Whats the best way of solving this ratio problem?
$begingroup$
I have a question relating to a school report that I submitted, essentially I was asked to prove why a certain value ($x_1$) had a large increase (40%) as compared with a second value ($x_2$). I wanted to prove it using the ratio of a second variable $y$.
Lets say the function is,
$$
x = y + C
$$
where $y$ is a variable and $C$ is a constant.
e.g. lets say there are two cars, and their speed ($x$) is described as a function of car engine size ($y$) and a drag coefficient ($C$), such that: $x = y + C$. I want to describe that the ratio of change in $x$ of car1 and car2, i.e. $x_2/x_1$ is equal to the ratio of the cars engine sizes $y_2/y_1$ added to a constant.
How do i express that the ratio of $x$ is equal to the ratio of $y + C$. I was thinking about writing it as:
$$
frac{x_2}{x_1} = frac{y_2 + C}{y_1 + C}
$$
or possibly;
$$
Delta x = frac{y_2 + C}{y_1 + C}
$$
Furthermore, I wondered if there's a simple way to relate $Delta x$ and $Delta y$ such that;
$$
Delta x = k Delta y
$$
I went and did some research on some of my old math, and realized that:
$$
Delta x = x_2 - x_1
$$
So whats the best way to express ratios using algebra?
ratio
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
$endgroup$
add a comment |
$begingroup$
I have a question relating to a school report that I submitted, essentially I was asked to prove why a certain value ($x_1$) had a large increase (40%) as compared with a second value ($x_2$). I wanted to prove it using the ratio of a second variable $y$.
Lets say the function is,
$$
x = y + C
$$
where $y$ is a variable and $C$ is a constant.
e.g. lets say there are two cars, and their speed ($x$) is described as a function of car engine size ($y$) and a drag coefficient ($C$), such that: $x = y + C$. I want to describe that the ratio of change in $x$ of car1 and car2, i.e. $x_2/x_1$ is equal to the ratio of the cars engine sizes $y_2/y_1$ added to a constant.
How do i express that the ratio of $x$ is equal to the ratio of $y + C$. I was thinking about writing it as:
$$
frac{x_2}{x_1} = frac{y_2 + C}{y_1 + C}
$$
or possibly;
$$
Delta x = frac{y_2 + C}{y_1 + C}
$$
Furthermore, I wondered if there's a simple way to relate $Delta x$ and $Delta y$ such that;
$$
Delta x = k Delta y
$$
I went and did some research on some of my old math, and realized that:
$$
Delta x = x_2 - x_1
$$
So whats the best way to express ratios using algebra?
ratio
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
$endgroup$
1
$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06
add a comment |
$begingroup$
I have a question relating to a school report that I submitted, essentially I was asked to prove why a certain value ($x_1$) had a large increase (40%) as compared with a second value ($x_2$). I wanted to prove it using the ratio of a second variable $y$.
Lets say the function is,
$$
x = y + C
$$
where $y$ is a variable and $C$ is a constant.
e.g. lets say there are two cars, and their speed ($x$) is described as a function of car engine size ($y$) and a drag coefficient ($C$), such that: $x = y + C$. I want to describe that the ratio of change in $x$ of car1 and car2, i.e. $x_2/x_1$ is equal to the ratio of the cars engine sizes $y_2/y_1$ added to a constant.
How do i express that the ratio of $x$ is equal to the ratio of $y + C$. I was thinking about writing it as:
$$
frac{x_2}{x_1} = frac{y_2 + C}{y_1 + C}
$$
or possibly;
$$
Delta x = frac{y_2 + C}{y_1 + C}
$$
Furthermore, I wondered if there's a simple way to relate $Delta x$ and $Delta y$ such that;
$$
Delta x = k Delta y
$$
I went and did some research on some of my old math, and realized that:
$$
Delta x = x_2 - x_1
$$
So whats the best way to express ratios using algebra?
ratio
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
$endgroup$
I have a question relating to a school report that I submitted, essentially I was asked to prove why a certain value ($x_1$) had a large increase (40%) as compared with a second value ($x_2$). I wanted to prove it using the ratio of a second variable $y$.
Lets say the function is,
$$
x = y + C
$$
where $y$ is a variable and $C$ is a constant.
e.g. lets say there are two cars, and their speed ($x$) is described as a function of car engine size ($y$) and a drag coefficient ($C$), such that: $x = y + C$. I want to describe that the ratio of change in $x$ of car1 and car2, i.e. $x_2/x_1$ is equal to the ratio of the cars engine sizes $y_2/y_1$ added to a constant.
How do i express that the ratio of $x$ is equal to the ratio of $y + C$. I was thinking about writing it as:
$$
frac{x_2}{x_1} = frac{y_2 + C}{y_1 + C}
$$
or possibly;
$$
Delta x = frac{y_2 + C}{y_1 + C}
$$
Furthermore, I wondered if there's a simple way to relate $Delta x$ and $Delta y$ such that;
$$
Delta x = k Delta y
$$
I went and did some research on some of my old math, and realized that:
$$
Delta x = x_2 - x_1
$$
So whats the best way to express ratios using algebra?
ratio
ratio
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
edited Dec 5 '18 at 7:59
Daniele Tampieri
1,8891619
1,8891619
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
asked Dec 5 '18 at 7:48
AfroeskimoAfroeskimo
114
114
1
$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06
add a comment |
1
$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06
1
1
$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06
$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06
add a comment |
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$begingroup$
$Delta x = x_2-x_1 = (y_2+C)-(y_1+C) = y_2-y_1 = Delta y$. However, I would call this the change in speed, not ratio of speeds. Ratio of speed would simply be the first expression you wrote. Also, assuming $y_2>y_1$, and $C$ positive, $x_2/x_1$ would always be less than $y_2/y_1$. I dont think anything more can be said about this generally
$endgroup$
– Anvit
Dec 5 '18 at 8:06