Q on polar co-ordinates

[Sonal7]

This is the question, part I is easy. as sin is an odd function and when you put in sin[MATH]\theta[/MATH], you can replace these by -sin[MATH]\theta[/MATH] and the rest is easy to prove using properties of sin function. I am confused about the graph iii

 

[LCKurtz]

What confuses you about the graph in iii? Have you tried plotting a few points?

 

[Sonal7]

I am also confused about part ii, as it should say y[MATH]\geq1[/MATH] since when [MATH]sin\theta =0[/MATH], y=1
This when you consider the equation y=[MATH]sin^2+1[/MATH], if one were plotting a polar graph it would be infinity when sin [MATH]\theta[/MATH]=0

 

[LCKurtz]

But [MATH]\sin\theta = 0[/MATH] doesn't give a point on the graph. For example, [MATH]\theta \ne 0,\pi[/MATH].

 

[Steven G]

part I is easy. as sin is an odd function and when you put in sinθ, you can replace these by -sinθ
Any change that you explain what that means? Where are you putting sinθ into? What are the these that can be replaced with sinθ?

 

[Sonal7]

What confuses you about the graph in iii? Have you tried plotting a few points?

I can make much sense of this. When sin [MATH]\theta =0[/MATH] when [MATH]\theta[/MATH]=[MATH]\pi,0[/MATH], the coordinates at [MATH]\theta =0[/MATH] should be equal to 1. I suppose they are plotting the r value so its infinity when [MATH]\theta=0[/MATH]. Yes that makes more sense.

 

[Sonal7]

part I is easy. as sin is an odd function and when you put in sinθ, you can replace these by -sinθ
Any change that you explain what that means? Where are you putting sinθ into? What are the these that can be replaced with sinθ?

 

[Sonal7]

i hope you dont mind me copying the answer to save the time it might take to type it using latex. This is A level maths, fairly easy.

 

[Sonal7]

But [MATH]\sin\theta = 0[/MATH] doesn't give a point on the graph. For example, [MATH]\theta \ne 0,\pi[/MATH].

What ! I thought they wanted us to plot a graph of the function for [MATH]0<\theta<\pi[/MATH] as quoted in the question.

 

[Steven G]

0<θ<π means that θ is in between 0 and π but not equal to the endpoints. It does NOT say that 0<θ<π. Is this clear?

 

[LCKurtz]

What ! I thought they wanted us to plot a graph of the function for [MATH]0<\theta<\pi[/MATH] as quoted in the question.

[MATH]0 < \theta < \pi[/MATH] does not include the end points. Do you understand that the graph is plotted in polar coordinates? They are plotting [MATH]r[/MATH] against [MATH]\theta[/MATH].

 

[Sonal7]

0<θ<π means that θ is in between 0 and π but not equal to the endpoints. It does NOT say that 0<θ<π. Is this clear?

Oh i see, how silly of me not to think of that. so the inequality is correct.

 

[Sonal7]

[MATH]0 < \theta < \pi[/MATH] does not include the end points. Do you understand that the graph is plotted in polar coordinates? They are plotting [MATH]r[/MATH] against [MATH]\theta[/MATH].

Yes I now gathered that it must be r plotted for various values of theta, the range was specified. So when [MATH]\theta =\pi /2[/MATH], r =2.
When theta is [MATH]\pi /2<\theta<0[/MATH] the r is increasing from infinity to 2. I dont think the asymptote is correct.
In my defence, they should not be putting a y on the vertical axis on the graph!

 

[Steven G]

Oh i see, how silly of me not to think of that. so the inequality is correct.

Is which inequality correct? If you mean the one that was given to you we must assume yes unless it contradicts something.

 

[LCKurtz]

I dont think the asymptote is correct.
In my defence, they should not be putting a y on the vertical axis on the graph!

There is nothing wrong with having [MATH]y[/MATH] labeled on the graph. The asymptote is correctly labeled as [MATH]y=1[/MATH]. It might have been helpful had they drawn a polar radius to a point on the graph and labeled [MATH]r[/MATH] and [MATH]\theta[/MATH]. You can verify that [MATH]y=1[/MATH] is the horizontal asymptote because [MATH]y = r\sin\theta = \sin^2\theta + 1 \to 1[/MATH] as [MATH]\theta \to 0[/MATH].

 

[Sonal7]

There is nothing wrong with having [MATH]y[/MATH] labeled on the graph. The asymptote is correctly labeled as [MATH]y=1[/MATH]. It might have been helpful had they drawn a polar radius to a point on the graph and labeled [MATH]r[/MATH] and [MATH]\theta[/MATH]. You can verify that [MATH]y=1[/MATH] is the horizontal asymptote because [MATH]y = r\sin\theta = \sin^2\theta + 1 \to 1[/MATH] as [MATH]\theta \to 0[/MATH].

Yes its not a polar graph then, I dont think they are plotting r. They are plotting y. its actually the function y=[MATH]sin^2\theta +1[/MATH]

 

[LCKurtz]

No. You are not understanding what I have told you.

 

[Sonal7]

Is which inequality correct? If you mean the one that was given to you we must assume yes unless it contradicts something.

Sorry I was rather talking to myself. It says in part ii show that y>1. I had thought y[MATH]\geq 1[/MATH]But in my defence, they didn't mention the range until part iii. when they state [MATH]0<\theta<pi[/MATH]

 

[Sonal7]

 

[Sonal7]

No. You are not understanding what I have told you.

I understand it now but very confusing as they left me clueless what i was meant to sketching, r or y.