Question

Why is y(t) = x^2(t) not linear?

rule:
ax1(t) + bx2(t) -> ay1(t) + by2(t)

Answer 1

Just to clarify, is that (x^2)*(t) or x^(2t)?

Answer 2

its (x^2)(t)

Answer 3

Alright let me type this out, one sec...

Answer 4

oke thanks :)

Answer 5

if i use the rule:

a(x1^2)(t) + b(x2^2)(t) -> ay1(t) + by2(t)

BUT!

ay1(t) = a(x1^2)(t)
ay2(t) = a(x2^2)(t)

so its linear?! but ISNT.... am i doing something wrong?

Answer 6

Yeah i'll explain it. its just taking a bit to type.

Answer 7

okeee... thats why i am asking this question :) i will wait :)

Answer 8

\[y(t)=x^2(t)\]
\[ax_1(t) + bx_2(t) -> ay_1(t) + by_2(t)\]
\[ax_1(t)+bx_2(t)->ax_1^2(t)+bx_2^2(t)\]
its not linear because this is not true
\[x_1^2 \neq x_1 \text{ and } x_2^2 \neq x_2\]

Answer 9

So what I would do on this is take the derivative with respect to t.
The derivative gives you the equation for the slope of the original equation.
In order for an equation to be considered linear, its slope must be linear.

So for this equation, when you take the derivative with respect to t, you get:
\[\left(\begin{matrix}d \\ dt\end{matrix}\right)[y(t)=(x^2)(t)]\rightarrow y'(t)=x^2\]
A linear slope would be something like y'(t)=x or y'(t)=4x.
However the slope of this line is y'(t)=x^2 which is quadratic, so the initial equation as a whole would be considered quadratic.

Make sense??

Answer 10

The answer before mine also works but I just like this way better

Answer 11

wait guys... i have to use the rule ive mentioned

Answer 12

Alright well in that case, use the other person's answer. I could explain the other answer too if you don't get it

Answer 13

@myininaya @ jpomer325

you are saying that \[x1 ^{2}\neq x1\]

but i have to multiply my x1^2 with a but why is it still x1?

Answer 14

jpomer i'm really confused about your answer
why didn't you differentiate the other side
and y'=constant implies y is linear

Answer 15

i dont get it..

i already wrote what i did.. but whats wrong with my solution?

Answer 16

i dont have to differentiate anything... i only have to use the rule from above.. but i dont get the solution of jpomer325 :s

Answer 17

haha its alright i didnt use those rules anyway.
a and b are just constants so you can ingnore them for all practical purposes in this problem

Answer 18

myininaya i really hope you can help me with this question :)

Answer 19

\[y(t) = x ^{2}(t)\]

\[\alpha(x1^{2})(t) + \beta(x2^{2})(t) = \alpha(y1^{2})(t) +\beta(y2^{2})(t) \]

Answer 20

\[y(t)=(x^2)(t)\]

Using the above equation, she plugged in (x^2)(t) for all y(t)'s in this equation:
\[ax_1(t)+bx_2(t)−>ay_1(t)+by_2(t)\]
Resulting in the following:
\[ax_1(t)+bx_2(t)−>a(x_1)^{2}(t)+b(x_2)^{2}(t)\]
However, this statement is obviously not true, because by definition:
\[x _{1}\neq(x _{1})^{2}\]
and
\[x _{2}\neq(x _{2})^{2}\]
So the rule for linearity is broken by the given equation y(t)=(x^2)(t).

Answer 21

but why dont i have to apply ax1(t) as a(x1^2)(t)?

Answer 22

Well that is what you are doing, comparing them to see if the statement is true. But for simplicity's sake, you can leave out the coefficients "a" and "b" because they just cancel out.
\[[a(x)=a(x^2)]/a \rightarrow x=x^2\]

Answer 23

Thats just an example of how it cancels

Answer 24

okeee... but could you explain why

y(t) = 2x(t) + 3 is also not linear?

Answer 25

is there a rule with that one too or not

Answer 26

the same rule has to be applied

Answer 27

Alright well for this one, I don't see how its not linear either, because the quick check I was always taught was that as long as the y(t) and x(t) terms were linear (and not (x^3)(t) or (y^2)(t) for example), then the equation is linear.

Answer 28

it is not linear :S i have to apply the same rule. but i dont get it at all -.-

Answer 29

Well let me see what happens if we plug in the same way we did on the first problem...

Answer 30

\[y(t)=2x(t)+3\]

Using the above equation, we plug in 2x(t)+3 for all y(t)'s in this equation:
\[ax_1(t)+bx_2(t)−>ay_1(t)+by_2(t)\]

Resulting in the following:
\[a(x_1)(t)+b(x_2)(t)−>a*(2x_1(t)+3)+b*(2x_2(t)+3)\]

However, this statement is obviously not true, because by definition:
\[x_1(t)≠2x_1(t)+3\]

and
\[x_2(t)≠2x_2(t)+3\]

So the rule for linearity is broken by the given equation y(t)=2x(t)+3.

Answer 31

Pretty much the same as the first one. But I'm also still not entirely sure thats all there is to it. Might be about that time to say f*ck it. You'll go crazy otherwise haha

Answer 32

haha thats funny it changes f word to flutter

Answer 33

hmm.. that makes sence.. but this one?

y(t) = Ax(t)

this one is linear

Answer 34

my teachers said that i had to fill the formula also in for ax1(t) + bx2(t)

Answer 35

Well since I would assume A is a coefficient, y(t) would vary linearly with x(t), so its linear

Answer 36

Where did they say you plug that in?

Answer 37

i dont know.. but i had to use this rule... and use it for these examples..

Answer 38

Yeah I really dont know. I'd have to see the whole problem or the textbook its based out of

Answer 39

whats the question?

Answer 40

why y(t) = Ax(t) is linear

Answer 41

use your definition for linearity

Answer 42

and why are you only filling in for ay1(t) + by2(t) and not the ax1(t) + bx2(t)

Answer 43

my teacher said that i also had to fill in the ax1(t) + bx2(t)

Answer 44

\[y(t)=x^2(t)\]
ok maybe you like this more:
we have
\[y_1(t)=x_1^2(t)\]
\[y_2(t)=x_2^2(t)\]
\[\text{ Let } x_3(t)=ax_1(t)+bx_2(t) \text{ then } y_3(t)=x_3^2(t)=(ax_1(t)+bx_2(t))^2\]
\[=a^2x_1^2(t)+2abx_1(t)x_2(t)+b^2x^2_2(t)\]

\[\text{ but } ay_1(t)+by_2(t) = ax_1^2(t)+bx_2^2(t)\]

Answer 45

so the anwser on your question is : why ... ? so because we know that every quadratic equation has graph parabola what not equal linear ...