Which is the quadratic variation equation for the relationship?

y varies directly with x^2 and y = 72 when x = 6.

A.
y = 2x^2

B.
y = 4x^2

C.
y = 12x

D.
y = x^2 + 25

Question

Which is the quadratic variation equation for the relationship?

y varies directly with x^2 and y = 72 when x = 6.
 
 
 A.
y = 2x^2
 
 B.
y = 4x^2
 
 C.
y = 12x
 
 D.
y = x^2 + 25

keeli_grace · Answer

@ParthKohli , could you please help me? :)

parthkohli · Answer

$$y = kx^2$$The above is what a quadratic variation looks like.  $k$ is the constant of variation.  Let's try to find out the constant of variation.

It is given that when $y =72$, $x = 6$.  Let's plug that in to our equation.$$y = kx^2 \implies 72 = k(6)^2$$

keeli_grace · Answer

I'm sorry, I'm absolutely terrible at math. Do you think you could give me a two-second math course? :/

parthkohli · Answer

Well, I'm sorry about that.  Do you know what a quadratic variation really is?

keeli_grace · Answer

Um, no.. :(

anonymous · Answer

@ParthKohli will help :)

keeli_grace · Answer

But I am a quick learner, so I'm sure I could figure it out fairly quickly if you helped me to understand! :)

parthkohli · Answer

A "variation" is a relation between two quantities.  It's like a pattern in how one quantity is, when you play around with another.

If you look at a circle, the radius of the circle and the area are related.  There is an equation between the two:$$\rm area = \pi \times (radius)^2$$If you make a new circle with twice the radius than one circle, then the new circle's area would not be twice the original circle's area: it would be FOUR times the first circle's area!  If you make the radius thrice, the area will become NINE times the original.

This is an example of a quadratic variation.  When you make one quantity in a quadratic equation $n$ times its original, the dependent quantity would be $n^2$ times the original one.  You can see why this happens from the equation of a variation.

keeli_grace · Answer

Okay, so far so good. You seem very passionate about algebra. :)

mathslover · Answer

Remember this : Parth is the great young Mathematician. Feel lucky to get help from him.

parthkohli · Answer

Now, now, now - if $y$ is a quantity that depends on $x^2$, then as you increase $x^2$, the more you will see an increase in $y$.

It is obvious why $y = x^2$ is a quadratic variation.

Now, for example, see $y = 2x^2$.  You will wonder why this is a quadratic variation at first.  We saw how, from my last article, that if you multiply $x$ by $n$ in a variation, you will see that your new $y$ is $n^2$ times the original one.  Let's try to see if the same happens in this case.$$y = 2x^2$$Multiply $x$ by $n$.  The new quantity is $nx$.  Plug that for $x$ in $y = 2x^2$.

You will get $\text{the new }y = 2(nx)^2 $ by plugging in $nx$ for $x$.  This simplifies to $$\text{the new y } = 2n^2 x^2 \ ~~~~~~~ \quad ~~\quad = n^2 \times 2x^2 = n^2 \times y$$The new quantity $y$  is $n^2$ times the old $y$ quantity.

parthkohli · Answer

You can now see why this is called a "quadratic relationship" - quadratic literally means "the power two."

Now, as I hope, you're pretty clear about what a quadratic relationship is.  Let's now try to analyze many elements.

Before I start, remind yourself that a quadratic variation is in the *form* $y = kx^2$.  It is not $y = x^2$ itself.  This means that all of the variations such as $p^2 = 4q$ or $\rm area = \pi  r^2$ etc. are quadratic variations even though they don't initially appear to be.

- - -

○ What is the difference between $p = x^2$ and $q = 2x^2$?
A: Let's try to see how this works.  If you plug in $x=1$ in both of the variations, we see that $p = 1$ and $q = 2$.  When $x = 2$, we see $p = 4$ and $q = 8$.  If you continue, you will see how $q$ is always twice $p$.  This means that $q$ will be twice $p$ for all $x.$

We can extend this logic to variations $y_1 = x^2$ and $y_2 = kx^2$.  For all $x$, $y_2 $  is $k$ times $y_1.$

parthkohli · Answer

The point I stated above is very, very obvious, but I felt a need to state it.

---

Anyway.  Now, when you have questions that tell you about a direct variation between two quantities $y$ and $x^2$, always try to see the information you have.

Also remember that if you are given a direct variation, $k$ represents a constant.  If you know a value of $k$, the same value of $k $ will apply to all $x$ and $y$.

parthkohli · Answer

One more thing:

If $x$ is related with $y$, then $y$ is related with $k$.  (Stating the obvious!)

However, there is one small difference.  When you are given $y = k\cdot x$, then $x = \frac{1}{k} y$.  If $x$ varies with $y$ by a constant $k$, then $y$ varies with $x$ by a constant $\frac{1}k{}$.

parthkohli · Answer

From my last post, just note one thing:  if $k$ is a constant, $\frac{1}{k}$ is also a constant.

Basically, to sum it all up, if $\rm blah$ is in direct relation with $\rm meh$, $\rm bleh = k \cdot meh$.

- - - - -

Practice problem:

$y$ is directly related to $2x^3$.  When $y = 32$, $x$ is found to be $1$.  What is the constant of variation?