Ho Ho... No

[Peanut Butter]

If there is no number to take away or add to both sides should I just try to get sin and cos alone? so the first one would become x/3 = cos(2t) and second y/3 = sin(2t) and then square both?

Yep that's correct

[blacksanta62]

Okay ) (triple chin)

[blacksanta62]

Hey PB I got: x^2 + y^2 = 3^2
I believe I got one part of the domain and range: y = 3sin(0) and x = 3cos(0) resulting in 0 for y and 3 for x. When subbing in pi I got:
y = 3sin(2pi) and x = 3cos(2pi), got confused around this part  Do I treat the 2 pi as exact values for both which would result in :
y = 3sin(0) and x = 3cos(1)

[Peanut Butter]

Hey PB I got: x^2 + y^2 = 3^2
I believe I got one part of the domain and range: y = 3sin(0) and x = 3cos(0) resulting in 0 for y and 3 for x. When subbing in pi I got:
y = 3sin(2pi) and x = 3cos(2pi), got confused around this part  Do I treat the 2 pi as exact values for both which would result in :
y = 3sin(0) and x = 3cos(1)

Yes, use exact values

You might notice that this question is a bit tricky. You can find the domain easily, however the range is hard to find (you can't really use the method I used previously).

There is an easier way to do this. A general rule for a circle is: when x = acos(t) the domain is [-a,a] and when y = bcos(t) the range is [-b,b] when t = [0, 2pi]

However , our t is only pi. But since our 'theta' (what's inside the sin/cos brackets) is 2t. We can take this as 2pi.

Therefore, both the domain and range is [-3,3]

[blacksanta62]

I'm going to take some time to try to understand this 
Was the first part right?

[Peanut Butter]

I'm going to take a while to try to understand this :/
Was the first part right?

Yes. Both your Cartesian equation and the values for x and y when t = 0 is correct.

When subbing t=Pi:

x = 3cos(2pi)
cos(2pi) = 1
Therefore, x = 3

y = 3sin(2pi)
Sin(2pi) = 0
Therefore, y = 0

Does that part make sense?


(maybe there is a wayyyy easier way to do this than what I do... haha)

[blacksanta62]

Yes. Both your Cartesian equation and the values for x and y when t = 0 is correct.

When subbing t=Pi:

x = 3cos(2pi)
cos(2pi) = 1
Therefore, x = 3

y = 3sin(2pi)
Sin(2pi) = 0
Therefore, y = 0

Does that part make sense?


(maybe there is a wayyyy easier way to do this than what I do... haha)

That part made sense to me in your other post I was confused on how you got the range and domain as [-3,3]. Where did the 0 go?

[Peanut Butter]

That part made sense to me in your other post I was confused on how you got the range and domain as [-3,3]. Where did the 0 go?

Which 0 are you talking about?

If it's the one we found above (y = 0), it's still included in the range we found [-3,3] however it just means that when t = 0 or t = Pi it is not at the lowest or highest y value is that what you mean?

[blacksanta62]

Ahhhh. I gotcha  I messed up my own mind

What about this: So if given the equation, say: (x-2)^2/4 + (y+3)^2/9 = 1 and told to sketch it, the radius would be 1, centre (2,-3), a^2 = 4 ---> 2, b^2 = 9 ----> 3

I would then find vertices (for the x axis) via: (-2 + 2, -4) and ( 2 + 2, -4) and for the y (2, -3 + (-3)) and (2, 3 +(-3)
Then find x and y intercepts and sketch right. This question is about an ellipse but the formula looks so similar

Edit: The above question is wrong and so is the working. The correct question is: (x-2)^2/4 + (y-3)^2/9 = 1

[Peanut Butter]

Ahhhh. I gotcha  I messed up my own mind

What about this: So if given the equation, say: (x-2)^2/4 + (y+3)^2/9 = 1 and told to sketch it, the radius would be 1, centre (2,-3), a^2 = 4 ---> 2, b^2 = 9 ----> 3

I would then find vertices (for the x axis) via: (-2 + 2, -4) and ( 2 + 2, -4) and for the y (2, -3 + (-3)) and (2, 3 +(-3)
Then find x and y intercepts and sketch right. This question is about an ellipse but the formula looks so similar

Yep the coordinates are all correct

Not sure if this helps... but when I do questions on ellipses I find the separate horizontal and vertical radius'.

For instance, with the question above, since a^2 = 4 the horizontal radius is 2 and since b^2 = 9 the vertical radius is 3. Then just add those values to the centre.

This is essentially what you've done but I just find it easier to process if I've written it out

[blacksanta62]

Cool, just edited my post above
Going to take a brake from specialist homework for now and do some methods Probably post questions PB if I get stuck and need help

[blacksanta62]

Sketch the graph of the following functions:
(x-2)^2/9 + (y+4)^2/4 = 1

I'm having trouble finding the x and y intercepts. The textbook says to make y = 0 & x = 0 (which I could normally do but I'm just lost with how the textbook got it's answer) but how would I do it with such a question? The question recommends no CAS be used. If some steps can be shown it will greatly appreciated. Thank you
 

[pi]

You're going to have to use the quadratic formula to get the intercepts. Gonna get a tad ugly and tedious, but do-able by hand (although you shouldn't be finding any x-intercepts ).

[blacksanta62]

Along with that question: This is the equation for a graph- (x+3)^2/25 + (y-2)^2/16 = 1
The centre of the equation is (-3,2) but how does one find the min & max of the ellipse? Do I have to sketch it first? This is a multiple choice question so a CAS can be used but does the calculator have the ability to graph ellipses?

[Peanut Butter]

Along with that question: This is the equation for a graph- (x+3)^2/25 + (y-2)^2/16 = 1
The centre of the equation is (-3,2) but how does one find the min & max of the ellipse? Do I have to sketch it first? This is a multiple choice question so a CAS can be used but does the calculator have the ability to graph ellipses?

Yes, you can graph on the CAS!

Open a graph page > menu > graph entry (3) > equation (2) > ellipse (4)

Then just enter all your values in

btw that's for the TI-nspire