Parametrizing semi-circle in clockwise orientation
$begingroup$
Please have a look at the curve below:
This is part of a line integral question. The solution to the parametrization of the curve from $(-3,0)$ to $(3,0)$ is $r(t) = (3sint, 3cost)$, $t in [frac{pi}{2}, frac{-pi}{2}]$.
How is this true? Could someone please help me understand this parametrization? Thanks
line-integrals parametrization
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
$endgroup$
add a comment |
$begingroup$
Please have a look at the curve below:
This is part of a line integral question. The solution to the parametrization of the curve from $(-3,0)$ to $(3,0)$ is $r(t) = (3sint, 3cost)$, $t in [frac{pi}{2}, frac{-pi}{2}]$.
How is this true? Could someone please help me understand this parametrization? Thanks
line-integrals parametrization
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
$endgroup$
add a comment |
$begingroup$
Please have a look at the curve below:
This is part of a line integral question. The solution to the parametrization of the curve from $(-3,0)$ to $(3,0)$ is $r(t) = (3sint, 3cost)$, $t in [frac{pi}{2}, frac{-pi}{2}]$.
How is this true? Could someone please help me understand this parametrization? Thanks
line-integrals parametrization
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
$endgroup$
Please have a look at the curve below:
This is part of a line integral question. The solution to the parametrization of the curve from $(-3,0)$ to $(3,0)$ is $r(t) = (3sint, 3cost)$, $t in [frac{pi}{2}, frac{-pi}{2}]$.
How is this true? Could someone please help me understand this parametrization? Thanks
line-integrals parametrization
line-integrals parametrization
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
asked May 2 '16 at 9:25
abruzzi26abruzzi26
941213
941213
add a comment |
add a comment |
1 Answer
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$begingroup$
First, since $(3sin(t))^2+(3cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=frac{pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-frac{pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $frac{pi}{2}$ and decreases to $-frac{pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $sin$ and $cos$ to check that we're on the right parts of the curve. For example, for $tin(0,frac{pi}{2})$, both $3sin(t)$ and $3cos(t)$, so we are in the first quadrant. For $tin(-frac{pi}{2},0)$, $3sin(t)$ is negative, but $3cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
First, since $(3sin(t))^2+(3cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=frac{pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-frac{pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $frac{pi}{2}$ and decreases to $-frac{pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $sin$ and $cos$ to check that we're on the right parts of the curve. For example, for $tin(0,frac{pi}{2})$, both $3sin(t)$ and $3cos(t)$, so we are in the first quadrant. For $tin(-frac{pi}{2},0)$, $3sin(t)$ is negative, but $3cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
First, since $(3sin(t))^2+(3cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=frac{pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-frac{pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $frac{pi}{2}$ and decreases to $-frac{pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $sin$ and $cos$ to check that we're on the right parts of the curve. For example, for $tin(0,frac{pi}{2})$, both $3sin(t)$ and $3cos(t)$, so we are in the first quadrant. For $tin(-frac{pi}{2},0)$, $3sin(t)$ is negative, but $3cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
First, since $(3sin(t))^2+(3cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=frac{pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-frac{pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $frac{pi}{2}$ and decreases to $-frac{pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $sin$ and $cos$ to check that we're on the right parts of the curve. For example, for $tin(0,frac{pi}{2})$, both $3sin(t)$ and $3cos(t)$, so we are in the first quadrant. For $tin(-frac{pi}{2},0)$, $3sin(t)$ is negative, but $3cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
$endgroup$
First, since $(3sin(t))^2+(3cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=frac{pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-frac{pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $frac{pi}{2}$ and decreases to $-frac{pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $sin$ and $cos$ to check that we're on the right parts of the curve. For example, for $tin(0,frac{pi}{2})$, both $3sin(t)$ and $3cos(t)$, so we are in the first quadrant. For $tin(-frac{pi}{2},0)$, $3sin(t)$ is negative, but $3cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
answered May 2 '16 at 10:56
Michael BurrMichael Burr
27k23262
27k23262
add a comment |
add a comment |
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