Partial Fractions and Parametric Equations

In summary, the conversation revolves around three problems relating to finding A and B constants in a given equation, expressing a parametric equation as a cartesian equation and finding the t value for the maximum point on a graph. In order to solve these problems, the individual asking for help has attempted to work through the first problem by equating terms and canceling denominators. For the second question, they have suggested using a trigonometric identity to simplify the equation. For the third problem, they are unsure about the concept of "expansion" and have asked for more information.
  • #1
Timiop2008
31
0
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
 
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  • #2
What have you tried? You need to show that you have made an effort at working these before you get any help.
 
  • #3
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
 
  • #4
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?
 
  • #5
Timiop2008 said:
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?
 
  • #6
Timiop2008 said:
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:

[tex]3x=Ax[/tex]

[tex]-1=3A+B[/tex]

Now, solve.
 
  • #7
Timiop2008 said:
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

[tex]x=cost[/tex]

[tex]y=2-cos2t[/tex]

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?
 
  • #8
Timiop2008 said:
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?
 
  • #9
For 3, I'm reasonably sure that the OP needs to expand a(1 + bx)^(-3) as a binomial series.
[tex](1 + x)^n~=~1~+~nx~+~\frac{n(n - 1)x^2}{2!}~+~\frac{n(n -1)(n - 2)x^3}{3!}~ +~ ...[/tex]

More information here.
 

Related to Partial Fractions and Parametric Equations

What are partial fractions?

Partial fractions are an algebraic method used to break down a complex rational expression into simpler fractions. This allows for easier integration and simplification of equations.

How do you solve partial fractions?

To solve partial fractions, you must first factor the denominator of the rational expression. Then, use the method of undetermined coefficients to determine the numerator of each fraction. Finally, equate the coefficients of the like terms and solve for the unknown variables.

What are parametric equations?

Parametric equations are a set of equations that express a set of quantities as functions of one or more independent variables, known as parameters. They are often used to describe the motion of objects in space or in other mathematical models.

How do you graph parametric equations?

To graph parametric equations, you can plot points by plugging in different values for the parameter. Another method is to eliminate the parameter by solving for one variable in terms of the other and then graphing the resulting equation. You can also use a graphing calculator or computer program to graph parametric equations.

What are the advantages of using parametric equations?

Parametric equations allow for the representation of complex curves and shapes that cannot be easily described by traditional equations. They also allow for the analysis of motion and behavior of objects in space. Additionally, parametric equations can be used to solve problems in many fields, including physics, engineering, and economics.

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