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Cartesian Equations with Trig Help
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nnnabil
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#1
Hi can someone explain to me the domain on this equation please. If you input the limits of -pi/2 and pi/2 into t, x is 0 on both ends? thank you
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nnnabil
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#2
(Original post by nnnabil)
Hi can someone explain to me the domain on this equation please. If you input the limits of -pi/2 and pi/2 into t, x is 0 on both ends? thank you
Hi can someone explain to me the domain on this equation please. If you input the limits of -pi/2 and pi/2 into t, x is 0 on both ends? thank you
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mqb2766
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#3
(Original post by nnnabil)
below
below
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nnnabil
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#4
(Original post by mqb2766)
Thats correct, its 0 on both ends. What does a cos curve look like in the middle and hence whats its max value?
Thats correct, its 0 on both ends. What does a cos curve look like in the middle and hence whats its max value?
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mqb2766
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#5
(Original post by nnnabil)
1.. but thats the y value so doesnt that refer to the range and not the domain? I get that the range is 0 to 8 but the domain isn't
1.. but thats the y value so doesnt that refer to the range and not the domain? I get that the range is 0 to 8 but the domain isn't
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nnnabil
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#6
(Original post by mqb2766)
Youve not posted the full question, but for the t->x map, the domain is -pi/2..pi/2 and the range is 0..8. For the x->y map, the domain is 0..8 and the range is ?
Youve not posted the full question, but for the t->x map, the domain is -pi/2..pi/2 and the range is 0..8. For the x->y map, the domain is 0..8 and the range is ?
so is the range of the 'x = 8 cos t' part the domain of f(x) and vice versa?
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mqb2766
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#7
(Original post by nnnabil)
sorry, the full question was about curve C which has the following properties: x = 8 cos t, y= 1/4sec^2t, -pi/2 < t < pi/2.
so is the range of the 'x = 8 cos t' part the domain of f(x) and vice versa?
sorry, the full question was about curve C which has the following properties: x = 8 cos t, y= 1/4sec^2t, -pi/2 < t < pi/2.
so is the range of the 'x = 8 cos t' part the domain of f(x) and vice versa?
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nnnabil
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#8
(Original post by mqb2766)
Can you post (a picture of) the full question pls?
Can you post (a picture of) the full question pls?
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mqb2766
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#9
(Original post by nnnabil)
my wifi is acting up sorry, but its edxcl pure year 2 chp8 ex 8b, q6 thank you
my wifi is acting up sorry, but its edxcl pure year 2 chp8 ex 8b, q6 thank you
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nnnabil
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#10
6 The curve C has parametric equations
x = 8 cos t, y = 1/4sec2 t, -pi/2 < t < pi/2
a) Find a Cartesian equation of C
b) Sketch the curve C on the appropriate domain
Thanks
x = 8 cos t, y = 1/4sec2 t, -pi/2 < t < pi/2
a) Find a Cartesian equation of C
b) Sketch the curve C on the appropriate domain
Thanks
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mqb2766
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#11
(Original post by nnnabil)
6 The curve C has parametric equations
x = 8 cos t, y = 1/4sec2 t, -pi/2 < t < pi/2
a) Find a Cartesian equation of C
b) Sketch the curve C on the appropriate domain
Thanks
6 The curve C has parametric equations
x = 8 cos t, y = 1/4sec2 t, -pi/2 < t < pi/2
a) Find a Cartesian equation of C
b) Sketch the curve C on the appropriate domain
Thanks
For t->x, the domain is -pi/2..pi/2 and the range is 0..8.
For t->y, the domain is -pi/2..pi/2 and the range is 1/4..inf
So the domain and range of the cartesian equation x->y is 0..8 and 1/4..inf
Last edited by mqb2766; 1 month ago
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nnnabil
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#12
(Original post by mqb2766)
So youre after the domain/range of the Cartesian equation x->y.
For t->x, the domain is -pi/2..pi/2 and the range is 0..8.
For t->y, the domain is -pi/2..pi/2 and the range is 1/4..inf
So the domain and range of the cartesian equation x->y is 0..8 and 1/4..inf
So youre after the domain/range of the Cartesian equation x->y.
For t->x, the domain is -pi/2..pi/2 and the range is 0..8.
For t->y, the domain is -pi/2..pi/2 and the range is 1/4..inf
So the domain and range of the cartesian equation x->y is 0..8 and 1/4..inf
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nnnabil
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#13
(Original post by nnnabil)
does that mean for y=f(x) that they've effectively switched? because the range for t -> x is now the new domain for the whole cartesian equation? sorry i just find this really confusing
does that mean for y=f(x) that they've effectively switched? because the range for t -> x is now the new domain for the whole cartesian equation? sorry i just find this really confusing
lets say you've been given a non-cartesian equation with x parts and y parts and also a range for t. to find the domain, you'd sub the t limits into the x part right?
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#14
(Original post by nnnabil)
does that mean for y=f(x) that they've effectively switched? because the range for t -> x is now the new domain for the whole cartesian equation? sorry i just find this really confusing
does that mean for y=f(x) that they've effectively switched? because the range for t -> x is now the new domain for the whole cartesian equation? sorry i just find this really confusing
Last edited by mqb2766; 1 month ago
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