Q on polar curves

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Sonal7

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I can't understand the Q, let along work out the ans. I worked out the area of the whole curve, which is 1/a2
I think as thats not an option, then i need to use calculus to work the min squares, I really don't know what I am suppose to do.Screenshot 2020-04-06 at 21.43.08.png
 
Unless you think that the polar curve you were given is a square why on earth would you think that the area of the smallest square that the curve would fit into would and the area of the curve would be equal???

Did you draw the polar curve, try to fit the smallest square around it and then try to find the area of that smallest square?

Please show us (and yourself) a picture of the polar curve and draw the smallest small around it. We will proceed from there.
 
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I can see the area they want us to work out on a graph.
Also there was a similar problem in a textbook so I know the r when [MATH]dx/d\theta=0[/MATH]
20200408_195018 (1).jpg
 
I am having problems working out y when the tangent is parallel to the curve as I am getting theta is 0 so y =0. I am attaching my workings below.20200408_205852.jpg
 
Sonal7 said:
I can see the area they want us to work out on a graph.
Also there was a similar problem in a textbook so I know the r when [MATH]dx/d\theta=0[/MATH]
View attachment 17783
Click to expand...
Sonal7 said:
I can see the area they want us to work out on a graph.
Also there was a similar problem in a textbook so I know the r when [MATH]dx/d\theta=0[/MATH]
View attachment 17783
Click to expand...
The square that you shaded in clearly does not contain the whole curve. I take it you are confused about what the whole curve is. Just think about what it means and do not assume for a second that it is some friendly part of the whole curve.
 
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It seems that you are looking for a formula or equation to work this problem. I'll be honest I do not see how to figure out the correct answer, but you are giving choices and I can easily rule out three of the choices. So in the end I can pick the correct choice.

Looking at the graph do you see that line (not drawn) of length 2a that goes from one tip of the curve to the other tip? Let's just concentrate on that line for a moment and nothing else. You want to draw the smallest square that can contain a line of length 2a. The area for that square will be the answer of the answer will be larger. Do you see why? Think about it. Please reply back?
 
I thought I had decomposed the problem but this is stupid mistake. It wont be square if you split the area into two and then find 2 squares that cover each petal. I was silly.
 
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Jomo said:
It seems that you are looking for a formula or equation to work this problem. I'll be honest I do not see how to figure out the correct answer, but you are giving choices and I can easily rule out three of the choices. So in the end I can pick the correct choice.

Looking at the graph do you see that line (not drawn) of length 2a that goes from one tip of the curve to the other tip? Let's just concentrate on that line for a moment and nothing else. You want to draw the smallest square that can contain a line of length 2a. The area for that square will be the answer of the answer will be larger. Do you see why? Think about it. Please reply back?
Click to expand...
I think i got it. I might not need to work out the y co ordinates as a square has equal sides.
 
I still made the mistake, I will double the x value and try.
 
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