Parametrization of a circle at (-1,-8) with Radius 9
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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
edited Nov 19 at 23:04
bjcolby15
1,1381916
asked Nov 19 at 22:42
Daniel2233
133
add a comment |
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
edited Nov 19 at 23:04
bjcolby15
1,1381916
asked Nov 19 at 22:42
Daniel2233
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
Nov 20 at 0:33
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
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
Nov 21 at 20:41
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
edited Nov 19 at 23:04
bjcolby15
1,1381916
asked Nov 19 at 22:42
Daniel2233
133
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 Nov 19 at 23:04
bjcolby15
1,1381916
asked Nov 19 at 22:42
Daniel2233
133
edited Nov 19 at 23:04
bjcolby15
1,1381916
asked Nov 19 at 22:42
Daniel2233
133
edited Nov 19 at 23:04
bjcolby15
1,1381916
edited Nov 19 at 23:04
bjcolby15
1,1381916
edited Nov 19 at 23:04
bjcolby15
1,1381916
1,1381916
asked Nov 19 at 22:42
Daniel2233
133
asked Nov 19 at 22:42
Daniel2233
133
asked Nov 19 at 22:42
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
Nov 20 at 0:33
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
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
Nov 21 at 20:41
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
Nov 20 at 0:33
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
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
Nov 21 at 20:41
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
Nov 20 at 0:33
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
Nov 20 at 0:33
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
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
Nov 21 at 20:41
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
Nov 21 at 20:41
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 $9$?
answered Nov 21 at 20:42
bjcolby15
1,1381916
add a comment |
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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 $9$?
answered Nov 21 at 20:42
bjcolby15
1,1381916
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 $9$?
answered Nov 21 at 20:42
bjcolby15
1,1381916
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 $9$?
answered Nov 21 at 20:42
bjcolby15
1,1381916
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 Nov 21 at 20:42
bjcolby15
1,1381916
answered Nov 21 at 20:42
bjcolby15
1,1381916
answered Nov 21 at 20:42
bjcolby15
1,1381916
answered Nov 21 at 20:42
bjcolby15
1,1381916
1,1381916
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
Nov 20 at 0:33
Try to write it around the (0,0) and than transkate
– Moti
Nov 20 at 7:37
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
Nov 21 at 20:41