Mixing
How can I Plot “Change Rate Graph” of Sine Graph
$begingroup$
Imagine I have a Temperature / Time graph which is plotted by using the function y=sin(x).
In this graph X axes represents 'the time'(in seconds) and Y axes represents 'Temperature'(calcius).
How can i draw temperature change rate graph which is X axes represents 'the time' (in seconds) again but Y axes will represent Temperature/Seconds.
How can i find the function which will draw this temperature change rate graph?
graphing-functions
asked Jan 25 at 17:44
user2617750user2617750
1011
$endgroup$
add a comment |
$begingroup$
Imagine I have a Temperature / Time graph which is plotted by using the function y=sin(x).
In this graph X axes represents 'the time'(in seconds) and Y axes represents 'Temperature'(calcius).
How can i draw temperature change rate graph which is X axes represents 'the time' (in seconds) again but Y axes will represent Temperature/Seconds.
How can i find the function which will draw this temperature change rate graph?
graphing-functions
asked Jan 25 at 17:44
user2617750user2617750
1011
$endgroup$
1
$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01
add a comment |
$begingroup$
Imagine I have a Temperature / Time graph which is plotted by using the function y=sin(x).
In this graph X axes represents 'the time'(in seconds) and Y axes represents 'Temperature'(calcius).
How can i draw temperature change rate graph which is X axes represents 'the time' (in seconds) again but Y axes will represent Temperature/Seconds.
How can i find the function which will draw this temperature change rate graph?
graphing-functions
asked Jan 25 at 17:44
user2617750user2617750
1011
$endgroup$
Imagine I have a Temperature / Time graph which is plotted by using the function y=sin(x).
In this graph X axes represents 'the time'(in seconds) and Y axes represents 'Temperature'(calcius).
How can i draw temperature change rate graph which is X axes represents 'the time' (in seconds) again but Y axes will represent Temperature/Seconds.
How can i find the function which will draw this temperature change rate graph?
graphing-functions
graphing-functions
asked Jan 25 at 17:44
user2617750user2617750
1011
asked Jan 25 at 17:44
user2617750user2617750
1011
asked Jan 25 at 17:44
user2617750user2617750
1011
asked Jan 25 at 17:44
user2617750user2617750
1011
asked Jan 25 at 17:44
user2617750user2617750
1011
1011
1
$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01
add a comment |
1
$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01
1
1
$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01
$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm guessing that you're not yet familiar with calculus.
A basic operation of calculus is differentiation. This takes a function $f(x)$ and produces another function $f'(x) $ for the slope of a graph of $y=f(x)$ at each point. This slope can also be seen as the rate of change of $f(x)$. It's known as the derivative of $f(x)$, and often represented as $frac{dy}{dx}$.
Originally $dy$ and $dx$ were thought of as infinitely small changes in $y$ and $x$, but nowadays $frac{dy}{dx}$ is defined slightly differently to stop it meaning $frac00$.
The first example you encounter is usually $y=x^2$,
for which $frac{dy}{dx}=2x$.
What you're asking for is the derivative—the rate of change—of $sin(x)$. This turns out to be $cos(x)$, so that's what your rate-of-change graph would need to plot.
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
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add a comment |
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1 Answer
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$begingroup$
I'm guessing that you're not yet familiar with calculus.
A basic operation of calculus is differentiation. This takes a function $f(x)$ and produces another function $f'(x) $ for the slope of a graph of $y=f(x)$ at each point. This slope can also be seen as the rate of change of $f(x)$. It's known as the derivative of $f(x)$, and often represented as $frac{dy}{dx}$.
Originally $dy$ and $dx$ were thought of as infinitely small changes in $y$ and $x$, but nowadays $frac{dy}{dx}$ is defined slightly differently to stop it meaning $frac00$.
The first example you encounter is usually $y=x^2$,
for which $frac{dy}{dx}=2x$.
What you're asking for is the derivative—the rate of change—of $sin(x)$. This turns out to be $cos(x)$, so that's what your rate-of-change graph would need to plot.
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
I'm guessing that you're not yet familiar with calculus.
A basic operation of calculus is differentiation. This takes a function $f(x)$ and produces another function $f'(x) $ for the slope of a graph of $y=f(x)$ at each point. This slope can also be seen as the rate of change of $f(x)$. It's known as the derivative of $f(x)$, and often represented as $frac{dy}{dx}$.
Originally $dy$ and $dx$ were thought of as infinitely small changes in $y$ and $x$, but nowadays $frac{dy}{dx}$ is defined slightly differently to stop it meaning $frac00$.
The first example you encounter is usually $y=x^2$,
for which $frac{dy}{dx}=2x$.
What you're asking for is the derivative—the rate of change—of $sin(x)$. This turns out to be $cos(x)$, so that's what your rate-of-change graph would need to plot.
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
I'm guessing that you're not yet familiar with calculus.
A basic operation of calculus is differentiation. This takes a function $f(x)$ and produces another function $f'(x) $ for the slope of a graph of $y=f(x)$ at each point. This slope can also be seen as the rate of change of $f(x)$. It's known as the derivative of $f(x)$, and often represented as $frac{dy}{dx}$.
Originally $dy$ and $dx$ were thought of as infinitely small changes in $y$ and $x$, but nowadays $frac{dy}{dx}$ is defined slightly differently to stop it meaning $frac00$.
The first example you encounter is usually $y=x^2$,
for which $frac{dy}{dx}=2x$.
What you're asking for is the derivative—the rate of change—of $sin(x)$. This turns out to be $cos(x)$, so that's what your rate-of-change graph would need to plot.
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
$endgroup$
I'm guessing that you're not yet familiar with calculus.
A basic operation of calculus is differentiation. This takes a function $f(x)$ and produces another function $f'(x) $ for the slope of a graph of $y=f(x)$ at each point. This slope can also be seen as the rate of change of $f(x)$. It's known as the derivative of $f(x)$, and often represented as $frac{dy}{dx}$.
Originally $dy$ and $dx$ were thought of as infinitely small changes in $y$ and $x$, but nowadays $frac{dy}{dx}$ is defined slightly differently to stop it meaning $frac00$.
The first example you encounter is usually $y=x^2$,
for which $frac{dy}{dx}=2x$.
What you're asking for is the derivative—the rate of change—of $sin(x)$. This turns out to be $cos(x)$, so that's what your rate-of-change graph would need to plot.
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
edited Jan 25 at 19:10
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
answered Jan 25 at 18:50
timtfjtimtfj
2,468420
2,468420
add a comment |
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$begingroup$
differentiate $sin x$ to get $cos x$ and plot that
$endgroup$
– David Quinn
Jan 25 at 18:01