The intercepts on the coordinate axes of the straight line with the given equation $4x-3y=12.$

Marvin Mccormick
2021-02-25
Answered

The intercepts on the coordinate axes of the straight line with the given equation $4x-3y=12.$

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asked 2022-09-05

Is the differential $\mathrm{d}\overrightarrow{r}$ a sensible mathematical object?

When doing differential geometry, physicists often use

$\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\text{}{\overrightarrow{e}}_{i}$

for many different things. For instance, they define the holonomic basis $\{{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}\}$ relative to a coordinate system $\{{x}^{\prime a}\}$ by imposing

$\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{\prime a}\text{}{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}=\frac{\mathrm{\partial}\overrightarrow{r}}{\mathrm{\partial}{x}^{\prime a}}$

and they compute the quadratic form of the metric $\mathrm{d}{s}^{2}$ as $\mathrm{d}\overrightarrow{r}\cdot \mathrm{d}\overrightarrow{r}$ .

Computing the differential of a vector field ($\overrightarrow{r}={x}^{i}{\overrightarrow{e}}_{i}$ i, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating k-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).

Not only that, the "true" definitions of holonomic bases and ds2 don't use this $\mathrm{d}\overrightarrow{r}$ at all.

EDIT: in fact, taking the derivative of $\overrightarrow{r}$, or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn't define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors $\mathrm{\partial}/\mathrm{\partial}{x}^{\prime a}$.

After thinking about it, I thought the differential of a vector field might just be

$\mathrm{d}\overrightarrow{\phi}=({\mathrm{\nabla}}_{i}{\phi}^{j})\text{}{\overrightarrow{e}}_{j}\otimes \mathrm{d}{x}^{i},$

so maybe $\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\text{}{\overrightarrow{e}}_{i}$ i means $\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\otimes {\overrightarrow{e}}_{i}$? How is $\mathrm{d}\overrightarrow{r}$ rigorously defined, otherwise?

When doing differential geometry, physicists often use

$\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\text{}{\overrightarrow{e}}_{i}$

for many different things. For instance, they define the holonomic basis $\{{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}\}$ relative to a coordinate system $\{{x}^{\prime a}\}$ by imposing

$\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{\prime a}\text{}{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\overrightarrow{e}}_{a}^{\text{}\mathrm{\prime}}=\frac{\mathrm{\partial}\overrightarrow{r}}{\mathrm{\partial}{x}^{\prime a}}$

and they compute the quadratic form of the metric $\mathrm{d}{s}^{2}$ as $\mathrm{d}\overrightarrow{r}\cdot \mathrm{d}\overrightarrow{r}$ .

Computing the differential of a vector field ($\overrightarrow{r}={x}^{i}{\overrightarrow{e}}_{i}$ i, in this case) feels strange, as in differential geometry differentials are usually considered to be alternating k-forms, so it would only make sense to talk about the differential of a scalar field (aka its exterior derivative).

Not only that, the "true" definitions of holonomic bases and ds2 don't use this $\mathrm{d}\overrightarrow{r}$ at all.

EDIT: in fact, taking the derivative of $\overrightarrow{r}$, or any other vector field, is something we are not allowed to do in a general differentiable manifold without a connection, so we obviously wouldn't define a holonomic basis like that. A holonomic basis would basically be the basis formed by the tangent vectors $\mathrm{\partial}/\mathrm{\partial}{x}^{\prime a}$.

After thinking about it, I thought the differential of a vector field might just be

$\mathrm{d}\overrightarrow{\phi}=({\mathrm{\nabla}}_{i}{\phi}^{j})\text{}{\overrightarrow{e}}_{j}\otimes \mathrm{d}{x}^{i},$

so maybe $\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\text{}{\overrightarrow{e}}_{i}$ i means $\mathrm{d}\overrightarrow{r}=\mathrm{d}{x}^{i}\otimes {\overrightarrow{e}}_{i}$? How is $\mathrm{d}\overrightarrow{r}$ rigorously defined, otherwise?

asked 2022-09-04

Using index notation to write ${d}^{2}=0$ in terms of a torsion free connection.

Let (M,g) be a Riemannian manifold and let $\omega $ be a 1-form on M. I want to rewrite ${d}^{2}\omega =0$ in terms of the Levi-Civita connection.

I can show the following:

$d\omega (X,Y)=({\mathrm{\nabla}}_{X}\omega )(Y)-({\mathrm{\nabla}}_{Y}\omega )(X),$

which in index notation reads

$(d\omega {)}_{ab}=2{\mathrm{\nabla}}_{[a}{\omega}_{b]}.$

Similiarly for a 2-form, $\mu $, we have:

$d\mu (X,Y,Z)=({\mathrm{\nabla}}_{X}\mu )(Y,Z)-({\mathrm{\nabla}}_{Y}\mu )(X,Z)+({\mathrm{\nabla}}_{Z}\mu )(X,Y),$

which in index notation reads

$(d\mu {)}_{abc}=3{\mathrm{\nabla}}_{[a}{\varphi}_{bc]}.$

Now plugging in $d\omega $ for $\mu $ we get

$0={d}^{2}\omega =(d(d\omega ){)}_{abc}={\mathrm{\nabla}}_{[a}(d\omega {)}_{bc]}.$

I want to plug in the above expression (in index notation) for dω but I'm not really sure how to handle the indices. Do I just get

$3{\mathrm{\nabla}}_{[a}2{\mathrm{\nabla}}_{[b}{\omega}_{c]]}=6{\mathrm{\nabla}}_{[a}{\mathrm{\nabla}}_{b}{\omega}_{c]}?$

Let (M,g) be a Riemannian manifold and let $\omega $ be a 1-form on M. I want to rewrite ${d}^{2}\omega =0$ in terms of the Levi-Civita connection.

I can show the following:

$d\omega (X,Y)=({\mathrm{\nabla}}_{X}\omega )(Y)-({\mathrm{\nabla}}_{Y}\omega )(X),$

which in index notation reads

$(d\omega {)}_{ab}=2{\mathrm{\nabla}}_{[a}{\omega}_{b]}.$

Similiarly for a 2-form, $\mu $, we have:

$d\mu (X,Y,Z)=({\mathrm{\nabla}}_{X}\mu )(Y,Z)-({\mathrm{\nabla}}_{Y}\mu )(X,Z)+({\mathrm{\nabla}}_{Z}\mu )(X,Y),$

which in index notation reads

$(d\mu {)}_{abc}=3{\mathrm{\nabla}}_{[a}{\varphi}_{bc]}.$

Now plugging in $d\omega $ for $\mu $ we get

$0={d}^{2}\omega =(d(d\omega ){)}_{abc}={\mathrm{\nabla}}_{[a}(d\omega {)}_{bc]}.$

I want to plug in the above expression (in index notation) for dω but I'm not really sure how to handle the indices. Do I just get

$3{\mathrm{\nabla}}_{[a}2{\mathrm{\nabla}}_{[b}{\omega}_{c]]}=6{\mathrm{\nabla}}_{[a}{\mathrm{\nabla}}_{b}{\omega}_{c]}?$

asked 2021-08-08

Discuss: Different Coordinate Systems As was noted in the overview of the chapter, certain curves are more naturally described in one coordinate system than in another. In each of the following situations, which coordinate system would be appropriate: rectangular or polar? Give reasons to support your answer.

a) You need to give directions to your house to a taxi driver.

b) You need to give directions to your house to a homing pigeon.

a) You need to give directions to your house to a taxi driver.

b) You need to give directions to your house to a homing pigeon.

asked 2022-07-02

Find the equation of the line that satisfies the given conditions:

Goes through $(-1,4)$, slope $-9$

Goes through $(-1,4)$, slope $-9$

asked 2021-03-02

To plot: Thepoints which has polar coordinate $(2,\frac{7\pi}{4})$ also two alternaitve sets for the same.

asked 2022-07-03

Find the Distance Between The Given Points:

$(1,1),(4,5)$

$(1,1),(4,5)$

asked 2021-06-05

The Cartesian coordinates of a point are given.

a)$(2,-2)$

b)$(-1,\sqrt{3})$

Find the polar coordinates$(r,\theta )$ of the point, where r is greater than 0 and 0 is less than or equal to $\theta $ , which is less than $2\pi $

Find the polar coordinates$(r,\theta )$ of the point, where r is less than 0 and 0 is less than or equal to $\theta $ , which is less than $2\pi $

a)

b)

Find the polar coordinates

Find the polar coordinates