gradient help needed

[dpt90]

first of this is part of my assignment , secondly please forgive me if i posted this in the wrong place.... this just seemd like the place to post this.

 

i have 2 curved graphs S and V..... and one is increasing more rapidly than the other over a distance at one points in the graph and the other after a few values starts to increase faster than the other one. how would i find the values for the distance where S Gradient is ascending faster than V, and when V Gradient is ascending faster than S...... i only need a push in the right direction please....

 

i hope i said that coherently . my teacher said to use change in y/ change in x or something but im lost really.... =[

[BigMoosie]

Your wording is a little confusing but this sounds like a simple calculus problem, here is what I assume you are asking:

 

S and V are single-valued functions defined over the reals. For what values of 'x' will the gradient at S(x) be greater than the gradient at V(x)?

 

d/dx F(x) will provide the gradient so this will occur when:

 

d/dx S(x) > d/dx V(x)

 

or:

 

d/dx {S(x) - D(x)} > 0

[dpt90]

S and V are single-valued functions defined over the reals. For what values of 'x' will the gradient at S(x) be greater than the gradient at V(x)?

 

you say it so much better than me .... i suck lol..... yes that is what im trying to do. and also finding the value of 'x' where the gradient of V(x) is greater than the gradient at S(x).

 

if you feel like it, would you please like to educate me on the lil formulas you gave me ..... they seem to be a bit foreign to me, making little sense......

 

i looked at the math tutorials on the site and i read up on this. http://www.scienceforums.net/forum/showpost.php?p=423206&postcount=6

 

it looks like what i need... but i need some one to dumb it out for me a little bit..... i am smart... just a little slow

[BigMoosie]

if you feel like it, would you please like to educate me on the lil formulas you gave me ..... they seem to be a bit foreign to me, making little sense......

 

i looked at the math tutorials on the site and i read up on this. http://www.scienceforums.net/forum/showpost.php?p=423206&postcount=6

 

it looks like what i need... but i need some one to dumb it out for me a little bit..... i am smart... just a little slow

 

d/dx f(x) means 'the derivative of f'. This is calculus. If I know what S and V are then I may be able to provide further help, otherwise all I can do is explain what calculus is. But if I were to provide a *perfect* explaination it would likely not be very different to whatever your textbook provides, so I will only give a brief overfiew.

 

When you find the gradient of a line, such as f(x)=2x (in this case the gradient is 2), what is it exactly that you have? You have a number which indicates how quickly the height of the function changes as you move along the x-axis. In this case, move across 1 unit and up 2 (rise/run).

 

Other ways to express rise/run:

 

(change of rise) / (change of run)

(change of y) / (change of x)

dy / dx

 

Here 'd' is not a variable but an operator, kind of like +-/* are operators, a variable that means 'very small change'. This can also be expressed as:

 

d f(x) / dx

 

But often the f(x) is moved to the right of the fraction so that it is clearer to see what is happening:

 

d/dx f(x)

 

So now 'd/dx' can stand alone and can be read as 'the derivative of'. Luckily with a straight line the gradient doesn't change, hence:

 

d/dx (2x) = 2

 

That is why there is a constant, it doesn't matter what 'x' is, it will always be 2. If your function is not a straight line (perhaps a porabola, hyperbola or whatever) then the gradient will be different depending on what part of the function you look at, so for derivatives of these functions the answer will be expressed in terms of 'x'.

 

The interesting thing about derivatives is that if you add two functions together, you can also add their derivatives together. For instance if you add 2x and 3x then the gradient is (2+3). This works for any function, not just lines. That is why I was able to move V(X) to the left hand side of the equation, you can add and subtract them fine. (sorry just notice I made a typo in my first post writing 'D(x)' instead of 'V(x)).

 

There are lots of tricks for remembering how to manipulate different expressions to find their derivatives, and there is also a general method that can be used for all functions but is more tedious. These you will need your textbook for.

[dpt90]

sry to keep nagging but here is a picture of what im trying to find.

 

 

 

at one point surface area lines gradient is going faster than the Volumes, i need to know the x value for the point where the volume takes over.... is what i looked at before relavent .... it looked a bit off....

Edited March 12, 2009 by dpt90

[the tree]

Well okay - knowing vaguely what S and V represent is a little more helpful. If we knew what they were the surface area and volume of we could basically do the whole problem.

 

All you really need is to find the solutions for:

 

[math]\frac{d}{dx}S(x)=\frac{d}{dx}V(x)[/math]

 

Which will give the points where their respective gradients match, although that means recalling the explicit formulae for the surface area and gradient of this mystery object, and doing basic differentiation on each.

[dpt90]

they are just quantities, i got by using the excel formula i got from the book. and the graph was made in excel.

 

Length	       Surface area	      Volume
0	        =6*(A2^2)              =A2^3
1	        "                       "
2       	"                       "
3	        "                       "
4	        "                       "
5	        "                       "
6	        "                       "
7	        "                       "
8	        "                       "
9	        "                       "
10	        "                       "

 

they are not the Volume or surface area of anything.

please do not do the whole thing. i want to do it myself.

Edited March 12, 2009 by dpt90

[the tree]

Okay those are the formulae for the surface area and volume of a cube - not "not of anything".

 

Now since you've revealed that this is a computing project, there's two ways you can approach this - but ideally you'd do both:

 

First, the analytic approach by differentiating both formulae and finding exactly where the gradients match - as per the equation in my last post.

 

Secondly, to do it numerically:

 

Create a fourth and fifth column in the spread sheet, change in surface area and change in volume.

 

Calculate each value in the change in surface area as the difference between the surface area currently, and previously - divided by the size of the increment in length*. Similarly for the change in volume. That will give you the gradient (as described by BigMoosie).

 

Do you want a computational approach to identify when one takes over the other, or to identify it yourself?

 

*currently 1, but you might want to improve on that.

[dpt90]

sorry for my late reply, but personal things kept me away from internet access. ok so now i have the gradients for each of the points.

im sorry but could you please explain to me how to do simple differentiation.

[the tree]

There's a satisfactory explanation here.

[dpt90]

thanx tree, sorry for the annoyence....