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How do I draw the graph for y = $fracKQx[R^2+x^2]^3/2$
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How do I draw the graph for y = $fracKQx[R^2+x^2]^3/2$.
Actually this is from a physics question which requires me to draw the graph of electric field (E or y) caused by uniformly charged ring against distance x from ring. Here k,Q and R are constants. The graph given in my book is similar to this :
I understand that maxima are at x= plus or minus R/√2, when dy/dx = 0. Also when x = 0 or infinity y will be zero. However why is the graph curved? Isn't it possible for it to be composed of straight lines? Am I missing out some major concept here? Please help me.
graphing-functions
asked Jul 26 at 12:09
Hema
486113
add a comment |Â
up vote
0
down vote
favorite
How do I draw the graph for y = $fracKQx[R^2+x^2]^3/2$.
Actually this is from a physics question which requires me to draw the graph of electric field (E or y) caused by uniformly charged ring against distance x from ring. Here k,Q and R are constants. The graph given in my book is similar to this : I understand that maxima are at x= plus or minus R/√2, when dy/dx = 0. Also when x = 0 or infinity y will be zero. However why is the graph curved? Isn't it possible for it to be composed of straight lines? Am I missing out some major concept here? Please help me.
graphing-functions
asked Jul 26 at 12:09
Hema
486113
1
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
1
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
1
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I draw the graph for y = $fracKQx[R^2+x^2]^3/2$.
Actually this is from a physics question which requires me to draw the graph of electric field (E or y) caused by uniformly charged ring against distance x from ring. Here k,Q and R are constants. The graph given in my book is similar to this : I understand that maxima are at x= plus or minus R/√2, when dy/dx = 0. Also when x = 0 or infinity y will be zero. However why is the graph curved? Isn't it possible for it to be composed of straight lines? Am I missing out some major concept here? Please help me.
graphing-functions
asked Jul 26 at 12:09
Hema
486113
How do I draw the graph for y = $fracKQx[R^2+x^2]^3/2$.
Actually this is from a physics question which requires me to draw the graph of electric field (E or y) caused by uniformly charged ring against distance x from ring. Here k,Q and R are constants. The graph given in my book is similar to this : I understand that maxima are at x= plus or minus R/√2, when dy/dx = 0. Also when x = 0 or infinity y will be zero. However why is the graph curved? Isn't it possible for it to be composed of straight lines? Am I missing out some major concept here? Please help me.
graphing-functions
asked Jul 26 at 12:09
Hema
486113
asked Jul 26 at 12:09
Hema
486113
asked Jul 26 at 12:09
Hema
486113
asked Jul 26 at 12:09
Hema
486113
486113
1
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
1
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
1
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52
add a comment |Â
1
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
1
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
1
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52
1
1
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
1
1
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
1
1
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Here is a quick way to get a feel for the graph:
Note the antisymmetry: If you replace $x$ by $-x$, $y$ changes sign but keeps its magnitude. So you only need to understand the part where $x>0$, say, and then reflect that through the origin.
When $x$ is small compared to $R$, then $x^2$ in the denominator is small compared to $R^2$, so you get an approximation by throwing away the $x^2$ term. Then the curve is simplified to $y approx KQx/R^3$ (a straight line through the origin).
Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $yapprox KQ/x^2$. I am sure you know what its graph looks like.
Below is a figure showing the original function, in red, and the two approximations in blue. Note that the red curve lies below the blue curves. This should not be surprising, given the terms thrown away. Note how the red curve starts close to one blue curve, then it has to bend somehow to start matching the other one.
To get further, you'd have to do the second derivative song and dance, but this simple reasoning gets you a good idea what to expect.
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here is a quick way to get a feel for the graph:
Note the antisymmetry: If you replace $x$ by $-x$, $y$ changes sign but keeps its magnitude. So you only need to understand the part where $x>0$, say, and then reflect that through the origin.
When $x$ is small compared to $R$, then $x^2$ in the denominator is small compared to $R^2$, so you get an approximation by throwing away the $x^2$ term. Then the curve is simplified to $y approx KQx/R^3$ (a straight line through the origin).
Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $yapprox KQ/x^2$. I am sure you know what its graph looks like.
Below is a figure showing the original function, in red, and the two approximations in blue. Note that the red curve lies below the blue curves. This should not be surprising, given the terms thrown away. Note how the red curve starts close to one blue curve, then it has to bend somehow to start matching the other one.
To get further, you'd have to do the second derivative song and dance, but this simple reasoning gets you a good idea what to expect.
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
add a comment |Â
up vote
1
down vote
accepted
Here is a quick way to get a feel for the graph:
Note the antisymmetry: If you replace $x$ by $-x$, $y$ changes sign but keeps its magnitude. So you only need to understand the part where $x>0$, say, and then reflect that through the origin.
When $x$ is small compared to $R$, then $x^2$ in the denominator is small compared to $R^2$, so you get an approximation by throwing away the $x^2$ term. Then the curve is simplified to $y approx KQx/R^3$ (a straight line through the origin).
Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $yapprox KQ/x^2$. I am sure you know what its graph looks like.
Below is a figure showing the original function, in red, and the two approximations in blue. Note that the red curve lies below the blue curves. This should not be surprising, given the terms thrown away. Note how the red curve starts close to one blue curve, then it has to bend somehow to start matching the other one.
To get further, you'd have to do the second derivative song and dance, but this simple reasoning gets you a good idea what to expect.
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here is a quick way to get a feel for the graph:
Note the antisymmetry: If you replace $x$ by $-x$, $y$ changes sign but keeps its magnitude. So you only need to understand the part where $x>0$, say, and then reflect that through the origin.
When $x$ is small compared to $R$, then $x^2$ in the denominator is small compared to $R^2$, so you get an approximation by throwing away the $x^2$ term. Then the curve is simplified to $y approx KQx/R^3$ (a straight line through the origin).
Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $yapprox KQ/x^2$. I am sure you know what its graph looks like.
Below is a figure showing the original function, in red, and the two approximations in blue. Note that the red curve lies below the blue curves. This should not be surprising, given the terms thrown away. Note how the red curve starts close to one blue curve, then it has to bend somehow to start matching the other one.
To get further, you'd have to do the second derivative song and dance, but this simple reasoning gets you a good idea what to expect.
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
Here is a quick way to get a feel for the graph:
Note the antisymmetry: If you replace $x$ by $-x$, $y$ changes sign but keeps its magnitude. So you only need to understand the part where $x>0$, say, and then reflect that through the origin.
When $x$ is small compared to $R$, then $x^2$ in the denominator is small compared to $R^2$, so you get an approximation by throwing away the $x^2$ term. Then the curve is simplified to $y approx KQx/R^3$ (a straight line through the origin).
Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $yapprox KQ/x^2$. I am sure you know what its graph looks like.
Below is a figure showing the original function, in red, and the two approximations in blue. Note that the red curve lies below the blue curves. This should not be surprising, given the terms thrown away. Note how the red curve starts close to one blue curve, then it has to bend somehow to start matching the other one.
To get further, you'd have to do the second derivative song and dance, but this simple reasoning gets you a good idea what to expect.
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
answered Jul 26 at 14:00
Harald Hanche-Olsen
27.2k23959
27.2k23959
add a comment |Â
add a comment |Â
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1
Of course you can approximate any function with linear partitions.
– Sven Krüger
Jul 26 at 12:26
1
Analyze the second derivative; it will be very helpful to understand the behavior of the function.
– Claude Leibovici
Jul 26 at 13:26
@ClaudeLeibovici is there an easier way to do it? These questions are JEE pattern and must be done in a few minutes at the most. Is there an intuitive way of knowing if the graph is curved?
– Hema
Jul 26 at 13:48
1
No algebraic function, other than the trivial cases $ax+b$, has any straight line segment in its graph. So you don't need to worry on that account.
– Harald Hanche-Olsen
Jul 26 at 13:52