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Circular measure with trigno. Wrong answer.

  • Thread starter SolCon
  • Start date
  • #1
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1
Hello again.

I've got 2 simple questions relating to trigonometry in circular measure. I've got answers to both but they are not satisfactory. So here goes.

1) This question requires that we show that A (which is the area of a circle) is a maximum and not a minimum, when r=3/2.

A=3r-r^2

I've got the answer coming as 9/4 after substituting the value of 'r' in the equation. However, I do not know how this proves the question's requirement.

2) This is slightly bigger. I have the following figure:

http://usera.ImageCave.com/biosyn/q1.PNG

Q: . We are required to find the perimeter of the shaded region.

The conditions:

. OPQ is an equilateral triangle.
. OS=PS=QS
. arc PXQ has centre O with radius 12 cm
. Answer must be in terms of pi and under-root 3

First of all, since OS,PS and QS are equal, I believe that the angles PSQ, PSO and QSO are equal and =360. This means each is 120 degrees or 2pi/3. I've also established the radius PS as 6/under-root 3/4. I've used s=r(theta) to find PXQ as 12pi/3[under-root(3/4)].
However, adding them all together is not giving me the correct answer.

What am I missing here?

Thanks for any help. :smile:
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
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1) This question requires that we show that A (which is the area of a circle) is a maximum and not a minimum, when r=3/2.

A=3r-r^2

I've got the answer coming as 9/4 after substituting the value of 'r' in the equation. However, I do not know how this proves the question's requirement.
What is the leading coefficient of
[tex]A = 3r - r^2[/tex]
?
And what does the sign of the leading coefficient tell us about how the parabola opens?

The conditions:

. OPQ is an equilateral triangle.
. OS=PS=QS
. arc PXQ has centre O with radius 12 cm
. Answer must be in terms of pi and under-root 3

First of all, since OS,PS and QS are equal, I believe that the angles PSQ, PSO and QSO are equal and =360. This means each is 120 degrees or 2pi/3. I've also established the radius PS as 6/under-root 3/4. I've used s=r(theta) to find PXQ as 12pi/3[under-root(3/4)].
However, adding them all together is not giving me the correct answer.

What am I missing here?

Thanks for any help. :smile:
(See bolded.) If I'm reading this right, the radius is NOT PS. You stated:
". arc PXQ has centre O with radius 12 cm"
Sounds like to me that the radius is OP (or OQ), not PS.
 
Last edited:
  • #3
verty
Homework Helper
2,164
198
I've got the answer coming as 9/4 after substituting the value of 'r' in the equation. However, I do not know how this proves the question's requirement.
It fails to prove the requirement. It does not prove that 9/4 is either the maximum or minimum value of the function on its domain.
 
  • #4
33
1
Thank you both for your response.

I'll try q.2 again with the side PO as radius and see what comes up.

However, I think I'll need some help with q.1.

Here's the complete question. It has 2 parts and also involves theta (hereby written as 0).

Q. A circle with Centre O has radius r cm. A sector of the circle, which as an angle of 0 radians at O, has perimeter 6 cm.

1) Show that 0=(6/r)-2, and express the area A cm^2 of the sector in terms of r.

I solved the first part (rearranging 6=2r + r0) and the second part came as A=3r-r^2.

2) Show that A is a maximum and not a minimum, when r=3/2, and calculate the corresponding value of 0.

I've solved the second part of this question, where 0=2 which is correct. But I don't know how I can solve the first part.
 
  • #5
eumyang
Homework Helper
1,347
10
I will repeat my question:
What is the leading coefficient of
[tex]A = 3r - r^2[/tex]
?
And what does the sign of the leading coefficient tell us about how the parabola opens?
The function A = 3r - r2 is a quadratic, so it's graph will a parabola. Are you familiar with graphs of parabolas and their characteristics?
 
  • #6
33
1
What is the leading coefficient of
A = 3r - r^2
?
And what does the sign of the leading coefficient tell us about how the parabola opens?
Thanks for letting me know about the equation and the parabola and the leading coefficient (I forgot about this at the time of doing the question). A quick review of parabolas led me to this:

The leading coefficient could be -1 for when A=3r-r^2 or 1 for when r^2-3r+A=0. I'm not sure, there isn't any condition of not being =0.

If we use the equation ax^2+bx=c, we will see that in the case of the former (-1), the parabola will open downwards and hence, will be a maximum (since a<0). For the latter (1), the parabola will open upwards and hence, will be a minimum (since a>0).

But which one is right?
 
  • #7
eumyang
Homework Helper
1,347
10
The leading coefficient could be -1 for when A=3r-r^2 or 1 for when r^2-3r+A=0. I'm not sure, there isn't any condition of not being =0.

If we use the equation ax^2+bx=c, we will see that in the case of the former (-1), the parabola will open downwards and hence, will be a maximum (since a<0). For the latter (1), the parabola will open upwards and hence, will be a minimum (since a>0).

But which one is right?
The former. Don't set the function equal to zero like that. A = -r2 + 3r will suffice. The general form of a quadratic is not ax2+bx=c, but f(x) = ax2 + bx + c. It is from this form that we can make observations about the leading coefficient a.
 
  • #8
verty
Homework Helper
2,164
198
The minimum or maximum value can be found by completing the square or checking the signs of the first derivative. The first method uses just algebra, the second requires calculus knowledge.
 
  • #9
33
1
Once again, thank you both for the help. Both questions have been solved with the correct solutions. :smile:
 

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