Partial Fractions and Parametric Equations

[Timiop2008]

Hi
Can anybody help me with these 3 problems?:

1)
Express (3x-1)/(x+3)^2 in the form A/(x+3) + B/((X+3)^2) where A and B are constants.

2)
A curve C has parametric equations:
x=cost and y=2-cos2t (between 0 and pi)
a)prove this can be expressed as the cartesian equation y=3-2x^2
b) sketch the graph of C
c) what is the t value for the maximum point on the graph

3)
The coefficients of x^2 and x in the expansion of a(1-bx)^-3 are 72 and 18 respectively.
Find a and b, given they are above 0.

I really need help with these 3. Thank You

 

[Mark44]

What have you tried? You need to show that you have made an effort at working these before you get any help.

 

[Timiop2008]

This is my attempt at question (1)

(3x-1)/(x+3)^2 = A/(x+3) + B/(x+3)^2
(3x-1)/(x+3)^2 = A(x+3)/(x+3)(x+3) + B/(x+3)^2
(3x-1)/(x+3)^2 = Ax+3A/(x+3)^2 + B/(x+3)^2
now cancel all the denominators to get
3x-1=Ax+3A+B
I don't know where to go from here

 

[Timiop2008]

For the Parametric Equations Question, can you do part (a) by thinking of x=cost as cos=k and y=2-cos2t as y=2-k2t where k is some constant?

Can you sketch the graph by calculating a table of values of x,y and t from -3 to 3?

 

[ideasrule]

For the Parametric Equations Question, can you do part (a) by thinking of x=cost as cos=k and y=2-cos2t as y=2-k2t where k is some constant?

What's the meaning of k2t? cos(2t) does not equal to 2cos(t), if that was what you were thinking. However, cos(2t) does equal 2cos^2(t) - 1, and you know that x=cost.

Can you sketch the graph by calculating a table of values of x,y and t from -3 to 3?

Sure, but the question tells you that y=3-2x^2 is equivalent to that pair of parametric equations. Isn't it much easier to sketch y=3-2x^2?

 

[singular]

This is my attempt at question (1)

(3x-1)/(x+3)^2 = A/(x+3) + B/(x+3)^2
(3x-1)/(x+3)^2 = A(x+3)/(x+3)(x+3) + B/(x+3)^2
(3x-1)/(x+3)^2 = Ax+3A/(x+3)^2 + B/(x+3)^2
now cancel all the denominators to get
3x-1=Ax+3A+B
I don't know where to go from here

You are so close. Now, you want to pull the terms with x out of the equation and make them into a new equation, leaving you with two separate equations. It should look like this:

$$3x=Ax$$

$$-1=3A+B$$

Now, solve.

 

[singular]

For the Parametric Equations Question, can you do part (a) by thinking of x=cost as cos=k and y=2-cos2t as y=2-k2t where k is some constant?

Can you sketch the graph by calculating a table of values of x,y and t from -3 to 3?

No, cosine is a function of t and can't be thought of as a constant, k.

Try to tackle the parametric equations by starting with the complicated part. The idea when dealing with trig functions in parametric equations is to get the relations x(t) and y(t) to contain similar terms so that you can eventually substitute one into the other forming one equation that eventually shouldn't contain t. The equation is

$$x=cost$$

$$y=2-cos2t$$

The x equation is very simple, just like we want it, so leave it alone. Look at the y equation. Can you think of a trigonometric identity that can help us get rid of the cosine double angle term and replace it with sines and/or cosines?

 

[singular]

3) The coefficients of x^2 and x in the expansion of a(1-bx)^-3 are 72 and 18 respectively. Find a and b, given they are above 0.

I'm not really sure what you mean by expansion. Are you wanting a partial fraction expansion? Can you elaborate a little bit and show some of your work on this problem?

 

[Mark44]

For 3, I'm reasonably sure that the OP needs to expand a(1 + bx)^(-3) as a binomial series.
$$(1 + x)^n~=~1~+~nx~+~\frac{n(n - 1)x^2}{2!}~+~\frac{n(n -1)(n - 2)x^3}{3!}~ +~ ...$$

More information here.