Equal area divided by line

  • Thread starter terryds
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  • #1
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Homework Statement



qppac1.png


In the picture above, line y = c intersects with parabola y = 6x-x^2 in the first quadrant.
If the gray area below line y = c and the gray area above line y=c are equal, then value of c is ...
A. 19/4
B.21/4
C.23/4
D.25/4
E. 27/4

Homework Equations



Area under parabola = 2/3* base * height
Area above parabola = 1/3 * base* height
Area = definite integral

The Attempt at a Solution


[/B]
Parabola y=6x-x^2 has peak point (3,9)

Suppose a is the x-intercept of the line and parabola.
At x = a, c equals 6a-a^2
So, the gray area below the line is 1/3 * a * (6a-a^2)

The base for the upper gray parabola area is 2(3) - 2a = 6-2a
So, the gray area above the line is 2/3 * (6-2a) * (9-(6a-a^2))

I get a = 1.592
and c = 6a-a^2 = 7.017

However, I'm sure the correct answer is 27/4 (I've tested it using integral method)
Please help...
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



qppac1.png


In the picture above, line y = c intersects with parabola y = 6x-x^2 in the first quadrant.
If the gray area below line y = c and the gray area above line y=c are equal, then value of c is ...
A. 19/4
B.21/4
C.23/4
D.25/4
E. 27/4

Homework Equations



Area under parabola = 2/3* base * height
Area above parabola = 1/3 * base* height
Area = definite integral

The Attempt at a Solution


[/B]
Parabola y=6x-x^2 has peak point (3,9)

Suppose a is the x-intercept of the line and parabola.
At x = a, c equals 6a-a^2
So, the gray area below the line is 1/3 * a * (6a-a^2)

The base for the upper gray parabola area is 2(3) - 2a = 6-2a
So, the gray area above the line is 2/3 * (6-2a) * (9-(6a-a^2))

I get a = 1.592
and c = 6a-a^2 = 7.017

However, I'm sure the correct answer is 27/4 (I've tested it using integral method)
Please help...
Your expression for the gray area below the line is incorrect.
 
Last edited:
  • #3
blue_leaf77
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Translate the parabola downward till the line y=c coincides with the x axis and then integrate the translated parabola from x=0 to the right intersection with the x axis. You will find c by making this integral vanish.
 
  • #4
blue_leaf77
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Alternatively you can work with the inverse function ##x=f^{-1}(y)##. I think this way can br easier.
 
  • #5
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13
Your expression for the gray area below the line is incorrect.
Why??
I think it's correct

Translate the parabola downward till the line y=c coincides with the x axis and then integrate the translated parabola from x=0 to the right intersection with the x axis. You will find c by making this integral vanish.
Okay, so the translated parabola become

y = -x^2+6x-c

But, I don't know the roots of the polynomial since there is an unknown c.
Even if I substitute it with ( -6 ± 36 - 4k ) / -2 , I see it'll be very messy and hard to solve

Alternatively you can work with the inverse function ##x=f^{-1}(y)##. I think this way can br easier.
The inverse function of the parabola is

f^-1(x) = √(9-x) + 3

But, what does the inverse have to do with the area ? I don't understand the method
 
  • #6
blue_leaf77
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I have tried following the method suggested in post #4. I ended up having to solve cubic polynomial in ##c##. However using graphical method I did find 27/4 as one of the solutions. Probably there are other members who can propose a more elegant method.
 
  • #7
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13
I have tried following the method suggested in post #4. I ended up having to solve cubic polynomial in ##c##. However using graphical method I did find 27/4 as one of the solutions. Probably there are other members who can propose a more elegant method.
Okay..
Thanks anyway
 
  • #8
ehild
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Why??
I think it's correct



Okay, so the translated parabola become

y = -x^2+6x-c

But, I don't know the roots of the polynomial since there is an unknown c.
Even if I substitute it with ( -6 ± 36 - 4k ) / -2 , I see it'll be very messy and hard to solve
Anyway, you can write the solutions x1 and x2 (x1<x2) of the quadratic equation in terms of c. ##\int _0 ^{x_1}(c-y)dx = \int _{x_1} ^{x_2}(y-c)dx ## means that ## \int _0 ^ {x_2} {(-x^2+6x-c)dx}=0##
Do the integration, substitute the limits, and solve the equation for c (it will become a quadratic one), not that terrible.
 
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  • #9
blue_leaf77
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the quadratic equation in terms of c
I haven't tried myself what I suggested in post #3 which seems to coincide to your proposed method, but is it really quadratic in ##c##?
I mean first we need to solve for ##x_2##, probably by using the usual quadratic root formula, and this will sure contain ##c## under a square root. As for the integral itself, there will be ##x^3## arising from the quadratic term in ##y(x)##.
 
  • #10
ehild
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I haven't tried myself what I suggested in post #3 which seems to coincide to your proposed method, but is it really quadratic in ##c##?
I mean first we need to solve for ##x_2##, probably by using the usual quadratic root formula, and this will sure contain ##c## under a square root. As for the integral itself, there will be ##x^3## arising from the quadratic term in ##y(x)##.
Yes, you proposed the same method.
It will be simplified to a quadratic, as all terms of the definite integral contain x2.
 
  • #11
Ray Vickson
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Why??
I think it's correct

*************************************
Because when I did it (or, rather, had Maple do it) I got an answer that is different from yours. Not just difference in appearance, but different in fact: having a non-zero deviation between yours and mine.

Anyway, you did not actually show the step-by-step details of your work, so I cannot say where you went astray.

***************************************


Okay, so the translated parabola become

y = -x^2+6x-c

But, I don't know the roots of the polynomial since there is an unknown c.
Even if I substitute it with ( -6 ± 36 - 4k ) / -2 , I see it'll be very messy and hard to solve



The inverse function of the parabola is

f^-1(x) = √(9-x) + 3

But, what does the inverse have to do with the area ? I don't understand the method
 
  • #12
392
13
Anyway, you can write the solutions x1 and x2 (x1<x2) of the quadratic equation in terms of c. ##\int _0 ^{x_1}(c-y)dx = \int _{x_1} ^{x_2}(y-c)dx ## means that ## \int _0 ^ {x_2} {(-x^2+6x-c)dx}=0##
Do the integration, substitute the limits, and solve the equation for c (it will become a quadratic one), not that terrible.

Thanks.. I've got it
 

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