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.










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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















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.










share|cite|improve this question
























  • 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













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.










share|cite|improve this question















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






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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


















  • 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










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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 $9$?






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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 $9$?






    share|cite|improve this answer

























      up vote
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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 $9$?






      share|cite|improve this answer























        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 $9$?






        share|cite|improve this answer












        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 $9$?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 20 hours ago









        bjcolby15

        1,0961916




        1,0961916






























             

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