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Cartesian Equations with Trig Help

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

Reply 1

nnnabil

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

below

Screenshot 2022-05-16 16.56.41.png4.0KB

Reply 2

mqb2766

Original post by nnnabil

below

Thats correct, its 0 on both ends. What does a cos curve look like in the middle and hence whats its max value?

Reply 3

nnnabil

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?

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

Reply 4

mqb2766

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

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 ?

Reply 5

nnnabil

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 ?

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?

Reply 6

mqb2766

Original post by nnnabil

Can you post (a picture of) the full question pls?

Reply 7

nnnabil

Original post by mqb2766

Can you post (a picture of) the full question pls?

my wifi is acting up sorry, but its edxcl pure year 2 chp8 ex 8b, q6 thank you

Reply 8

mqb2766

Original post by nnnabil

my wifi is acting up sorry, but its edxcl pure year 2 chp8 ex 8b, q6 thank you

Dont have it, sorry. Can you type up the relevant question part?

Reply 9

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 :smile:

Reply 10

mqb2766

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 :smile:

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

(edited 1 year ago)

Reply 11

nnnabil

Original post by mqb2766

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

Reply 12

nnnabil

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

so like to clarify....

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?

Reply 13

mqb2766

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

There are 3 functions here, t->x, t->y and x->y. Each has a domain (input) and a range (output). The domain and range of the Cartesian map x->y is given by the range of the t->x map and the range of the t->y map, respectively.