Math Question For Clive. (2 Viewers)

[evilc]

Yeah, it is partial differentiation but the lecturer didn't really explain what that was. So is it just differentiating one variable?

yes

Also, you know how you differentiate the equation and put it equal to zero to find the maximum, doesn't that also find the minimum and you have to test to find if it is a maximum or minimum?

yeah, i was just lazy you really should check the second derivative to determine whether it is a maximum or minimum.


∂F/∂x = 49.52 - 10.96x + 1.4y + 21.5912

∂²F/∂x² = -10.96


second derivative is < 0 so .: the function is strictly concave down, hence a maximum value of x will occur when ∂F/∂x = 0

 

[mrbassman]

Yeah, it is partial differentiation but the lecturer didn't really explain what that was. So is it just differentiating one variable? Also, you know how you differentiate the equation and put it equal to zero to find the maximum, doesn't that also find the minimum and you have to test to find if it is a maximum or minimum?

yes it does find both but since its got a -x^2 term and the polynomial is of order 2 you can be pretty sure that the turning point F' gives you will be a max

 

[baker182]

Shit I'am screwed for Business Economics then, I only did General Maths

 

[SoCal]

yes

Sorry for all these questions, so does that measure the rate at which one variable changes with respect to the other held constant? What are the practical applications of that?

yeah, i was just lazy you really should check the second derivative to determine whether it is a maximum or minimum.


∂F/∂x = 49.52 - 10.96x + 1.4y + 21.5912

∂²F/∂x² = -10.96


second derivative is < 0 so .: the function is strictly concave down, hence a maximum value of x will occur when ∂F/∂x = 0

Damn, I can't remember what the second derivate measures. The derivative measures the slope of the line while the second derivate measures the rate of change of the line, is that right?

yes it does find both but since its got a -x^2 term and the polynomial is of order 2 you can be pretty sure that the turning point F' gives you will be a max

Yeah, true, I actually remember that!

Shit I'am screwed for Business Economics then, I only did General Maths

I did Extension 1 and I still can't remember this basic stuff. You don't even do Calculus in General Maths do you? If so, I would hate to be you!

 

[evilc]

Sorry for all these questions, so does that measure the rate at which one variable changes with respect to the other held constant? What are the practical applications of that?

I like being asked questions



Partial differentiation is just Differentiation of a function of two or more variables with respect to one of the variables - ie. seeing how the function changes with respect to one variable when the other variable(s) is held constant.

Damn, I can't remember what the second derivate measures. The derivative measures the slope of the line while the second derivate measures the rate of change of the line, is that right?

Second derivative measures the rate at which the gradient changes on a graph - ie. the rate of change of the rate of change of the original graph

The larger the magnitude of the second derivative is, the steeper the graph will be.
A good physical example of second derivative is acceleration.
Acceleration is the rate of change of velocity with respect to time.
Velocity is the rate of change of position with respect to time.

for example.
(x = position, v = velocity, a = acceleration)

x = 10t²
v = dx/dt = 20t
a = dv/dt = d²x/dt² = 20


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EDIT: I attatched a document (an extract from MATH1120 course notes) which should explain partial derivatives a lot better than i can online it has diagrams as well which are a bonus.

 

[baker182]

I did Extension 1 and I still can't remember this basic stuff. You don't even do Calculus in General Maths do you? If so, I would hate to be you!

Nope General Maths donst do Calculus, I'am totally screwed I might have to ring up a mate from Boarding school, who got a band E4 in 4 unit maths, to tutor me in the wonders of Calculus

 

[SoCal]

The larger the magnitude of the second derivative is, the steeper the graph will be.

OK, I understand everything except this. So when the second derivative is negative, why does it mean it is concave down? I remember doing this but can't remember why. I don't really need to know this but I am just curious.

EDIT: I attatched a document (an extract from MATH1120 course notes) which should explain partial derivatives a lot better than i can online it has diagrams as well which are a bonus.

Wow, I just had a look at that and I think it is going to go way over my head. Nevertheless, I will give it a read and see how I go.

 

[SoCal]

Nope General Maths donst do Calculus, I'am totally screwed I might have to ring up a mate from Boarding school, who got a band E4 in 4 unit maths, to tutor me in the wonders of Calculus

You don't really need to know much about it. It is just a bit of basic differentiation. You could just memorise the above example and do exactly the same thing in every question. Most of what I am asking is just asking why you do whatever you are doing because I can't remember.

 

[evilc]

OK, I understand everything except this. So when the second derivative is negative, why does it mean it is concave down? I remember doing this but can't remember why. I don't really need to know this but I am just curious.

Wow, I just had a look at that and I think it is going to go way over my head. Nevertheless, I will give it a read and see how I go.

Don't worry too much about the theory, just look at the diagrams and see how they fix one of the variables as a constant and effectively take a cross section through the graph parallel to the axis of the constant variable.

hmm...forgot to mention that:

(see attatched graphs, the red lines represent the gradient at various points on the graph)


positive second derivative = concave up

note how the graph has a negative slope on the LHS of the curve, and a positive slope on the RHS of the curve. The 2nd derivative is positive in this case because going from left to right the slope of the graph increases,from a negative slope to zero slope, and then to a positive slope.
***********************************



***********************************
negative second derivative = concave down

note how the graph has a positive slope on the LHS of the curve, and a negative slope on the RHS of the curve. The 2nd derivative is negative in this case because going from left to right the slope of the graph decreases, to zero slope, and then to a negative slope.
***********************************

 

[evilc]

Nope General Maths donst do Calculus, I'am totally screwed I might have to ring up a mate from Boarding school, who got a band E4 in 4 unit maths, to tutor me in the wonders of Calculus

just go through what has been posted in these threads and try and follow the process. ask me/mrbassman/jumb if you have any questions

 

[mrbassman]

Merethrond, you ask about why negative second derivative means concave down and trust us you dont want to know the theory behind it. It is involved with first principle. The most general answer is that Newton decided it should

 

[jumb]

It's easy to see the pattern when you graph f(x), f'(x) and f''(x).

 

[um..]

this thread is so fucked

 

[SoCal]

Thanks for that explanation Clive. I understand everything now and I will leave it at that.

Merethrond, you ask about why negative second derivative means concave down and trust us you dont want to know the theory behind it. It is involved with first principle. The most general answer is that Newton decided it should

Yeah, actually I though I learnt it in High School but after reading through Clive's explanation I don't think I actually did!

 

[baker182]

It's already been done but..anyway

Easy bit of partial differentiation here

call Profit(2) F(x,y)
call P2 x
call P1 y

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)

F(x,y) = 49.52x - 5.48x^2 + 1.40xy + 21.5912x - 5.488y - 195.1088

Partially differentiating the expression with respect to x (we only need to find out how the expression changes with respect to x, as y is held constant)

∂F/∂x = 49.52 - 10.96x + 1.4y + 21.5912

solve ∂F/∂x = 0 to find maximum

49.52 - 10.96x + 1.4y + 21.5912 = 0

10.96x = 71.1112 + 1.4y

x = (71.1112 + 1.4y)/(10.96)

x = 6.488 + 0.1277y

I just went through your answer, then I attempted the question by myself, without looking at your method and I got the same answer

 

[jumb]

I just went through your answer, then I attempted the question by myself, without looking at your method and I got the same answer

I have no idea what partial differntiation is, however, my method works and I was firstest, so I'm the bestsest

 

[evilc]

I have no idea what partial differntiation is, however, my method works and I was firstest, so I'm the bestsest

you will hit it semester 2 just an extension of normal differentiation to functions of two or more variables

 

[PrincessSJ]

I have no idea what partial differntiation is, however, my method works and I was firstest, so I'm the bestsest

pity your english method sucks

 

[Emily.]

I have no idea what partial differntiation is, however, my method works and I was firstest, so I'm the bestsest

idiot .

 

[jumb]

pity your english method sucks

I'm almost an engineer. If you talk to some other engineers, you will soon discover that english isn't part of the job description.