## מישור משיק

guerin
הודעות: 106
הצטרף: 19:13 17/01/2011

### מישור משיק

מישהו יכול להסביר לי איך מוצאים מישור משיק לפונקציה?
יש דוגמא המסבירה את זה בשיפרין, אבל אני לא מצליח להבין למה זאת הרך למצוא מישור משיק..
קבצים מצורפים tangent.png (10.9 KiB) נצפה 1651 פעמים

avners
הודעות: 280
הצטרף: 01:41 20/04/2007

### Re: מישור משיק

אני מתכנן לדבר על הדוגמא הזאת מחר, בתקווה שזה יעזור
"התמדה זה הכול,
אין שום סיבה להפסיק..."
"איזה כיף", הדג נחש מסבירים את החוק הראשון של ניוטון

klgamit
הודעות: 55
הצטרף: 18:12 17/11/2009

### Re: מישור משיק

Ohai.

Suppose you have
$$z = f(x,y)$$
We want the tangent plane to the point $$(x_0,y_0,f(x_0,y_0))$$
We define the function:
$$g(x,y,z) = z-f(x,y)$$
The normal to the plane is given by the gradient of this function at $$(x_0,y_0)$$
$$\hat{n} = \nabla{g}(x_0,y_0) = (-\frac{\partial{f}}{\partial{x}}(x_0,y_0),-\frac{\partial{f}}{\partial{y}}(x_0,y_0),1)$$
The vector from $$(x_0,y_0,f(x_0,y_0))$$ to any point $$(x,y,z)$$ on the plane must be perpendicular to the normal (definition of a plane):
$$\nabla{g}(x_0,y_0)\cdot(x-x_0,y-y_0,z-f(x_0,y_0)) = 0$$
$$-\frac{\partial{f}}{\partial{x}}(x_0,y_0)(x-x_0)-\frac{\partial{f}}{\partial{y}}(x_0,y_0)(y-y_0)+z-f(x_0,y_0) = 0$$
$$\Rightarrow z=f(x_0,y_0) + \nabla{f(x_0,y_0)}\cdot(x-x_0,y-y_0)$$

VaultTec
הודעות: 47
הצטרף: 00:18 24/10/2010

### Re: מישור משיק

klgamit כתב:
Ohai.

Suppose you have
$$z = f(x,y)$$
We want the tangent plane to the point $$(x_0,y_0,f(x_0,y_0))$$
We define the function:
$$g(x,y,z) = z-f(x,y)$$
The normal to the plane is given by the gradient of this function at $$(x_0,y_0)$$
$$\hat{n} = \nabla{g}(x_0,y_0) = (-\frac{\partial{f}}{\partial{x}}(x_0,y_0),-\frac{\partial{f}}{\partial{y}}(x_0,y_0),1)$$
The vector from $$(x_0,y_0,f(x_0,y_0))$$ to any point $$(x,y,z)$$ on the plane must be perpendicular to the normal (definition of a plane):
$$\nabla{g}(x_0,y_0)\cdot(x-x_0,y-y_0,z-f(x_0,y_0)) = 0$$
$$-\frac{\partial{f}}{\partial{x}}(x_0,y_0)(x-x_0)-\frac{\partial{f}}{\partial{y}}(x_0,y_0)(y-y_0)+z-f(x_0,y_0) = 0$$
$$\Rightarrow z=f(x_0,y_0) + \nabla{f(x_0,y_0)}\cdot(x-x_0,y-y_0)$$
למה הגרדיאנט של הפונקציה בנקודה(Xo,Yo) הוא הנורמל למישור?

klgamit
הודעות: 55
הצטרף: 18:12 17/11/2009

### Re: מישור משיק

Ohai again.
While writing the above I did not include an explanation of why the gradient of the defined function $$g(x,y,z)$$ is indeed normal to the surface at the point $$(x_0,y_0,f(x_0,y_0))$$. Since I did not know the explanation at the time I had no way to include it. Now however, I think I found a suitable proof of this, hehe:

Let us look at very small changes $$\Delta x$$ and $$\Delta y$$ from the point $$(x_0,y_0,f(x_0,y_0))$$, and form 2 vectors that have their origins in this point and are pointing in the directions of change. Since we can assume $$f$$ is smooth enough we can understand that these two vectors will define together a tangent plane to the surface at the given point. In order to obtain the normal to the surface from it, all we need to do is to take their cross product. So let's do it:

$$\vec{v_1} = (x_0+\Delta x,y_0,f(x_0+\Delta x,y_0))-(x_0,y_0,f(x_0,y_0))=(\Delta x,0,f(x_0+\Delta x,y_0)-f(x_0,y_0))$$
$$\vec{v_2} = (x_0,y_0 + \Delta y,f(x_0,y_0+\Delta y))-(x_0,y_0,f(x_0,y_0))=(0,\Delta y,f(x_0,y_0+ \Delta y)-f(x_0,y_0))$$

Since $$\Delta x , \Delta y$$ are small enough we can write:

$$\vec{v_1} = (\Delta x,0,\frac{\partial f}{\partial x}(x_0,y_0)\Delta x)$$
$$\vec{v_2} = (0,\Delta y,\frac{\partial f}{\partial y}(x_0,y_0)\Delta y)$$

$$\vec{v_1} \times \vec{v_2} = (-\frac{\partial f}{\partial x}(x_0,y_0),-\frac{\partial f}{\partial y}(x_0,y_0),1)\Delta x \Delta y$$

Which you can see is proportional to $$\nabla g$$ from before and hence we've shown that $$\nabla g$$ is also normal to the surface at $$(x_0,y_0,f(x_0,y_0))$$.

VaultTec
הודעות: 47
הצטרף: 00:18 24/10/2010

### Re: מישור משיק

מעניין.. אני אצטרך להתעמק בזה עוד קצת

תודה רבה,

דן.