 My Math Forum Question about Gradient, Directional Derivative and normal line
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 January 22nd, 2017, 10:46 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Question about Gradient, Directional Derivative and normal line This question is about vector calculus, gradient, directional derivative and normal line. If the gradient is the direction of the steepest ascent: >> gradient(x, y) = [ derivative_f_x(x, y), derivative_f_y(x, y) ] Then it really confuse me as when calculating the normal line perpendicular to the tangent plane, the formula would be: >> normal line = (derivative_f_x(x, y), derivative_f_y(x, y), z), But both derivative_f_x(x,y) & derivative_f_y(x,y) are gradient (the slope of the tangent plane). I don't think the steepest ascent/descent is the slope of the normal line perpendicular to the tangent plane! For example Find a vector function for the line normal to x^2 + 2y^2 + 4z^2 = 26 at (2, -3, -1). Answer: (2 + 4t, -3 -12t, -1 - 8t). Anyone care to give it a shot and show me the step?? Any information would be much appreciated. Thanks. January 22nd, 2017, 06:07 PM #2 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 First write the function as \displaystyle \begin{align*} f\left( x,y,z \right) = x^2 + 2\,y^2 + 4\,z^2 - 26 = 0 \end{align*}, and then the normal vector can be found using \displaystyle \begin{align*} \nabla f &= \left( \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z} \right) \\ &= \left( 2\,x , 4\,y , 8\,z \right) \\ &= \left( 2\cdot 2 , 4 \left( -3 \right) , 8\left( -1 \right) \right) \textrm{ at } \left( 2, -3, -1 \right) \\ &= \left( 4, -12, -8 \right) \end{align*} So the direction vector for the normal line at \displaystyle \begin{align*} \left( 2, -3, -1 \right) \end{align*} is \displaystyle \begin{align*} \left( 4, -12, -8 \right) \end{align*}. It can be made infinitely long by multiplying by a parameter t, which can take on any real number, giving \displaystyle \begin{align*} \left( 4\,t , -12\,t , -8\,t \right) \end{align*}. Finally, this infinitely long vector can be positioned in the right spot as we know it goes through \displaystyle \begin{align*} \left( 2, -3, -1 \right) \end{align*}, so by shifting that many units in each direction, the correct line is \displaystyle \begin{align*} \left( 2 + 4\,t , -3 - 12\,t , -1 - 8\,t \right) \end{align*}. January 22nd, 2017, 06:17 PM   #3
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Thanks for the quick response. It would really help if you could explain:

\displaystyle \begin{align*} \nabla f \end{align*} is the directional vector of the normal line (which is perpendicular to the gradient \displaystyle \begin{align*} \nabla f )\end{align*} )

Quote:
 Originally Posted by Prove It First write the function as \displaystyle \begin{align*} f\left( x,y,z \right) = x^2 + 2\,y^2 + 4\,z^2 - 26 = 0 \end{align*}, and then the normal vector can be found using \displaystyle \begin{align*} \nabla f &= \left( \frac{\partial f}{\partial x} , \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z} \right) \\ &= \left( 2\,x , 4\,y , 8\,z \right) \\ &= \left( 2\cdot 2 , 4 \left( -3 \right) , 8\left( -1 \right) \right) \textrm{ at } \left( 2, -3, -1 \right) \\ &= \left( 4, -12, -8 \right) \end{align*} So the direction vector for the normal line at \displaystyle \begin{align*} \left( 2, -3, -1 \right) \end{align*} is \displaystyle \begin{align*} \left( 4, -12, -8 \right) \end{align*}. It can be made infinitely long by multiplying by a parameter t, which can take on any real number, giving \displaystyle \begin{align*} \left( 4\,t , -12\,t , -8\,t \right) \end{align*}. Finally, this infinitely long vector can be positioned in the right spot as we know it goes through \displaystyle \begin{align*} \left( 2, -3, -1 \right) \end{align*}, so by shifting that many units in each direction, the correct line is \displaystyle \begin{align*} \left( 2 + 4\,t , -3 - 12\,t , -1 - 8\,t \right) \end{align*}. January 22nd, 2017, 06:42 PM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Think of z=f(x,y) as the equation of a hill (surface). At any point x,y at the base of the hill the hill has a height f(x,y). Let dx,dy be a small excursion in the x,y plane from the point x,y. The corresponding change in z is: $\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\vec{\bigtriangledown} f\cdot d\vec{r}$ Directional derivative: $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}=\vec{\bigtriangledown} f\cdot \vec{n}$ What does dz=0 mean? No change in z, so $\displaystyle \vec{\bigtriangledown}f$ is perpenticular to the curve z=c going through the point x,y. It's like being on a horizantal trail on the side of the hill and $\displaystyle \vec{\bigtriangledown}f$ is a horizantal vector perpendicular to the path. $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in dir $\displaystyle \vec{\bigtriangledown}F$. __________________________________________________ _____ If you are walking on the side of the hill in an arbitrary direction, your coordinates satisfy F=z-f(x,y)=0. A step in any direction from a point on the side of the hill must satisfy $\displaystyle dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=\vec{\bigtriangledown}F\cdot d \vec{r}=0$ Therefore $\displaystyle \vec{\bigtriangledown}F$ is normal to the hill at x,y,z. The same is true if equation of the hill (surface) is given by F(x,y,z)=0. ================================================== ====== If w=F(x,y,z) $\displaystyle dw=\vec{\bigtriangledown}F\cdot d\vec{r}$, and directional derivative is: $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}= \vec{\bigtriangledown}F\cdot \vec{n}$ dw=0 for an excursion along the surface w=c through x,y,z and $\displaystyle \vec{\bigtriangledown}F$ is normal to this surface. $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in direction $\displaystyle \vec{\bigtriangledown}F$. January 23rd, 2017, 04:26 PM   #5
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Thanks! This is a very very good explanation for me.

Quote:
 Originally Posted by zylo Think of z=f(x,y) as the equation of a hill (surface). At any point x,y at the base of the hill the hill has a height f(x,y). Let dx,dy be a small excursion in the x,y plane from the point x,y. The corresponding change in z is: $\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\vec{\bigtriangledown} f\cdot d\vec{r}$ Directional derivative: $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}=\vec{\bigtriangledown} f\cdot \vec{n}$ What does dz=0 mean? No change in z, so $\displaystyle \vec{\bigtriangledown}f$ is perpenticular to the curve z=c going through the point x,y. It's like being on a horizantal trail on the side of the hill and $\displaystyle \vec{\bigtriangledown}f$ is a horizantal vector perpendicular to the path. $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in dir $\displaystyle \vec{\bigtriangledown}F$. __________________________________________________ _____ If you are walking on the side of the hill in an arbitrary direction, your coordinates satisfy F=z-f(x,y)=0. A step in any direction from a point on the side of the hill must satisfy $\displaystyle dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=\vec{\bigtriangledown}F\cdot d \vec{r}=0$ Therefore $\displaystyle \vec{\bigtriangledown}F$ is normal to the hill at x,y,z. The same is true if equation of the hill (surface) is given by F(x,y,z)=0. ================================================== ====== If w=F(x,y,z) $\displaystyle dw=\vec{\bigtriangledown}F\cdot d\vec{r}$, and directional derivative is: $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}= \vec{\bigtriangledown}F\cdot \vec{n}$ dw=0 for an excursion along the surface w=c through x,y,z and $\displaystyle \vec{\bigtriangledown}F$ is normal to this surface. $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in direction $\displaystyle \vec{\bigtriangledown}F$. January 23rd, 2017, 05:29 PM   #6
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Please correct me if I am wrong...

$\displaystyle \vec{\bigtriangledown}F$ is the gradient at the point (x, y), which is also the steepest ascent slope of (z) at the point (x, y). The directional vector is the one that is being compared against the gradient and see how much the directional vector is deviated from the gradient.

$\displaystyle \vec{\bigtriangledown}F * \vec v$ is maximum if both the gradient and the directional vector have the same direction.

Quote:
 Originally Posted by zylo Think of z=f(x,y) as the equation of a hill (surface). At any point x,y at the base of the hill the hill has a height f(x,y). Let dx,dy be a small excursion in the x,y plane from the point x,y. The corresponding change in z is: $\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\vec{\bigtriangledown} f\cdot d\vec{r}$ Directional derivative: $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}=\vec{\bigtriangledown} f\cdot \vec{n}$ What does dz=0 mean? No change in z, so $\displaystyle \vec{\bigtriangledown}f$ is perpenticular to the curve z=c going through the point x,y. It's like being on a horizantal trail on the side of the hill and $\displaystyle \vec{\bigtriangledown}f$ is a horizantal vector perpendicular to the path. $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in dir $\displaystyle \vec{\bigtriangledown}F$. __________________________________________________ _____ If you are walking on the side of the hill in an arbitrary direction, your coordinates satisfy F=z-f(x,y)=0. A step in any direction from a point on the side of the hill must satisfy $\displaystyle dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=\vec{\bigtriangledown}F\cdot d \vec{r}=0$ Therefore $\displaystyle \vec{\bigtriangledown}F$ is normal to the hill at x,y,z. The same is true if equation of the hill (surface) is given by F(x,y,z)=0. ================================================== ====== If w=F(x,y,z) $\displaystyle dw=\vec{\bigtriangledown}F\cdot d\vec{r}$, and directional derivative is: $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}= \vec{\bigtriangledown}F\cdot \vec{n}$ dw=0 for an excursion along the surface w=c through x,y,z and $\displaystyle \vec{\bigtriangledown}F$ is normal to this surface. $\displaystyle \frac{\mathrm{d} w}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in direction $\displaystyle \vec{\bigtriangledown}F$. January 24th, 2017, 04:19 AM   #7
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Quote:
 Please correct me if I am wrong... $\displaystyle \vec{\nabla}F$ is the gradient at the point (x, y), which is also the steepest ascent slope of (z) at the point (x, y).
I would not use the word "slope". The gradient is the vector pointing in the direction of steepest ascent and whose length is the rate of ascent in that direction.

Quote:
 The directional vector is the one that is being compared against the gradient and see how much the directional vector is deviated from the gradient.
The "directional vector" is simply a unit vector pointing in the given direction. It does not necessarily have anything to do with the gradient.

The rate of change in a given direction is the dot product of the gradient and the "directional vector" in that direction.

Quote:
 $\displaystyle \vec{\nabla}F$ is maximum if both the gradient and the directional vector have the same direction.
I'm not sure how to respond to this! You may have the right idea but this is very confusingly worded! In the first place the gradient is a vector and vectors cannot be ordered so the gradient cannot be "maximum". Further the length of the gradient is a fixed number and it does not change with the directional vector. What is true is that the derivative in the direction of the directional vector is maximum if the directional vector is in the direction of the gradient. January 24th, 2017, 08:39 AM   #8
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Quote:
 Originally Posted by zylo Think of z=f(x,y) as the equation of a hill (surface). At any point x,y at the base of the hill the hill has a height f(x,y). Let dx,dy be a small excursion in the x,y plane from the point x,y. The corresponding change in z is: $\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\vec{\bigtriangledown} f\cdot d\vec{r}$ Directional derivative: $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}=\vec{\bigtriangledown} f\cdot \vec{n}$ What does dz=0 mean? No change in z, so $\displaystyle \vec{\bigtriangledown}f$ is perpenticular to the curve z=c going through the point x,y. It's like being on a horizantal trail on the side of the hill and $\displaystyle \vec{\bigtriangledown}f$ is a horizantal vector perpendicular to the path. $\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in dir $\displaystyle \vec{\bigtriangledown}F$.* __________________________________________________ _____
* Sorry about typo. F should be f:
$\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ is a maximum when $\displaystyle \vec{n}$ is in dir $\displaystyle \vec{\bigtriangledown}f$.

$\displaystyle \vec{\bigtriangledown}f$ points in direction in x,y plane in which increase in z is the greatest. This is confusing part. If you are at the point x,y,z on the hill, you can't actually move in this direction, you can only move on hill. If $\displaystyle \vec{\bigtriangledown}f$ is due east, look at your compass and take a step east, along the hill (you can't step off the hill or walk into it). That is the direction of steepest ascent (descent). January 25th, 2017, 05:26 AM #9 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 z = f(x,y) 1) $\displaystyle dz = f_{x}dx + f_{y}dy = \bigtriangledown f\cdot$ dr 2) $\displaystyle dz/ds = \bigtriangledown f\cdot$ n 3) $\displaystyle f_{x}dx + f_{y}dy - dz = N\cdot$ dr = 0 1) Gradient. $\displaystyle \bigtriangledown f$ points in dir of max increase of z; see Eq 2). 2) Directional derivative in dir of unit vector n. n=dr/ds, ds=|dr| 3) Equation of tangent to surface. N is normal to surface. dr is defined by respective Eq. January 29th, 2017, 01:18 PM   #10
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I have been contemplating all of your responses

First... please forgive my stubbornness and my lack of deeper appreciation of the intricacy of the beauty of mathematics. I am doing my best to catch up...

Let me focus on the Gradient alone. The gradient is a set of (formulas/vectors) to describe the steepest ascent at *any* point on a surface. If I enter inputs (a,b): Gradient(a, b) = c, then c remains as c as long as inputs are a and b.

Another-word: c is a fixed scalar value to describe the steepest ascent, at point (a, b) on a surface.

When a direction vector 'dot product' to a gradient at point (a, b). I assume we are evaluating the direction vector and see how much the direction vector 'deviates' from the gradient.

However zylo's response causes me to rethink the nature of the gradient.....

Quote:
 What does dz=0 mean? No change in z, so ▽⃗ f▽→f is perpenticular to the curve z=c going through the point x,y. It's like being on a horizantal trail on the side of the hill and ▽⃗ f▽→f is a horizantal vector perpendicular to the path.
His reply suggests when the result of the dot product is 0, then the gradient is perpendicular to the curve. The gradient is now normal to the surface because the direction vector is in parallel with the horizontal path of the 'hill'.

It seems that the perception of gradient is also affected by the direction vector. Tags derivative, directional, gradient, line, normal, question ### mathematics forums

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