Finding a formula of a math problem

[laker_gurl3]

i don't know what to do for this question, what does it mean by A = 1 when x = 1 and y = 0 ? do i just sub it in or what? is this some sort of equation in an equation thingy?


The quantity A is proportional to X^2 and inversely proportional to (y-a)^3
Find a formula for A given that A = 1 when x = 1 and y = 0

 

[Gokul43201]

Please post such questions in the Coursework Forum, not here !

First write A as a function of (x,y) using arbitrary constants. For instance if told that V is proportional to I, you can write the equation V=kI, where k is some constant.

Next you are told that when you substitute the given values of x and y, you get a certain value of A. This helps determine the value of such a constant. In my example, if told that V = 5 when I = 2, I have 5 = 2k, which tells me that k=2.5. So my Original equation can be becomes V = 2.5I

 

[laker_gurl3]

okay after following ur example, i have A=x^2, it's the inversely proportional part that i can't understand...

 

[Gokul43201]

If you are told that A is inversely proportional to some z, then that is the same as saying that A is proportional to 1/z. Or A = k/z, where k is some constant.

 

[HallsofIvy]

"The quantity A is proportional to X^2 and inversely proportional to (y-a)^3"

A is "proportional" to B means A is just a constant times B: A= kB.

A is "inversely proportional" to B means A is a constant time 1/B or is a constant divided by B: A= k/B.

A is proportional to x² and inversely proportional to (y-a)³ gives both: A= kx²/(y-a)³.

" given that A = 1 when x = 1 and y = 0" means that if you replace A in the formula with 1, x with 1, and y with 0, it is still true. That is:
1= k(1²)/(0-a)³= -k/a³. You can use that to find k: k= -a³ (a was part of the original information but we are not given a specific value for a so all we can do is leave it in there.)
Putting that value for k back into the original formula,
A= -a³x²/(y-a)³.