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January 22nd, 2017, 10:46 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2  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  
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2 
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:
 
January 22nd, 2017, 06:42 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 
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=zf(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  
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2 
Thanks! This is a very very good explanation for me. Quote:
 
January 23rd, 2017, 05:29 PM  #6  
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2 
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:
 
January 24th, 2017, 04:19 AM  #7  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,240 Thanks: 884  Quote:
Quote:
The rate of change in a given direction is the dot product of the gradient and the "directional vector" in that direction. Quote:
 
January 24th, 2017, 08:39 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100  Quote:
$\displaystyle \frac{\mathrm{d} z}{\mathrm{d} s}$ 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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 
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  
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2  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. Anotherword: 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:
It seems that the perception of gradient is also affected by the direction vector.  

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