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How do I draw the graph for y = $\frac{KQx}{[R^2+x^2]^{3/2}}$. 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 :enter image description here 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.

Hema
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    Of course you can approximate any function with linear partitions. – Sven-Eric Krüger Jul 26 '18 at 12:26
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    Analyze the second derivative; it will be very helpful to understand the behavior of the function. – Claude Leibovici Jul 26 '18 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 '18 at 13:48
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    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 '18 at 13:52

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Here is a quick way to get a feel for the graph:

  1. 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.

  2. 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).

  3. Similarly, when $x$ is large compared to $R$, you can throw away $R^2$ instead, leading to $y\approx KQ/x^2$. 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.

enter image description here

Harald Hanche-Olsen
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