Partial derivatives & gradient

take... a... long... time...) you will reach an "extreme" of the function- a point where the gradient is (0,0) and that will be a local maximum or minimum.
  • #1
kingwinner
1,270
0
http://www.geocities.com/asdfasdf23135/advcal4.JPG

Let f(x,y)=depth.
What I've seen in the model solutions is that they used the estimate that
the partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100 = 1/4,
& the partial dervaitve of f with respect to y evaluate at (0,0) is equal to [f(0,100) - f(0,0)] / 100 =-1/2

Gradient of f at (0,0) = (1/4, -1/2)

So we suggest going in the direction (-1/4, 1/2) [answer] which is in the direction opposite to the gradient.

======================
Now there are three subtle points that I don't understand:

1. WHY can you use the estimate that partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100? This is like taking the secant line to be equal to the tangent line, but intution tells me that this estimate can be way way off...

2. Is there any possible way to find the formula for the function of the dome in question (half-sphere)?

3. (-1/4, 1/2) is in the direction opposite to the gradient, why does this give the answer? Pointing in the direction of maximum rate of decrease of f doesn't necessary mean that it's pointing to the absolute minimum of f (i.e. top of dome), right? If I am right, then (-1/4, 1/2) can't be correct...


Thanks for explaining! I really appreciate your help!
 
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  • #2
kingwinner said:
http://www.geocities.com/asdfasdf23135/advcal4.JPG

Let f(x,y)=depth.
What I've seen in the model solutions is that they used the estimate that
the partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100 = 1/4,
& the partial dervaitve of f with respect to y evaluate at (0,0) is equal to [f(0,100) - f(0,0)] / 100 =-1/2

Gradient of f at (0,0) = (1/4, -1/2)

So we suggest going in the direction (-1/4, 1/2) [answer] which is in the direction opposite to the gradient.

======================
Now there are three subtle points that I don't understand:

1. WHY can you use the estimate that partial dervaitve of f with respect to x evaluate at (0,0) is equal to [f(100,0) - f(0,0)] / 100? This is like taking the secant line to be equal to the tangent line, but intution tells me that this estimate can be way way off...
Yes, it might be- but do you have any more accurate information? That's the best you can do with the information given.

2. Is there any possible way to find the formula for the function of the dome in question (half-sphere)?
You know that the general equation of a sphere is [itex](x-x_0)^2+ (y-y_0)^2+ (z- z_0)^3= R^2[/itex]. You are given (x,y,z) for three points. Is that enough to determine the four unknown constants, [itex]x_0[/itex], [itex]y_0[/itex], [itex]z_0[/itex], and R?

3. (-1/4, 1/2) is in the direction opposite to the gradient, why does this give the answer? Pointing in the direction of maximum rate of decrease of f doesn't necessary mean that it's pointing to the absolute minimum of f (i.e. top of dome), right? If I am right, then (-1/4, 1/2) can't be correct...
No one said it was "correct" (would give the minimum depth)- it's the best you can do given the information. Derivatives can only give "local" information (what is true close to the given point) and can not directly give the "global" maximum or minimum. However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum.


Thanks for explaining! I really appreciate your help!
 
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  • #3
HallsofIvy said:
Yes, it might be- but do you have any more accurate information? That's the best you can do with the information given.


You know that the general equation of a sphere is [itex](x-x_0)^2+ (y-y_0)^2+ (z- z_0)^3= R^2[/itex]. You are given (x,y,z) for three points. Is that enough to determine the four unknown constants, [itex]x_0[/itex], [itex]y_0[/itex], [itex]z_0[/itex], and R?

3. (-1/4, 1/2) is in the direction opposite to the gradient, why does this give the answer? Pointing in the direction of maximum rate of decrease of f doesn't necessary mean that it's pointing to the absolute minimum of f (i.e. top of dome), right? If I am right, then (-1/4, 1/2) can't be correct...
No one said it was "correct" (would give the minimum depth)- it's the best you can do given the information. Derivatives can only give "local" information (what is true close to the given point) and can not directly give the "global" maximum or minimum. However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum.
2. So is it true that...
we need 4 points to determine a unique sphere, and that 3 points can NEVER determine a unique sphere?


3. "However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum" <-----I don't get this point. In our case, it's a dome, so there is one and only one local/absolute maximum, and therefore the local max is the absolute max...same thing...


Thanks a lot!
 
  • #4
kingwinner said:
2. So is it true that...
we need 4 points to determine a unique sphere, and that 3 points can NEVER determine a unique sphere?
Yes, that's true. Just as you need 3 points to determine a unique circle in the plane, you need 4 points to determine a unique sphere in three dimensions.


3. "However, "following" the direction of the gradient at different points will, eventually, lead to a local maximum" <-----I don't get this point. In our case, it's a dome, so there is one and only one local/absolute maximum, and therefore the local max is the absolute max...same thing...


Thanks a lot!
Yes, in this particular case, any "local" maximum is a "global" maximum. I wasn't referring to this situation only. You can "follow" the gradient by: choose a point. Calculate the gradient at that point. Move in the direction of the gradient a short distance. At this new point, repeat. Eventually (it can be implemented on a computer but tends to be slow) this will lead you to a local maximum but not necessarily a global maximum.
 
  • #5
Thanks for explaning! I am definitely understanding it better now.
 

1. What is a partial derivative?

A partial derivative is a mathematical concept that measures the rate of change of a function with respect to one of its variables, while keeping all other variables constant. It is denoted by ∂ (d in a stylized form) and is often used in multivariate calculus to analyze how a function changes in different directions.

2. How is a partial derivative different from a regular derivative?

A regular derivative measures the rate of change of a function with respect to a single variable, whereas a partial derivative measures the rate of change with respect to a specific variable while holding all other variables constant. In other words, a partial derivative takes into account the effects of other independent variables on the function.

3. What is the chain rule for partial derivatives?

The chain rule for partial derivatives is a rule in multivariate calculus that allows us to find the partial derivative of a function with respect to one variable in terms of the partial derivatives of other variables. It states that the partial derivative of a composite function is equal to the product of the derivative of the outer function and the partial derivative of the inner function.

4. What is the gradient of a function?

The gradient of a function is a vector that contains the partial derivatives of the function with respect to all of its variables. It represents the direction and magnitude of the steepest ascent of the function at a particular point. The gradient is often used in optimization problems to find the maximum or minimum value of a function.

5. How is the gradient used in machine learning?

In machine learning, the gradient is used to update the parameters of a model during the training process. It is calculated using the partial derivatives of the loss function with respect to the model's parameters. By following the direction of the gradient, the model can iteratively adjust its parameters to minimize the loss function and improve its performance.

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