# Parametric Intergration Question

P3 Jan 02 Question 8

A table top in the shape of a parallelogram, is made from two types of wood. The area inside the ellipse is made from one type of wood, and the surrounding area is made from a second type of wood.

The ellipse has parametric equation

x = 5 cos(a) y = 4 sin(a)

The parallelogram consists of four line segments, which are tangents to the ellipse at the points where (a) = (b), (a) = -(b), (a) = pi + (b) and (a) = pi-(b)*

ii) Find by intergration the area enclosed by the ellipse
iii) Hence show that the area enclosed between the ellipse and the parallelogram is:
80/sin 2(b) - 20pi

iii) Find the value of (b) for which the areas of the two types of wood are equal.

Sorry, i know this is a long question but i really want to know how i get the marks for this question incase the same sort of question comes up in the actual p3 exam on the 23rd may!

Thanks

*(a) and (b) are just angles
(i)
x = 5cost
y = 4sint

dx = -5sint dt

Ae = Area of ellipse

Consider the area in the +ve quadrant only

¼Ae = &#8747; y dx = &#8747; 4sint.(-5sint) dt [pi/2, 0]
¼Ae = 20 &#8747; sin²t dt [0, pi/2]
¼Ae = 10 &#8747; (1 - cos2t) dt [0, pi/2]
¼Ae = 10[t - ½sin2t] {0, pi/2}
¼Ae = 10(pi/2) = 5pi
Ae = 20pi
=======

(ii)
dy/dt = 4cost
dx/dt = -5sint

dy/dx = dy/dt / dx/dt = 4cost / -5sint = -(4/5)cott

at t = b, dy/dx = -(4/5)cotb
at t = -b, dy/dx = (4/5)cotb
at t = pi+b, dy/dx = (4/5)tanb
at t = -(pi+b), dy/dx = -(4/5)tanb

Line/tangent in +ve quadrant is l1.
l1: y = mx + c, where m = -(4/5)cotb

at t = b
x = 5cosb
y = 4sinb

This point is on both the ellipse and the tangent to it (l1).
:. (5cosb, 4sinb) satisfies the eqn of the line l1.

4sinb = -(4/5)cotb.5cosb + c
4sinb = -4cos²b/sinb + c
c = 4sin²b/sinb + 4cos²b/sinb
c = 4(sin²b + cos²b)/sinb
c = 4cosecb
=========

L1 intersects the x-axis when y = 0
0 = -(4/5)cotb.x + 4cosecb
cotb.x = 5cosecb
x = 5secb
========

Using c and x from above, we get
BC = 2c = 8cosecb
AO = x = 5secb

Area of triangle = A = ½BC.AO
A = ½.8cosecb.5secb
A = 20/(sinb.cosb)
A = 40/sin2b
=========
Ap = area of parallelogram
Ap = 2A
Ap = 80/sin2b
===========

Area enclosed between parallelogram and ellipse = Ap - Ae
= 80/sin2b - 20pi
==============

(iii)
Ap = Ae
80/sin2b - 20pi = 20pi
80/sin2b = 40pi
sin2b = 2/pi
2b = 39.54
b = 19.77 deg
==========
Edit: corrected the (order of the) limits in part(i)
:O! I wouldnt be able to do that in the exam!.......
Fermat
(i)
x = 5cost
y = 4sint

dx = -5sint dt

Ae = Area of ellipse

Consider the area in the +ve quadrant only

¼Ae = &#8747; y dx = &#8747; 4sint.(-5sint) dt [pi/2,0]
¼Ae = 20 &#8747; sin²t dt [pi/2,0]
¼Ae = 10 &#8747; (1 - cos2t) dt [pi/2,0]
¼Ae = 10[t - ½sin2t] {pi/2, 0}
¼Ae = 10(pi/2) = 5pi
Ae = 20pi
=======

(ii)
dy/dt = 4cost
dx/dt = -5sint

dy/dx = dy/dt / dx/dt = 4cost / -5sint = -(4/5)cott

at t = b, dy/dx = -(4/5)cotb
at t = -b, dy/dx = (4/5)cotb
at t = pi+b, dy/dx = (4/5)tanb
at t = -(pi+b), dy/dx = -(4/5)tanb

Line/tangent in +ve quadrant is l1.
l1: y = mx + c, where m = -(4/5)cotb

at t = b
x = 5cosb
y = 4sinb

This point is on both the ellipse and the tangent to it (l1).
:. (5cosb, 4sinb) satisfies the eqn of the line l1.

4sinb = -(4/5)cotb.5cosb + c
4sinb = -4cos²b/sinb + c
c = 4sin²b/sinb + 4cos²b/sinb
c = 4(sin²b + cos²b)/sinb
c = 4cosecb
=========

L1 intersects the x-axis when y = 0
0 = -(4/5)cotb.x + 4cosecb
cotb.x = 5cosecb
x = 5secb
========

Using c and x from above, we get
BC = 2c = 8cosecb
AO = x = 5secb

Area of triangle = A = ½BC.AO
A = ½.8cosecb.5secb
A = 20/(sinb.cosb)
A = 40/sin2b
=========
Ap = area of parallelogram
Ap = 2A
Ap = 80/sin2b
===========

Area enclosed between parallelogram and ellipse = Ap - Ae
= 80/sin2b - 20pi
==============

(iii)
Ap = Ae
80/sin2b - 20pi = 20pi
80/sin2b = 40pi
sin2b = 2/pi
2b = 39.54
b = 19.77 deg
==========

Thanks, i appreciate the effort
Fermat why did you use the limit as pi/2? Is it always pi/2 in the first quadrant?

Otherwise I thought the line cuts x axis when y=0, so 0=sint, .'. t =0, pi.
It's true that the lower limit is t=0 (since that's where y=0), but the upper limit is when x=0 (see Fermat's attachement).
x=0 => cost = 0
t = pi/2
Fermat

(ii)
dy/dt = 4cost
dx/dt = -5sint

dy/dx = dy/dt / dx/dt = 4cost / -5sint = -(4/5)cott

at t = b, dy/dx = -(4/5)cotb
at t = -b, dy/dx = (4/5)cotb
at t = pi+b, dy/dx = (4/5)tanb
at t = -(pi+b), dy/dx = -(4/5)tanb

Line/tangent in +ve quadrant is l1.
l1: y = mx + c, where m = -(4/5)cotb

at t = b
x = 5cosb
y = 4sinb

This point is on both the ellipse and the tangent to it (l1).
:. (5cosb, 4sinb) satisfies the eqn of the line l1.

4sinb = -(4/5)cotb.5cosb + c
4sinb = -4cos²b/sinb + c
c = 4sin²b/sinb + 4cos²b/sinb
c = 4(sin²b + cos²b)/sinb
c = 4cosecb
=========

L1 intersects the x-axis when y = 0
0 = -(4/5)cotb.x + 4cosecb
cotb.x = 5cosecb
x = 5secb
========

Using c and x from above, we get
BC = 2c = 8cosecb
AO = x = 5secb

Area of triangle = A = ½BC.AO
A = ½.8cosecb.5secb
A = 20/(sinb.cosb)
A = 40/sin2b
=========
Ap = area of parallelogram
Ap = 2A
Ap = 80/sin2b
===========

Area enclosed between parallelogram and ellipse = Ap - Ae
= 80/sin2b - 20pi
==========

Soz, am REALLY confused to what you've done in the second part here... Can someone explain. It's only a for 4 marks!!!
mockel
It's true that the lower limit is t=0 (since that's where y=0), but the upper limit is when x=0 (see Fermat's attachement).
x=0 => cost = 0
t = pi/2

I'm still unsure why it's when x=0 - is it because as you go from 0 to pi/2, the line curves around from the x axis towards the y axis?
Hmm. Maybe a better way to think about it is to consider the variable 't'.
t=0, y = 0
t = pi/2 , y = 4

So you should integrate from 0 to pi/2.
mockel
Hmm. Maybe a better way to think about it is to consider the variable 't'.
t=0, y = 0
t = pi/2 , y = 4

So you should integrate from 0 to pi/2.

Soz, fermat... i made u do all that excess work for nothing

In the first part of the question we worked out that the tangent to the ellipse can be written in the form (but i forgot to write it down)

5y sin (a) + 4x cos (a) = 20

so you can just put x=0 and get the y intercept and also put y=0 and get the x-intercept.

Will give you:

y = 4 / sin (a)
and x = 5 / cos (a)

I then got the area of the parallelogram to be
4 * 1/2 * 5 / cos (a) * 4 / sin (a)
= 40 / sin(a)cos(a)
which simplifies to 80 / sin (2a)

Minus 20pi, that you worked out from part (a) and you get your answer

I think my method is slightly easier but i think you just confused me since you had to work out an equation of the tangent which i didnt give you.
endeavour
Fermat why did you use the limit as pi/2? Is it always pi/2 in the first quadrant?

Otherwise I thought the line cuts x axis when y=0, so 0=sint, .'. t =0, pi.

When you're finding the area under a curve, you always work from left to right. i.e. from the lower x-coordinate to the higher x-coordinate.
The area I was considering was the ¼-area of the ellipse in the +ve quadrant only.
The cartesian form of the eqn is x²/25 + y²/16 = 1.
If we were using that form, we would write the area as.

¼A = &#8747; y dx

and put the limits as x1 = 0, x2 = 5

But x1 (the lower limit) is given by x = 5cost
i.e. 0 = 5cost => t = pi/2

Similarly, x2 (the higher limit) is given by x = 5cost
i.e. 5 = 5cost => t = 0
Revenged
Soz, fermat... i made u do all that excess work for nothing

In the first part of the question we worked out that the tangent to the ellipse can be written in the form (but i forgot to write it down)

5y sin (a) + 4x cos (a) = 20

so you can just put x=0 and get the y intercept and also put y=0 and get the x-intercept.

Will give you:

y = 4 / sin (a)
and x = 5 / cos (a)

I then got the area of the parallelogram to be
4 * 1/2 * 5 / cos (a) * 4 / sin (a)
= 40 / sin(a)cos(a)
which simplifies to 80 / sin (2a)

Minus 20pi, that you worked out from part (a) and you get your answer

I think my method is slightly easier but i think you just confused me since you had to work out an equation of the tangent which i didnt give you.

No problem revenged.
I'm glad to find out that you didn't have to do all that work for just 4 marks. It's a lot of work to do in an exam.
The stuff I did fills up a lot of space, but hopefully anyone reading it will be able to follow my method and understand it more clearly. Hence all the individual steps. In an exam, If you understand the method, you can just write down only the relevant lines and do intermediate steps in your head - saves time !
Fermat
When you're finding the area under a curve, you always work from left to right. i.e. from the lower x-coordinate to the higher x-coordinate.
The area I was considering was the ¼-area of the ellipse in the +ve quadrant only.
The cartesian form of the eqn is x²/25 + y²/16 = 1.
If we were using that form, we would write the area as.

¼A = &#8747; y dx

and put the limits as x1 = 0, x2 = 5

But x1 (the lower limit) is given by x = 5cost
i.e. 0 = 5cost => t = pi/2

Similarly, x2 (the higher limit) is given by x = 5cost
i.e. 5 = 5cost => t = 0

Sorry to dwel on this question Fermat, but another question

if the lower limit is when x1=0, and t=pi/2; and x2=5 and t=0
does this mean the you actually integrate wrt t from 0 to 5 ie
0pi/2&#8747;-20sint.sint) dt
which gives you a positive area?

(or alternatively, pi/20&#8747;(20sint.sint) dt )

Otherwise, if you integrate from pi/2 to 0 you get a negative area.
Yes, you integrate from 0 to 5 (wrt x) which is the same as from pi/2 to 0 (wrt t).
Changing the order of the limits changes the sign of the result.
Which I forgot to do (change the order) when writing out my original post (corrected now).