d}{ds}|alpha(s)|^2=2alpha(s)*alpha'(s)$
and thus by the product rule $f''(s)=2(alpha'(s)^2+alpha(s)alpha''(s))$.
So, i'm suppose to see something by examining this function apparently, but I am not. Is there something obvious i'm missing?
differential-geometry
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Suppose $alpha$ is an arclength-parameterized curve with the property that $|alpha(s)| leq |alpha(s_o)| = R,forall s$ sufficiently close to $s_o$. Prove that $kappa(s) geq 1/R$, where $kappa(s)$ is the curvature of $alpha$ at $s$.
A hint is given to define a function $f(s)=|alpha(s)|^2$ and to consider it's second derivative at $s_o$
$frac{d}{ds} f(s)=frac{d}{ds}|alpha(s)|^2=2alpha(s)*alpha'(s)$
and thus by the product rule $f''(s)=2(alpha'(s)^2+alpha(s)alpha''(s))$.
So, i'm suppose to see something by examining this function apparently, but I am not. Is there something obvious i'm missing?
differential-geometry
$endgroup$
add a comment |
$begingroup$
Suppose $alpha$ is an arclength-parameterized curve with the property that $|alpha(s)| leq |alpha(s_o)| = R,forall s$ sufficiently close to $s_o$. Prove that $kappa(s) geq 1/R$, where $kappa(s)$ is the curvature of $alpha$ at $s$.
A hint is given to define a function $f(s)=|alpha(s)|^2$ and to consider it's second derivative at $s_o$
$frac{d}{ds} f(s)=frac{d}{ds}|alpha(s)|^2=2alpha(s)*alpha'(s)$
and thus by the product rule $f''(s)=2(alpha'(s)^2+alpha(s)alpha''(s))$.
So, i'm suppose to see something by examining this function apparently, but I am not. Is there something obvious i'm missing?
differential-geometry
$endgroup$
Suppose $alpha$ is an arclength-parameterized curve with the property that $|alpha(s)| leq |alpha(s_o)| = R,forall s$ sufficiently close to $s_o$. Prove that $kappa(s) geq 1/R$, where $kappa(s)$ is the curvature of $alpha$ at $s$.
A hint is given to define a function $f(s)=|alpha(s)|^2$ and to consider it's second derivative at $s_o$
$frac{d}{ds} f(s)=frac{d}{ds}|alpha(s)|^2=2alpha(s)*alpha'(s)$
and thus by the product rule $f''(s)=2(alpha'(s)^2+alpha(s)alpha''(s))$.
So, i'm suppose to see something by examining this function apparently, but I am not. Is there something obvious i'm missing?
differential-geometry
differential-geometry
asked Jan 21 at 15:44
user624065
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1 Answer
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By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) leq 0$. Moreover, $|alpha'(s)| = 1$ for every $s$ (since $alpha$ is parametrized by arc-length).
Since
$$
f''(s) = 2 |alpha'(s)|^2+ 2 alpha(s)cdot alpha''(s)
= 2( 1 + alpha(s)cdot alpha''(s)),
$$
the condition $f''(s_0) leq 0$ gives
$$
alpha(s_0) cdot alpha''(s_0) leq -1,
$$
so that, by the Cauchy-Schwarz inequality,
$$
R |alpha''(s_0)| = |alpha(s_0)|, |alpha''(s_0)|
geq - alpha(s_0) cdot alpha''(s_0) geq 1.
$$
answered Jan 21 at 16:13
RigelRigel
11.3k11320
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and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
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Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
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1 Answer
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1 Answer
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active
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$begingroup$
By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) leq 0$. Moreover, $|alpha'(s)| = 1$ for every $s$ (since $alpha$ is parametrized by arc-length).
Since
$$
f''(s) = 2 |alpha'(s)|^2+ 2 alpha(s)cdot alpha''(s)
= 2( 1 + alpha(s)cdot alpha''(s)),
$$
the condition $f''(s_0) leq 0$ gives
$$
alpha(s_0) cdot alpha''(s_0) leq -1,
$$
so that, by the Cauchy-Schwarz inequality,
$$
R |alpha''(s_0)| = |alpha(s_0)|, |alpha''(s_0)|
geq - alpha(s_0) cdot alpha''(s_0) geq 1.
$$
answered Jan 21 at 16:13
RigelRigel
11.3k11320
$endgroup$
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
add a comment |
$begingroup$
By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) leq 0$. Moreover, $|alpha'(s)| = 1$ for every $s$ (since $alpha$ is parametrized by arc-length).
Since
$$
f''(s) = 2 |alpha'(s)|^2+ 2 alpha(s)cdot alpha''(s)
= 2( 1 + alpha(s)cdot alpha''(s)),
$$
the condition $f''(s_0) leq 0$ gives
$$
alpha(s_0) cdot alpha''(s_0) leq -1,
$$
so that, by the Cauchy-Schwarz inequality,
$$
R |alpha''(s_0)| = |alpha(s_0)|, |alpha''(s_0)|
geq - alpha(s_0) cdot alpha''(s_0) geq 1.
$$
answered Jan 21 at 16:13
RigelRigel
11.3k11320
$endgroup$
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
add a comment |
$begingroup$
By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) leq 0$. Moreover, $|alpha'(s)| = 1$ for every $s$ (since $alpha$ is parametrized by arc-length).
Since
$$
f''(s) = 2 |alpha'(s)|^2+ 2 alpha(s)cdot alpha''(s)
= 2( 1 + alpha(s)cdot alpha''(s)),
$$
the condition $f''(s_0) leq 0$ gives
$$
alpha(s_0) cdot alpha''(s_0) leq -1,
$$
so that, by the Cauchy-Schwarz inequality,
$$
R |alpha''(s_0)| = |alpha(s_0)|, |alpha''(s_0)|
geq - alpha(s_0) cdot alpha''(s_0) geq 1.
$$
answered Jan 21 at 16:13
RigelRigel
11.3k11320
$endgroup$
By assumption, the function $f(s)$ has a maximum at $s = s_0$, hence $f'(s_0) = 0$ and $f''(s_0) leq 0$. Moreover, $|alpha'(s)| = 1$ for every $s$ (since $alpha$ is parametrized by arc-length).
Since
$$
f''(s) = 2 |alpha'(s)|^2+ 2 alpha(s)cdot alpha''(s)
= 2( 1 + alpha(s)cdot alpha''(s)),
$$
the condition $f''(s_0) leq 0$ gives
$$
alpha(s_0) cdot alpha''(s_0) leq -1,
$$
so that, by the Cauchy-Schwarz inequality,
$$
R |alpha''(s_0)| = |alpha(s_0)|, |alpha''(s_0)|
geq - alpha(s_0) cdot alpha''(s_0) geq 1.
$$
answered Jan 21 at 16:13
RigelRigel
11.3k11320
answered Jan 21 at 16:13
RigelRigel
11.3k11320
answered Jan 21 at 16:13
RigelRigel
11.3k11320
answered Jan 21 at 16:13
RigelRigel
11.3k11320
11.3k11320
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
add a comment |
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
and $|alpha''(s_o)|=kappa(s_o)$ because $alpha(s)$ is arc-length parameterized?
$endgroup$
– user624065
Jan 21 at 16:29
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
$begingroup$
Yes, that's right.
$endgroup$
– Rigel
Jan 21 at 16:32
add a comment |
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