Mixing
parametrization of a circle given a property
$begingroup$
Provide a parametrization with the given properties:
The curve is circled at point $(a, b)$. It is traced once counterclockwise, starting at the point $(a+r, b)$ with $t in[0,2π]$
attempt:
$x = a + r cos(t), y = a+rsin(t)$ for $t in [0, 2pi]$
is above not right?
multivariable-calculus
asked Jan 22 at 0:59
TinlerTinler
558314
$endgroup$
add a comment |
$begingroup$
Provide a parametrization with the given properties:
The curve is circled at point $(a, b)$. It is traced once counterclockwise, starting at the point $(a+r, b)$ with $t in[0,2π]$
attempt:
$x = a + r cos(t), y = a+rsin(t)$ for $t in [0, 2pi]$
is above not right?
multivariable-calculus
asked Jan 22 at 0:59
TinlerTinler
558314
$endgroup$
$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05
add a comment |
$begingroup$
Provide a parametrization with the given properties:
The curve is circled at point $(a, b)$. It is traced once counterclockwise, starting at the point $(a+r, b)$ with $t in[0,2π]$
attempt:
$x = a + r cos(t), y = a+rsin(t)$ for $t in [0, 2pi]$
is above not right?
multivariable-calculus
asked Jan 22 at 0:59
TinlerTinler
558314
$endgroup$
Provide a parametrization with the given properties:
The curve is circled at point $(a, b)$. It is traced once counterclockwise, starting at the point $(a+r, b)$ with $t in[0,2π]$
attempt:
$x = a + r cos(t), y = a+rsin(t)$ for $t in [0, 2pi]$
is above not right?
multivariable-calculus
multivariable-calculus
asked Jan 22 at 0:59
TinlerTinler
558314
asked Jan 22 at 0:59
TinlerTinler
558314
asked Jan 22 at 0:59
TinlerTinler
558314
asked Jan 22 at 0:59
TinlerTinler
558314
asked Jan 22 at 0:59
TinlerTinler
558314
558314
$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05
add a comment |
$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05
$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have a circle of equation:
$$(x-a)^2+(y-b)^2=r^2$$
Setting $x=a+rcos(t)$ and $y=b+rsin t$ gives:
$$(rcos(t))^2+(rsin(t))^2=r^2$$
$$to r^2(cos^2(t)+sin^2(t))=r^2$$
which holds because $cos^2(t)+sin^2(t)=1$
So your mistake was simply you need $y=b+...$ rather than $y=a+...$
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have a circle of equation:
$$(x-a)^2+(y-b)^2=r^2$$
Setting $x=a+rcos(t)$ and $y=b+rsin t$ gives:
$$(rcos(t))^2+(rsin(t))^2=r^2$$
$$to r^2(cos^2(t)+sin^2(t))=r^2$$
which holds because $cos^2(t)+sin^2(t)=1$
So your mistake was simply you need $y=b+...$ rather than $y=a+...$
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
We have a circle of equation:
$$(x-a)^2+(y-b)^2=r^2$$
Setting $x=a+rcos(t)$ and $y=b+rsin t$ gives:
$$(rcos(t))^2+(rsin(t))^2=r^2$$
$$to r^2(cos^2(t)+sin^2(t))=r^2$$
which holds because $cos^2(t)+sin^2(t)=1$
So your mistake was simply you need $y=b+...$ rather than $y=a+...$
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
We have a circle of equation:
$$(x-a)^2+(y-b)^2=r^2$$
Setting $x=a+rcos(t)$ and $y=b+rsin t$ gives:
$$(rcos(t))^2+(rsin(t))^2=r^2$$
$$to r^2(cos^2(t)+sin^2(t))=r^2$$
which holds because $cos^2(t)+sin^2(t)=1$
So your mistake was simply you need $y=b+...$ rather than $y=a+...$
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
$endgroup$
We have a circle of equation:
$$(x-a)^2+(y-b)^2=r^2$$
Setting $x=a+rcos(t)$ and $y=b+rsin t$ gives:
$$(rcos(t))^2+(rsin(t))^2=r^2$$
$$to r^2(cos^2(t)+sin^2(t))=r^2$$
which holds because $cos^2(t)+sin^2(t)=1$
So your mistake was simply you need $y=b+...$ rather than $y=a+...$
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 1:15
Rhys HughesRhys Hughes
6,9741530
6,9741530
add a comment |
add a comment |
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$begingroup$
I believe your second equation should be $y = b + rsinleft(tright)$, but otherwise everything else looks fine to me. Also, with your second line starting with "The curve is circled at ...", I assume "circled" is meant to be "centered".
$endgroup$
– John Omielan
Jan 22 at 1:05
$begingroup$
Yes it is correct if you change a to b in y.
$endgroup$
– lightxbulb
Jan 22 at 1:05