How to convert this parametric equation into a Cartesian equation?
Clash Royale CLAN TAG#URR8PPP
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Sketch the curve by using the parametric equation to plot points. Indicate with arrow the direction in which the curve is traced as $t$ increases.
$$x=t^2+t$$
$$y=t^2-t$$
I tried to convert this parametric equation into a Cartesian equation, but I didn't know how. Please help.
parametric curves
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
asked Dec 27 '13 at 12:08
user32104
1972518
 |Â
show 1 more comment
up vote
0
down vote
favorite
Sketch the curve by using the parametric equation to plot points. Indicate with arrow the direction in which the curve is traced as $t$ increases.
$$x=t^2+t$$
$$y=t^2-t$$
I tried to convert this parametric equation into a Cartesian equation, but I didn't know how. Please help.
parametric curves
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
asked Dec 27 '13 at 12:08
user32104
1972518
1
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
1
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Sketch the curve by using the parametric equation to plot points. Indicate with arrow the direction in which the curve is traced as $t$ increases.
$$x=t^2+t$$
$$y=t^2-t$$
I tried to convert this parametric equation into a Cartesian equation, but I didn't know how. Please help.
parametric curves
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
asked Dec 27 '13 at 12:08
user32104
1972518
Sketch the curve by using the parametric equation to plot points. Indicate with arrow the direction in which the curve is traced as $t$ increases.
$$x=t^2+t$$
$$y=t^2-t$$
I tried to convert this parametric equation into a Cartesian equation, but I didn't know how. Please help.
parametric curves
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
asked Dec 27 '13 at 12:08
user32104
1972518
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
edited Aug 24 at 8:03
Rodrigo de Azevedo
12.7k41751
12.7k41751
asked Dec 27 '13 at 12:08
user32104
1972518
asked Dec 27 '13 at 12:08
user32104
1972518
asked Dec 27 '13 at 12:08
user32104
1972518
1972518
1
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
1
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25
 |Â
show 1 more comment
1
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
1
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25
1
1
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
1
1
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
3
down vote
HINT:
On addition we have $displaystyle 2t^2=x+yiff t^2=fracx+y2$
On subtraction, $displaystyle 2t=x-yiff t=fracx-y2$
Can you eliminate $t$ from here?
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
add a comment |Â
up vote
2
down vote
HINT :
$t^2=x-t=y+t$.
So, you can represent $t$ by $x,y$.
Since we have $t=fracx-y2$, we have
$$x=left(fracx-y2right)^2+fracx-y2.$$
We can get the following equation with $x$ and $y$ :
$$x^2-2xy+y^2-2x-2y=0.$$
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
HINT:
On addition we have $displaystyle 2t^2=x+yiff t^2=fracx+y2$
On subtraction, $displaystyle 2t=x-yiff t=fracx-y2$
Can you eliminate $t$ from here?
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
add a comment |Â
up vote
3
down vote
HINT:
On addition we have $displaystyle 2t^2=x+yiff t^2=fracx+y2$
On subtraction, $displaystyle 2t=x-yiff t=fracx-y2$
Can you eliminate $t$ from here?
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
add a comment |Â
up vote
3
down vote
up vote
3
down vote
HINT:
On addition we have $displaystyle 2t^2=x+yiff t^2=fracx+y2$
On subtraction, $displaystyle 2t=x-yiff t=fracx-y2$
Can you eliminate $t$ from here?
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
HINT:
On addition we have $displaystyle 2t^2=x+yiff t^2=fracx+y2$
On subtraction, $displaystyle 2t=x-yiff t=fracx-y2$
Can you eliminate $t$ from here?
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
answered Dec 27 '13 at 12:09
lab bhattacharjee
216k14153265
216k14153265
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
add a comment |Â
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
1
1
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
Thank you, do you mean $((x-y)/2)^2=((x+y)2)$?
– user32104
Dec 27 '13 at 12:12
1
1
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
@user32104, my pleasure. Nice to hear that you could complete
– lab bhattacharjee
Dec 27 '13 at 14:44
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
how about x = 4 + 2t and y = 5 + 2t?
– Hasan Iqbal
Apr 4 '16 at 11:01
add a comment |Â
up vote
2
down vote
HINT :
$t^2=x-t=y+t$.
So, you can represent $t$ by $x,y$.
Since we have $t=fracx-y2$, we have
$$x=left(fracx-y2right)^2+fracx-y2.$$
We can get the following equation with $x$ and $y$ :
$$x^2-2xy+y^2-2x-2y=0.$$
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
add a comment |Â
up vote
2
down vote
HINT :
$t^2=x-t=y+t$.
So, you can represent $t$ by $x,y$.
Since we have $t=fracx-y2$, we have
$$x=left(fracx-y2right)^2+fracx-y2.$$
We can get the following equation with $x$ and $y$ :
$$x^2-2xy+y^2-2x-2y=0.$$
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT :
$t^2=x-t=y+t$.
So, you can represent $t$ by $x,y$.
Since we have $t=fracx-y2$, we have
$$x=left(fracx-y2right)^2+fracx-y2.$$
We can get the following equation with $x$ and $y$ :
$$x^2-2xy+y^2-2x-2y=0.$$
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
HINT :
$t^2=x-t=y+t$.
So, you can represent $t$ by $x,y$.
Since we have $t=fracx-y2$, we have
$$x=left(fracx-y2right)^2+fracx-y2.$$
We can get the following equation with $x$ and $y$ :
$$x^2-2xy+y^2-2x-2y=0.$$
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
edited Dec 28 '13 at 8:55
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
answered Dec 27 '13 at 12:09
mathlove
87.1k877208
87.1k877208
add a comment |Â
add a comment |Â
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1
How did you try? What manipulations of it did you use? Did any of them get you a weird result? What have you found?
– JMCF125
Dec 27 '13 at 12:14
1
I don't think converting to Cartesian coordinates is what you were intended to do. Rather, select values of $t$ and then plot the points $(x,y)$ given by the parameterization. Then, informally deduce what the curve is. Note if you just look at the Cartesian equation, you won't be able to determine the direction in which the curve is traced out.
– David Mitra
Dec 27 '13 at 12:16
I have tried this x=t(t+1) and y=t(t-1), then t=y/(t-1)=x/(t+1) but I didn't know how to eliminate t
– user32104
Dec 27 '13 at 12:18
@DavidMitra thank you but how to know the shape without use points (because our teacher says we will not be able to draw table for points in the exam)
– user32104
Dec 27 '13 at 12:21
Well, the problem is oddly phrased then, it clearly indicates (to me at least) to use a "table method".
– David Mitra
Dec 27 '13 at 12:25