Integration (2 Viewers)

[FDownes]

Can anyone give me a bit of help with this integration question? The first part I can figure out but I don't know how to do the second. It asks;

The points A (3, 9) and B (-2, 4) lie on the parabola y = x². The line y = x + 6 joins A and B. The point P (p, p²) is a variable point on the parbola below the line.

a) Find the are of the parabolic segment APB, i.e. the area below the line and above the parabola.

= 125/6 units²

b) Show that the greatest possible area of the triangle APB is three-quaters of the area of the parabolic segment APB.

 

[lyounamu]

Can anyone give me a bit of help with this integration question? The first part I can figure out but I don't know how to do the second. It asks;

The points A (3, 9) and B (-2, 4) lie on the parabola y = x². The line y = x + 6 joins A and B. The point P (p, p²) is a variable point on the parbola below the line.

a) Find the are of the parabolic segment APB, i.e. the area below the line and above the parabola.

= 125/6 units²

b) Show that the greatest possible area of the triangle APB is three-quaters of the area of the parabolic segment APB.

Use the simultaneous equation to find the common points:
y=x^2
y=x+6
Therefore, x^2 = x+6
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or x= -2.

Therefore, the points of intersection are x = -2 and x=3.

y=x^2
integral A = x^3/3
y = x + 6
integral B = x^2/2 + 6x

A = integral B - integral A (since y = x+6 is situated above y=x^2)
= x^2/2 + 6x - x^3/3

Now substitute, x=3 and x-2 separetely, (like you do the integrals)
A = 3^2/2 + 6.3 - 3^3/3 - ((-2)^2/2 + 6.-2 - (-2)^3/3)
= 125/6

2nd question:

Using the distance formula, I can work out the distance AB to be root 50.

And then, I used the perpendicular distance formula.

So, I got PD = Absolute value of (p-p^2 + 6)/root2
Find the derivative, I got 1/root 2 - 2p/root2 = 0 (you want to work out the maximum PD)
so p = 1/2

Substitute that back in and you will find that the PD = 4.41941738...

A = PD . AB divided by 2 = 125/8 = 3.4 of 125/6

Mission Completed

 

[FDownes]

Thanks. You didn't need to solve the first part though, I had that covered.

 

[lyounamu]

Thanks. You didn't need to solve the first part though, I had that covered.

Yeah, I realised that I didn't need to do that question when I finished the q1, lol.

But I had to do Q1 anyway. It was quite a challenging question. Thanks for the great question.

 

[FDownes]

No problem, have another ;

a) Prove that the line y = x - 4 is a tangent to the parabola y = x² - 5x + 5.

a) Let P be the point where the line y = x - 4 touches the parabola y - x² - 5x + 5. Show that the normal to the parabola at P is y = - x + 2.

The first part has me stumped. Should I just guess that the point of contact is (4, 0) and that the gradient is 1 (assuming the equation is in the form
y - y₁ = m(x - x₁)), or is there a more correct way of approaching this?

 

[Slidey]

a) x-4=x^2-5x+5
0=x^2-6x+9=(x-3)^2
Only one (repeated) root, therefore a tangent.

a number 2)
from above, point of contact at (3,-1)
y'=2x-5
y'(3)=2*3-5=1
gradient of the tangent is 1 (but we already knew this)
normal gradient = -1 (negative inverse of tangent gradient)
normal cuts at (3,-1), so
y=-x+p
-1=-3+p
p=2
Normal: y=-x+2

 

[lyounamu]

No problem, have another ;

a) Prove that the line y = x - 4 is a tangent to the parabola y = x² - 5x + 5.

a) Let P be the point where the line y = x - 4 touches the parabola y - x² - 5x + 5. Show that the normal to the parabola at P is y = - x + 2.

The first part has me stumped. Should I just guess that the point of contact is (4, 0) and that the gradient is 1 (assuming the equation is in the form
y - y₁ = m(x - x₁)), or is there a more correct way of approaching this?

Since slidey kindly did the first part I shall move on to the second one.

dy/dx = m1 = 1
and the point is (3,-1)

dy/dx = m2 = -1 (the gradient of normal)

so the equation is

y + 1 = -1(x-3)
y = -x + 2

 

[FDownes]

Thanks, but there's a third part to the question I forgot to ask. It should be pretty straight-forward, but my working doesn't seem to be producing the right answer.

c) Find the area of the region enclosed between the parabola and the line y = -x + 2.

 

[lolokay]

Thanks, but there's a third part to the question I forgot to ask. It should be pretty straight-forward, but my working doesn't seem to be producing the right answer.

c) Find the area of the region enclosed between the parabola and the line y = -x + 2.

is it 1 + 1/3?
which is (I think) the integral from 1->3 of [(x^2 - 5x + 5) - (-x + 2)]

 

[FDownes]

Okay, I've gone through this question several times and just can't seem to get the right answer. Can someone point out what I'm doing wrong?

= Intercepts = x = 1 and x = 3
= ₁³∫(x² - 5x + 5) - ₁³∫(-x + 2)
= [(x³)/3 - (5x²)/2 + 5x]₁³ - [(-x²)/2 + 2x]₁³
= [((3)³)/3 - (5(3)²)/2 + 5(3) - ((1)³)/3 + (5(1)²)/2 - 5(1)] - [(-(3)²)/2 + 2(3) + (-(1)²)/2 - 2(1)]

 

[lyounamu]

Okay, I've gone through this question several times and just can't seem to get the right answer. Can someone point out what I'm doing wrong?

= Intercepts = x = 1 and x = 3
= ₁³∫(x² - 5x + 5) - ₁³∫(-x + 2)
= [(x³)/3 - (5x²)/2 + 5x]₁³ - [(-x²)/2 + 2x]₁³
= [((3)³)/3 - (5(3)²)/2 + 5(3)]₁³ - x³)/3 + (5x²)/2 + 5x]₁³ [(-x²)/2 + 2x]₁³

By the way, are you saying that 1+1/3 is not the answer?

 

[FDownes]

1 and 1/3 is the answer, I just can't seem to get it through my own working. Could you show me yours so I can see where I'm going wrong?

 

[lolokay]

you just find "the integral from 1->3 of [(x^2 - 5x + 5) - (-x + 2)]"
ie, ∫₁³ [x^2 - 4x + 3]dx
giving a primitive of 1/3x^3 - 2x^2 + 3x (+C)
so just sub in for 1 and 3 and you get 1+1/3

 

[FDownes]

Thanks guys, I'm gonna put this down to calculator troubles. I moved some of the brackets around and finally got the right answer. I'll have to be more careful in the future.

 

[FDownes]

ARG! Stuck on yet another question. I'm sure it's just because I'm entering the equation in to my calculator incorrectly, but I just can't get it. Can someone help me out? The question asks;

Find the volume of the solid of revolution formed by rotating the curve y = x - 1/x about the x-axis between x = 1 and x = 4.

= y² = (x - 1/x)²
= y² = x² - 2 + 1/x²
= pi ₁⁴∫(x² - 2 + 1/x²)
= pi [x² - 2 + 1/x²]₁⁴
= pi [((4)³ - 1/(4) - 2(4)) - (1³ - 1/(1) - 2(1))]
= pi [(64 - 1/4 - 8) - (1 - 1 - 2)]

Where have I gone wrong?

 

[lyounamu]

ARG! Stuck on yet another question. I'm sure it's just because I'm entering the equation in to my calculator incorrectly, but I just can't get it. Can someone help me out? The question asks;

Find the volume of the solid of revolution formed by rotating the curve y = x - 1/x about the x-axis between x = 1 and x = 4.

= y² = (x - 1/x)²
= y² = x² - 2 + 1/x²
= pi ₁⁴∫(x² - 2 + 1/x²)
= pi [x² - 2 + 1/x²]₁⁴
= pi [((4)³ - 1/(4) - 2(4)) - (1³ - 1/(1) - 2(1))]
= pi [(64 - 1/4 - 8) - (1 - 1 - 2)]

Where have I gone wrong?

You didn't integrate it properly. (3-5 line)
From the third line V = pi . 4 -> 1 integral of (x^2 - 2 + 1/x^2)
= pi . 4->1 (x^3/3 - 2x - 1/x)
= pi . (4^3/3 - 2 . 4 - 1/4 - (1^3/3 - 2 . 1 - 1/1))
= 63/4 pi

 

[FDownes]

Thanks, I can't believe I didn't see such a stupid mistake... Got time for one more? I'm stuck on the first part of this question;

a) Find the points of intersection of the curve y = 2 + √(4x) with the y-axis and the line y = 4.

The answers for this one should be y = 2 (y-axis) and x = 1 (y = 4), but I'm not sure how to get them.

b) The area enclosed by the curve y = 2 + √(4x), the line y = 4 and the y-axis is rotated about the y-axis. Find the volume of the solid of revolution formed.

 

[lyounamu]

Thanks, I can't believe I didn't see such a stupid mistake... Got time for one more? I'm stuck on the first part of this question;

a) Find the points of intersection of the curve y = 2 + √

incomplete question?

y = 2 + square root (4x)

By using simultaneous equation, we can find the intersecting points.

Substitute x = 0 into y = 2 + square root of (4x)
y = 2 + square root of (4 . 0 ) = 2
You substitute x = 0 here because x = 0 is the equation for the y-axis

Now, y = 2 + square root (4x) and y = 4 (another equation)
4 = 2 + square root of (4x)
2 = square root of 4x
so x = 1

So the intersecting points are x=1 (y=4) and y=2

Second Question:

y = 2 + square root of (4x)
square root of (4x) = y -2
4x = y^2 - 4y + 4
x = y^2/4 - y + 1
x^2 = y^4 + y^2 +1 + y^2/4 - y - y^3/4 - y + y^2/4 - y^3/4
= y^4 - y^3/2 +3y^2/2 -2y +1
V = pi 4->2 I(y^4/16 + y^3/2 - 3y^2/2 - 2y + 1)
= pi 4->2(y^5/80 - y^4/8 + y^3/2 - y^2 + y)
= pi(4/5 - 2/5) = 2/5pi

 

[Doctor Jolly]

All your working out scares me .. I've just started on Integration

 

[FDownes]

D'oh! For some reason I was substituting y = 0 instead of x = 0. Thanks.

EDIT: I think you may have made a mistake. Did you square x before integrating it?