How do I determine how much does a variable "vary" by looking at other variables it depends on?

53 Views Asked by At

I'm aware this is pretty basic, I'm having trouble with directly and inversely proportional variations.

I have this equation

$$ A = \frac{BC^2\sqrt{D}}{E.F^3} $$

So, as I understand it, if $B$ gets bigger, $A$ should also get bigger as they are directly proportional because they follow the $y=kx$ pattern.

Also if $E$ gets bigger $A$ should get smaller as they are inversely proportional because they follow the $y=\frac{k}{x}$ pattern.

So far so good, however now I'm asked to know how much A will vary depending of what I do to the other unknowns.

For example:

What would happen to $A$ if $B$ was doubled?

I understand $A$ would grow porportionally, as stated above, but It also asks for the specific amount it will grow, for example, if $B$ was doubled then $A$ will also double.

And I cannot comprehend how to do this, or what if $E$ were halved, I know $A$ would grow, but how much?!

2

There are 2 best solutions below

4
On BEST ANSWER

@Nilknarf has dealt with the example for $B$ in the other answer. I thought I could add to what happens with $F$.

Suppose $F$ is doubled. Then the step to take is to replace the place where $F$ was with $2F$. So you end up with $$\text{something}=\frac{BC^2\sqrt D}{E(2F)^3}=\frac{BC^2\sqrt D}{8EF^3}=\frac18\frac{BC^2\sqrt D}{EF^3}=\frac18 A$$So $A$ would decrease by a factor of $8$.

This method is better in my opinion, since there is no "guesswork" involved (you said you would have divided both sides by $2$ instead).


B doubled:

$$\frac{(2B)C^2\sqrt D}{EF^3}=2\times\frac{BC^2\sqrt D}{EF^3}=2A$$So $A$ doubles.


C doubled: $$\frac{B(2C)^2\sqrt D}{EF^3}=\frac{4BC^2\sqrt D}{EF^3}=4A$$ So $A$ quadruples.


D doubled: $$\frac{BC^2\sqrt {2D}}{EF^3}=\sqrt2\times\frac{BC^2\sqrt D}{EF^3}=\sqrt2\times A$$So $A$ increases by factor of $\sqrt 2$.


E doubled: $$\frac{BC^2\sqrt D}{(2E)F^3}=\frac12\frac{BC^2\sqrt D}{EF^3}=\frac12 A$$So $A$ halves.

2
On

It is given that $$A=\frac{BC^2\sqrt D}{EF^3}$$ By multiplying both sides by $2$, it follows that $$2A=\frac{2BC^2\sqrt D}{EF^3}$$ or $$(2A)=\frac{(2B)C^2\sqrt D}{EF^3}$$ You can see now that replacing $A$ by $2A$ and $B$ by $2B$ results in the equation remaining solved. Thus, when $B$ is doubled, or replaced by $2B$, $A$ must be replaced by $2A$, or doubled.

Here's another example. Suppose that $F$ is doubled. By dividing both sides by $8$, it follows that $$A/8=\frac{BC^2\sqrt D}{8EF^3}$$ or $$(A/8)=\frac{BC^2\sqrt D}{E(2F)^3}$$ Thus, since $A\to A/8$ and $F\to 2F$ satisfy the equation, it follows that if $F$ is doubled, then $A$ must by reduced by a factor of $8$.

Here's yet another example. Suppose that $D$ is doubled. Then, by multiplying both sides by $\sqrt 2$, we have $$A\sqrt 2=\frac{\sqrt{2}BC^2\sqrt D}{EF^3}$$ or $$(A\sqrt 2)=\frac{BC^2\sqrt{2D}}{EF^3}$$ So when $D$ is doubled, $A$ is increased by a factor of $\sqrt 2$.