Parametrization of a circle at (-1,-8) with Radius 9
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The circle centered at $(−1,−8)$ with radius $9$ can be parametrized in many ways, this still happens even if we impose the extra constraint that the circle must be traversed in the counter-clockwise direction. This problem asks you to work with two of those counter-clockwise parametrizations.
a). If $x = -1 + 9 cos(t)$
b). If $x = -1 + 9 sin(t)$
So this is what i have so far,
i know that equation of a circle is $x^2+y^2=r^2$
I put the points into the equation $(x+1)^2+(y+8)^2=9^2$
Then I solved for $y$ and plugged in a). as a substitute to get the answer, which I believe is: $y=1-9cos(t)$
But i get a wrong answer, I am not to sure where to I am messing up, I think I am missing a couple of steps, but i just don't know how to get there.
calculus parametric parametrization
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up vote
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down vote
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Question:
The circle centered at $(−1,−8)$ with radius $9$ can be parametrized in many ways, this still happens even if we impose the extra constraint that the circle must be traversed in the counter-clockwise direction. This problem asks you to work with two of those counter-clockwise parametrizations.
a). If $x = -1 + 9 cos(t)$
b). If $x = -1 + 9 sin(t)$
So this is what i have so far,
i know that equation of a circle is $x^2+y^2=r^2$
I put the points into the equation $(x+1)^2+(y+8)^2=9^2$
Then I solved for $y$ and plugged in a). as a substitute to get the answer, which I believe is: $y=1-9cos(t)$
But i get a wrong answer, I am not to sure where to I am messing up, I think I am missing a couple of steps, but i just don't know how to get there.
calculus parametric parametrization
If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question:
The circle centered at $(−1,−8)$ with radius $9$ can be parametrized in many ways, this still happens even if we impose the extra constraint that the circle must be traversed in the counter-clockwise direction. This problem asks you to work with two of those counter-clockwise parametrizations.
a). If $x = -1 + 9 cos(t)$
b). If $x = -1 + 9 sin(t)$
So this is what i have so far,
i know that equation of a circle is $x^2+y^2=r^2$
I put the points into the equation $(x+1)^2+(y+8)^2=9^2$
Then I solved for $y$ and plugged in a). as a substitute to get the answer, which I believe is: $y=1-9cos(t)$
But i get a wrong answer, I am not to sure where to I am messing up, I think I am missing a couple of steps, but i just don't know how to get there.
calculus parametric parametrization
Question:
The circle centered at $(−1,−8)$ with radius $9$ can be parametrized in many ways, this still happens even if we impose the extra constraint that the circle must be traversed in the counter-clockwise direction. This problem asks you to work with two of those counter-clockwise parametrizations.
a). If $x = -1 + 9 cos(t)$
b). If $x = -1 + 9 sin(t)$
So this is what i have so far,
i know that equation of a circle is $x^2+y^2=r^2$
I put the points into the equation $(x+1)^2+(y+8)^2=9^2$
Then I solved for $y$ and plugged in a). as a substitute to get the answer, which I believe is: $y=1-9cos(t)$
But i get a wrong answer, I am not to sure where to I am messing up, I think I am missing a couple of steps, but i just don't know how to get there.
calculus parametric parametrization
calculus parametric parametrization
edited 2 days ago
bjcolby15
1,0961916
1,0961916
asked 2 days ago


Daniel2233
133
133
If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago
add a comment |
If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago
If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago
add a comment |
1 Answer
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Hint: Recall for a circle $(x-a)^2 + (y-b)^2 = r^2$, we can convert it into the parameterized equation $$begin {matrix} x = a + r cos t \ y = b + r sin t end {matrix}$$ What is the equation given your points $(-1, -8)$ and radius $9$?
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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up vote
0
down vote
Hint: Recall for a circle $(x-a)^2 + (y-b)^2 = r^2$, we can convert it into the parameterized equation $$begin {matrix} x = a + r cos t \ y = b + r sin t end {matrix}$$ What is the equation given your points $(-1, -8)$ and radius $9$?
add a comment |
up vote
0
down vote
Hint: Recall for a circle $(x-a)^2 + (y-b)^2 = r^2$, we can convert it into the parameterized equation $$begin {matrix} x = a + r cos t \ y = b + r sin t end {matrix}$$ What is the equation given your points $(-1, -8)$ and radius $9$?
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: Recall for a circle $(x-a)^2 + (y-b)^2 = r^2$, we can convert it into the parameterized equation $$begin {matrix} x = a + r cos t \ y = b + r sin t end {matrix}$$ What is the equation given your points $(-1, -8)$ and radius $9$?
Hint: Recall for a circle $(x-a)^2 + (y-b)^2 = r^2$, we can convert it into the parameterized equation $$begin {matrix} x = a + r cos t \ y = b + r sin t end {matrix}$$ What is the equation given your points $(-1, -8)$ and radius $9$?
answered 20 hours ago
bjcolby15
1,0961916
1,0961916
add a comment |
add a comment |
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If you want someone to check your work then it would make sense to actually show all the steps in as much detail as you can, including explanation. Your answer says that $y$ ranges from $-8$ to $10,$ which does not fit with the equation at all, so clearly there was some error along the way but it's hard to guess what steps you took.
– David K
2 days ago
Try to write it around the (0,0) and than transkate
– Moti
2 days ago
Can you show us what you did? $y=1-9 cos t$ is defintely incorrect...I would think for at least (a) would always be counterclockwise, but (b) might require a translation of in one of the terms so that $t$ can traverse counterclockwise. See also my hint below, which might help explain why.
– bjcolby15
20 hours ago