More Integration =/. (1 Viewer)

[independantz]

These questions are from page 422, Year 11 Cambridge book, Ex 11F, Q8 and 9
Q8)
a. On the same number plane sketch the graphs of the functions y=x^2 and x=y^2. clearly indicating points of intersection (this i can do)
b. Find the area bound by the curves.

Q9) Consider the function x^2=8y. Tangents are drawn at the points A(4,2) and B (-4,2) and intersect on the y-axis. Find the area bounded by the curve and the tangents.

Thanks in Advance

 

[undalay]

Q8

Find the points of intersection of the two curves.
Let them be 0 and @

Find the area under the y= x^2 between 0 and @
Double this value and subtract it from x times y of @ (e.g. the square where 0 is one corner, and @ is the opposite corner) to get the area between the two curves. Refer to your diagram and it will make more sense.

edit
Q9.
x^2 = 8y is an even function.

Differentiate and plug in x = 4, to get the gradient of the left tangent.
Get the equation of the left tangent.

Find the area between this line and the circle and the x axis. Double this area for the other half.

 

[sle3pe3bumz]

Q8.

note: int. = integral sign.

well you have the two curves and intercepts. @ 0 & 1.

int. (with 1 on top and 0 on bottom) x^1/2 - x^2 dx
= [ 2/3 x^3/2 - 1/3 x^3 ](1 on top, 0 on bottom)
sub in 1 and 0. and you get ..
= (2/3 - 1/3) - 0
= 1/3 u^2

 

[Slidey]

These questions are from page 422, Year 11 Cambridge book, Ex 11F, Q8 and 9
Q8)
a. On the same number plane sketch the graphs of the functions y=x^2 and x=y^2. clearly indicating points of intersection (this i can do)
b. Find the area bound by the curves.

Q9) Consider the function x^2=8y. Tangents are drawn at the points A(4,2) and B (-4,2) and intersect on the y-axis. Find the area bounded by the curve and the tangents.

Thanks in Advance

Man, it's been a while.

Set up simultaneous equations:

{ y=x^2
{ y^2=x (it's a square root function, inverse of y=x^2)
y=x^2
y=+sqrt(x) (now, take positive sqrt by looking at the graph)
0=x^2-sqrt(x)
x values of the pts of intersection are x=0, 1

The integration will thus occur between x=0 and x=1, and we'll integrate y w.r.t. x since that's what you are used to:

y=sqrt(x)-x^2

Int y dx from 0 to 1 =
Int sqrt(x)-x^2 dx from 0 to 1 =
[(2/3)*x^(3/2)-x^3/3] from 0 to 1
= 2/3-1/3 = 1/3 units squared


Next Question
Find the equation of the tangents:
y=x^2/8
dy/dx = x/4
Gradient at x=4 is 1, at x=-4 is -1
Two tangents:
(1) y=x+b -> sub pt (4,2) in to find b
2=4+b so y=x-2
(2) y=-x+b -> sub pt (-4,2)
So y=-x-2

Since this function is symmetric, we will only integrate from x=0 to x=4, then double this.

This question is confusing. Are the axes included as bounds? Hmm anyway I'll assume you want the area above and below the x-axis - from y=-2 to y=2:

Intersect pts are (4,2) and (-4,2)
y=x^2/8
y=x-2

Function to integrate: y=x^2/8 - x +2, from 0 to 4
Ans: [x^3/24 - x^2/2 + 2x] from 0 to 4
= 8/3 units squared
Multiply by two due to symmetry:
Final answer is 16/3 units squared

independantz: Sorry if you were reading the deleted post, as I made small errors in both questions. Fixed now.
Undalay: It's not a circle, it's a parabola.