Grapher- Implicit differentiation (how to graph derivatives?)

[Feefemon]

Hello,

In calc class the other day we learned implicit differentiation and I want to be able to graph some of the relations and their derivatives but have not figured out the proper notation in grapher.

I want to graph x^2+y^2=1 (a circle)

and its derivative, dy/dx=-x/y (has vertical asymptotes at 1 and -1 and has four branches starting at the origin and expanding in each quadrant)

The closest I've gotten is graphing y=-x/y and that got me what appears to be the graph of the relation y=-x^(1/2)

I want to be able to do this with other relations as well.


Thanks in advance.

 

[YB24]

Hello,

In calc class the other day we learned implicit differentiation and I want to be able to graph some of the relations and their derivatives but have not figured out the proper notation in grapher.

Hi Feefemon,

Your circle : x^2 + y^2 = 1 or y = ± sqrt (1 - x^2)
The derivative is dy / dx = ± x / y only if y is the function y(x), but not the name of the coordinate y
When you wrote y = ± x / y Grapher sees it as an implicit equation giving same graph as y^2 = ± x ( a double parabol Ox axis)
The derivative is the functon of x : ± x / sqrt(1 - x^2)
its graph is the one of the equation y = x / sqrt(1 - x^2)

What to do :
x^2 + y^2 = 1 circle, implicit equation
f(t) = {-1, 1}*sqrt(1 - t^2) the function
y = x / f(x) the derivative dy / dx

Just tried with Grapher 2.1 (Snow leopard) and Grapher 2.2 (Lion)

Enjoy,
YB24

 

[MisterMe]

You are going about this all wrong. When you enter the derivative, Grapher attempts to integrate the derivative and graph the function. Furthermore, the implicit derivative of x^2 + y^2=1 is dy/dx = -x/y, not dy/dx = ±x/y. However, to get the derivative, you want to plot a vector field of -x/y.

 

[YB24]

You are going about this all wrong. When you enter the derivative, Grapher attempts to integrate the derivative and graph the function. Furthermore, the implicit derivative of x^2 + y^2=1 is dy/dx = -x/y, not dy/dx = ±x/y. However, to get the derivative, you want to plot a vector field of -x/y.

Hi,
1) Sorry for the ±x/y it's -x/y as you wrote, the "±" is included in the fonction y(x) : + for a semi-circle, - for the other one.
2) You wrote "enter the derivative" about entering "dy/dx=-x/y" ; this is not a derivative but an equation, a differential equation. The implicit derivative is the function of x : "-x/y" where y is y(x). Similar : the derivative of the function of x "x^4" is the function of x "4x^3" and not the equation "y or y'=4x^3"
3) By the way the Grapher syntax for resolving the above differential equation is :
" y' = -x/y, y(0)=1, x=-1…1 " (semi-circle y > 0), then y(0)=-1 (semi-circle y<0).
4) What do you mean by "to get the derivative" ? A derivative (implicit or not) is a function, do you want to plot this function, or to plot the slope of the tangent (as vectors) on a few points on the circle ?
So long,
YB24

 

[MisterMe]

You are all over the map, pardon the pun. An explicit derivative is a function of x only. An implicit derivative is a function of x and y. You are absolutely correct that dy/dx = -x/y is a differential equation. So too is y' = -x/y. Grapher treats the two forms the same. What it does is to solve the differential equation and plot its solution. It does not plot the derivative.

You seem to be laboring under the misconception that y'=-x/y is a single-valued function. It is not. y' is defined for every point on the x-y plane except those on the x-axis. You have two choices:

1. Map y' into the third-dimension z. (Not recommended)
2. Plot a vector field of the derivative on a fine grid. This is a standard technique that is used widely in mathematics textbooks.