How to find the vector equation of the line segment.
$begingroup$
Let $C$ be the line segment from $(0,0)$ to $(2,2)$, and let $f(x,y)=x^2+y$.
Write down a vector equation $r(t)$ of the line segment, that is, find a parametrization of $C$.
The answer given is $r(t)=langle t,trangle,0le tle 2$.
My Question: How did they get $boldsymbol{r(t)=langle t,trangle}$?
calculus multivariable-calculus parametrization line-integrals
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
$endgroup$
add a comment |
$begingroup$
Let $C$ be the line segment from $(0,0)$ to $(2,2)$, and let $f(x,y)=x^2+y$.
Write down a vector equation $r(t)$ of the line segment, that is, find a parametrization of $C$.
The answer given is $r(t)=langle t,trangle,0le tle 2$.
My Question: How did they get $boldsymbol{r(t)=langle t,trangle}$?
calculus multivariable-calculus parametrization line-integrals
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
$endgroup$
$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55
add a comment |
$begingroup$
Let $C$ be the line segment from $(0,0)$ to $(2,2)$, and let $f(x,y)=x^2+y$.
Write down a vector equation $r(t)$ of the line segment, that is, find a parametrization of $C$.
The answer given is $r(t)=langle t,trangle,0le tle 2$.
My Question: How did they get $boldsymbol{r(t)=langle t,trangle}$?
calculus multivariable-calculus parametrization line-integrals
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
$endgroup$
Let $C$ be the line segment from $(0,0)$ to $(2,2)$, and let $f(x,y)=x^2+y$.
Write down a vector equation $r(t)$ of the line segment, that is, find a parametrization of $C$.
The answer given is $r(t)=langle t,trangle,0le tle 2$.
My Question: How did they get $boldsymbol{r(t)=langle t,trangle}$?
calculus multivariable-calculus parametrization line-integrals
calculus multivariable-calculus parametrization line-integrals
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
edited Dec 13 '18 at 12:09
Rócherz
2,7862721
2,7862721
asked Dec 13 '18 at 12:03
user982787user982787
1117
asked Dec 13 '18 at 12:03
user982787user982787
1117
asked Dec 13 '18 at 12:03
user982787user982787
1117
1117
$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55
add a comment |
$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55
$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55
$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
begin{eqnarray}
x(t) &=& t \
y(t) &=& x(t) = t
end{eqnarray}
For $0 leq t leq 2$ a parametrization of $C$ is
$$
{bf r}(t) = (t, t)~~~mbox{for}~~ 0leq t leq 2
$$
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$langle x,yrangle = langle x_0,y_0rangle +lambda langle x_1-x_0,y_1-y_0rangle, qquad 0 leq lambda leq 1.$$ After plugging in your numbers, you end up with $langle x,yrangle = lambda langle 2,2rangle = langle 2lambda,2lambdarangle$. Letting $t:=2lambda$ gives the desired parametrization, but that last step is more of a personal choice.
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
$endgroup$
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
$begingroup$
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
begin{eqnarray}
x(t) &=& t \
y(t) &=& x(t) = t
end{eqnarray}
For $0 leq t leq 2$ a parametrization of $C$ is
$$
{bf r}(t) = (t, t)~~~mbox{for}~~ 0leq t leq 2
$$
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
begin{eqnarray}
x(t) &=& t \
y(t) &=& x(t) = t
end{eqnarray}
For $0 leq t leq 2$ a parametrization of $C$ is
$$
{bf r}(t) = (t, t)~~~mbox{for}~~ 0leq t leq 2
$$
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
begin{eqnarray}
x(t) &=& t \
y(t) &=& x(t) = t
end{eqnarray}
For $0 leq t leq 2$ a parametrization of $C$ is
$$
{bf r}(t) = (t, t)~~~mbox{for}~~ 0leq t leq 2
$$
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
$endgroup$
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
begin{eqnarray}
x(t) &=& t \
y(t) &=& x(t) = t
end{eqnarray}
For $0 leq t leq 2$ a parametrization of $C$ is
$$
{bf r}(t) = (t, t)~~~mbox{for}~~ 0leq t leq 2
$$
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
edited Dec 13 '18 at 12:33
Rócherz
2,7862721
2,7862721
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
answered Dec 13 '18 at 12:16
caveraccaverac
14.6k31130
14.6k31130
add a comment |
add a comment |
$begingroup$
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$langle x,yrangle = langle x_0,y_0rangle +lambda langle x_1-x_0,y_1-y_0rangle, qquad 0 leq lambda leq 1.$$ After plugging in your numbers, you end up with $langle x,yrangle = lambda langle 2,2rangle = langle 2lambda,2lambdarangle$. Letting $t:=2lambda$ gives the desired parametrization, but that last step is more of a personal choice.
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
$endgroup$
add a comment |
$begingroup$
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$langle x,yrangle = langle x_0,y_0rangle +lambda langle x_1-x_0,y_1-y_0rangle, qquad 0 leq lambda leq 1.$$ After plugging in your numbers, you end up with $langle x,yrangle = lambda langle 2,2rangle = langle 2lambda,2lambdarangle$. Letting $t:=2lambda$ gives the desired parametrization, but that last step is more of a personal choice.
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
$endgroup$
add a comment |
$begingroup$
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$langle x,yrangle = langle x_0,y_0rangle +lambda langle x_1-x_0,y_1-y_0rangle, qquad 0 leq lambda leq 1.$$ After plugging in your numbers, you end up with $langle x,yrangle = lambda langle 2,2rangle = langle 2lambda,2lambdarangle$. Letting $t:=2lambda$ gives the desired parametrization, but that last step is more of a personal choice.
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
$endgroup$
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$langle x,yrangle = langle x_0,y_0rangle +lambda langle x_1-x_0,y_1-y_0rangle, qquad 0 leq lambda leq 1.$$ After plugging in your numbers, you end up with $langle x,yrangle = lambda langle 2,2rangle = langle 2lambda,2lambdarangle$. Letting $t:=2lambda$ gives the desired parametrization, but that last step is more of a personal choice.
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
answered Dec 13 '18 at 12:38
RócherzRócherz
2,7862721
2,7862721
add a comment |
add a comment |
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$begingroup$
What does $f$ have to do with anything?
$endgroup$
– amd
Dec 13 '18 at 21:55