Calc AB to BC: What is a Gradient?

  • Thread starter thharrimw
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In summary: no?well then you might want to look into it. vectors can be a little more complicated than gradients.i don't really know what vectors are. so you're telling me that gradients and vectors are two different things?yes.
  • #1
thharrimw
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I am a high school senior and I taught myself calc AB and got a 4 on the test last year but now I'm in my high school's calc class and i already know everything that is in the couse so I'm perparing to take the BC test and in my calc book it dosn't give a good explanation of what a gradent is or where it is used so firts of all what is a gradent? and then are they on the BC calc test? where are they used?
 
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  • #2
without getting into formulae, the general concept is that a gradient is any kind of slope. slope of a line is a gradient. it is commonly applied to more complex relationships. in engineering it is commonly used to describe the rate of change of some parameter (temperature, pressure) over a distance. also, it is easily viewed on a topographic map when you have high gradients. I'll let others get into the more complex discussion.
 
  • #3
so a gradent is simmler to a deritive.
 
  • #4
it is a form of the derivitive. if you calculate the gradient for a function f(x,y,z) (a 3-d surface) you basically get a vector in terms of the unit vectors i,j,k and variables x,y,z. when you plug in a point (x,y,z) on the surface into the gradient vector, you get a vector that points in the direction of steepest ascent and the magnitude of the vector is the value of the slope of the surface in that direction at that point (x,y,z).
 
  • #5
so the partial deritives that I've been learning how to find are used to find gradients?
 
  • #6
yes. and to correct myself above a 3-d surface would be z=f(x,y) not f(x,y,z)
 
  • #7
how do you use partial deritives to find a gradient? then what are gradents used for (pratical applacations)?
 
  • #8
  • #9
In England, it is common to refer to the derivative of a function of a single variable as the 'graident'. In the United States, that is normally reserved for functions of several variables. If f is a function of x, y,z, the "gradient" of f, sometimes abbreviated grad f, sometimes symbolized as [itex]\nabla f[/itex], is the vector each of whose components is a partial derivative of f: [itex]<\partial f/\partial x, \partial f/\partial y, \partial f/\partial z>[/itex]. You are correct in that it is the derivative. It is possible for a function which is NOT differentiable at a point to have partial derivatives there but the gradient cannot exist where the function is not differentiable.

If f(x,y,z) is differentiable at a point, then it has a derivative in any direction- and the derivative in the direction of the unit vector [itex]\vec{v}[/itex] is [itex]\nabla f\cdot\vect{v}[/itex]

The gradient of f always points in the direction of fastest increase (the derivative is largest) and its length is that largest derivative.
 
  • #10
so you add up all the partials to find the gradient of a function?
do i need to know how to use poler equations?
gradient is just another word for slope of grade?
 
  • #11
Where do you see anything about adding up the partials? Gradient is a vector thus it has components, is that what you are referring to? When you do "add up partials", are you thinking of divergence? In which case you have [tex] \nabla \cdot \mathbf{F} = \frac{\partial \mathbf{F}}{\partial x} + \frac{\partial \mathbf{F} }{\partial y} + \frac{\partial \mathbf{F}}{\partial z} [/tex] assuming [tex] \mathbf{F} = F_{x} \mathbf{i} + F_{y} \mathbf{j}+ F_{z} \mathbf{k} [/tex] is a cont. diff. vector field
 
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  • #12
yah so
[tex] \nabla \cdot \mathbf{F} = \frac{\partial \mathbf{F}}{\partial x} + \frac{\partial \mathbf{F} }{\partial y} + \frac{\partial \mathbf{F}}{\partial z} [tex]
is just stating that each parital is a just one component of a vector... i wasn't thinking about a 3D vector... I am still getting used to dealing with 3D equations.so for a 2D vector you could write it as [tex] \nabla \cdot \mathbf{F} = \frac{\partial \mathbf{F}}{\partial x} + \frac{\partial \mathbf{F} }{\partial y} [tex]
 
  • #13
sory i don't know how to use LaTeX but Grad F(x)= dF/dx+dF/dy
Right?
 
  • #14
No, that's divergence i.e. the grad operator dotted with F where "dotted" means dot product.
 
  • #15
how do you find a dot product and what is a dot product?
 
  • #17
i've done vectors but only in physics. I've herd of dot product but i only have a book and the internet to learn from and i haden't heard of it until now. I am alredy over my high school calc teachers head. she can't do anything with 3D.I'm also working on diffy Q's right now and basic limits for my actual calc class. I don't know what i should know or where to go next.
 
  • #19
I think it's admirable that you are trying to get ahead of the class, but you really need to do so while following some kind of a curriculum or study guide. If you say you read about gradients in your calc book they had to have discussed vectors on some level (or it's a really terrible book and you should get another one). Assuming they have, you should know vector notation. In Hall's post, he didn't mention sums anywhere, I'm still confused where you got sums of anything which led me to think you were talking about divergence if you just mistook what he wrote to mean sum instead of a vector with 3 components, then ignore what I said about divergence and consequently dot product. But as I said earlier and malawi said again just now, you should be able to look up definitions in your book, google, wiki, mathworld, planetmath, etc. IF a definition is not clear, ask a specific question and we can try to help.
 
  • #20
the vector notation I'm used to if V(x,y)=(cos(F(t)),sin(F(t))) i used that in physics but I've just got into 3D graphs and 3D vectors and i assumed the notation would be V(x,y,z)=(DF/Dx,DF/Dy,DF/Dz). i didn't know that vector notation used + i assumed that it ment to add the vectors componets
 
  • #21
Vector notation doesn't use +! Where do you see Halls using + when defining gradient?
 
  • #23
thharrimw said:
sory i don't know how to use LaTeX but Grad F(x)= dF/dx+dF/dy
Right?

This is what you wrote and this is what I was responding to, the wikipedia article uses versors and it explains that either form is correct you are missing the i and j on your partials, do you see the difference?
 
  • #24
oh now i see i mussed of missed that... so that was in the Quaternion format sorry.
 

FAQ: Calc AB to BC: What is a Gradient?

What is the difference between Calc AB and BC?

Calc AB and BC are different levels of calculus courses. Calc AB covers the fundamentals of single-variable calculus, while Calc BC goes into more depth and covers topics such as integration by parts, parametric equations, and polar coordinates.

What is a gradient in Calc AB or BC?

A gradient in calculus is a vector that points in the direction of the steepest increase of a function at a specific point. It is calculated by finding the partial derivatives of the function with respect to each variable and combining them into a vector.

How is the concept of gradient used in real life?

The concept of gradient is used in various fields such as physics, engineering, and economics. For example, in physics, the gradient is used to calculate the electric field around a charged object. In economics, it is used to determine the optimal production levels for a company.

What are some common applications of gradient in calculus?

The gradient is commonly used in calculus to optimize functions. For example, it can be used to find the maximum or minimum values of a function, or to find the direction of steepest descent for a multivariable function. It is also used in vector calculus to calculate line integrals and surface integrals.

Is it necessary to understand gradient in Calc AB before moving on to Calc BC?

Yes, understanding the concept of gradient is crucial in both Calc AB and BC. It is a fundamental concept in calculus and is used in many different applications. Without a solid understanding of gradient, it can be difficult to grasp more advanced topics in Calc BC.

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