Parabola? (1 Viewer)

[Lemiixem]

The Points A(3,9) and B(-2,4) lie on the parabola y=x^2. The line y=x+6 joins A and B. The Point P(p,p^2) is a variable on the parabola below the line. Find the area of the parabolic segment APB, i.e. the area below the line and above the parabola.

Stacking Logs, 2 in the top row, 3 in the second row, and so on, having one more log in each row. If there are 170 logs altogether, how many should there be in the bottom row.

 

[deswa1]

Uh the line y=x+6 only passes through A so it can't connect A and B. The line y=5x-6 does. Is that what you mean?

 

[Lemiixem]

Woops, co-ordinates for B is (-2,4)

 

[Timske]

B(2,4) y=x+6

4 = 2 + 6
nope.

 

[Carrotsticks]

The Points A(3,9) and B(2,4) lie on the parabola y=x^2. The line y=x+6 joins A and B. The Point P(p,p^2) is a variable on the parabola below the line. Find the area of the parabolic segment APB, i.e. the area below the line and above the parabola.

Stacking Logs, 2 in the top row, 3 in the second row, and so on, having one more log in each row. If there are 170 logs altogether, how many should there be in the bottom row.

First question:

Correction, B should be (-2,4) not (2,4).



Then using perpendicular distance formula with the line x-y+6=0 and the point P(p,p^2), we acquire the length:



Use the formula for area triangle = 1/2 * base * height to acquire the solution



Interesting note to confirm our answer: We note the two degenerate cases when p=-2 and p=3. When p is equal to the values -2 or 3, it coincides with the point A and B, so the area is 0.

By factorising the expression for the area, we get A = (p-3)(p+2), which is clearly equal to 0 when p is... you guessed it... -2 or 3!

Second question:



Using sum of AP formula, we have:



Solving this quadratic, we acquire two solutions n=-19 and n=18.

However we obviously cannot have a negative number of logs, so n=18.

Therefore, there are 18 logs in the last row.

 

[Lemiixem]

Is the Area
= 1/2 * Root50 * (|p-p^2+6|)/(Root2)
= Root50/2 * [(p-3)(p+2)]/Root5
= Root10[(p-3)(p+2)]/2

 

[Carrotsticks]

Is the Area
= 1/2 * Root50 * (|p-p^2+6|)/(Root2)
= Root50/2 * [(p-3)(p+2)]/Root5
= Root10[(p-3)(p+2)]/2

Oh my I just realised I made an arithmetic error. I accidentally substituted 2sqrt2 instead of 5sqrt2 into my solution.

The answer you acquired is close, but not correct.

But your final answer should be:

 

[Lemiixem]

I have misread the question. The point P is a moving point so they wanted the area inside the parabola that is closed off by the line y=x+6.
Please tell me if my working is correct.
A = [Integral of -2 to 3 of (x+6)dx] - [Integral of -2 to 3 of (x^2)dx]
= [x^2/2 +6x] - [x^3/3]
= [9/2+18-(2-12)] - [27/3+8/3]
= [24]-[35/3]
= 37/3 units^2

 

[Carrotsticks]

Seems correct.

However next time Lemiixem, please make sure you get the question right before you post it.

If we spend our time writing solutions for you, the least you can do is get the question right so we don't waste our time writing solutions to the wrong question.

 

[Timske]

[Integral of -2 to 3 of (x+6)dx] - [Integral of -2 to 3 of (x^2)dx]

[x^2/2 + 6x] - [x^3/3]
[[(9/2 + 18)] - [(2 -12)]] - [(9 + 8/3)]
[[(22.5) - (-10)]] - [35/3]
[[22.5 + 10]] - [35/3]
32.5 - 35/3 = 20.83 units squared

thats what i got.

 

[nightweaver066]

[Integral of -2 to 3 of (x+6)dx] - [Integral of -2 to 3 of (x^2)dx]

[x^2/2 + 6x] - [x^3/3]
[[(9/2 + 18)] - [(2 -12)]] - [(9 + 8/3)]
[[(22.5) - (-10)]] - [35/3]
[[22.5 + 10]] - [35/3]
32.5 - 35/3 = 20.83 units squared

thats what i got.

Second line. Supposed to be -8/3.

 

[Timske]

Second line. Supposed to be -8/3.

[Integral of -2 to 3 of (x^2)dx]
[x^3/3]
= [3^3/3] - [-2^3/3]
= [9] - [-8/3]
= 9 + 8/3

sure?? unless i did something wrong

Format is messy so hard to read :S

 

[Carrotsticks]

 

[nightweaver066]

[Integral of -2 to 3 of (x^2)dx]
[x^3/3]
= [3^3/3] - [-2^3/3]
= [9] - [-8/3]
= 9 + 8/3

sure?? unless i did something wrong

Format is messy so hard to read :S

mm yeah sorry about that, forgot there was another negative.