Finding the average slope of a parable

[mimi.janson]

Homework Statement

A parable is f(x) = 0,01x²-2x+100

it beginswith the cut with the y-axis and ends when it cuts the x-axis.

Questions :
a) find the average slope of the parable.
b) Find the place on the parable where the average slope is present.

Homework Equations

here i know i have to find two places on the parable and dfferentiate them and lastly divide by 2 so i can find the average.

The Attempt at a Solution

The problem is that i am not sure in which order to do the things.

I did start and differentiated it:

f(x)=0,01x²-2x+100
f'(x)=0,02x-2

then i wasnt sure because i could use the one not defferiantated and say that when my x is zero my y will be 100 couse f(0)=0,01*0²-2*0+100 is 100

and after that i didnt know how to find the one for x. besides if i do it this way i don't get what there is to differentiate...couse i end up with one number...like 100.

anyway i tried to draw it too so that it would be easier...i marked the two places where i wanted to find the slope with black...i had thought of finding it and then adding them and dividing by two ...

 

[NascentOxygen]

This is an interesting question. We need to work out what is meant by "average slope". Have you encountered "average slope" in any other questions recently?

 

[mimi.janson]

This is an interesting question. We need to work out what is meant by "average slope". Have you encountered "average slope" in any other questions recently?

oh did i translate it wrong maybe ?
Sorry if my english is a big fail this time. but i will try to show you what i meant on a picture...


when you differentiate you get how much they tipping ...the tinny lines i made on the picture on the function which i think is called a parable in english. I thought the tinny lines are what is called a slope. so i had to add all the tinny lines and divided by the number of lines that i added...my problem is I don't get how i can find the tinny lines in so many points on the parable whne i only have that single f(x) because then i will also have only 1 f '(x)

I hope you understand me better now and I am really sorry about my english...

I really really hope you will help me

 

[NascentOxygen]

I understood you perfectly well. The curve is a "parabola".

I can see the difficulty, and I'm still pondering the answer.

 

[NascentOxygen]

Maybe we could say "average slope" is the slope of a triangle's hypotenuse when that triangle has the same area as the area under the curve, and is on the same base? The result would be the same as finding the average height of the curve, i.e., area under the curve ÷ length of x-interval.

 

[gneill]

Suppose you measured the slope at n locations along the curve, equally spaced along the x-axis between 0 and 100 (where the portion of the curve under consideration ends). An average would be the sum of those slopes divided by n. Can you think of a way to do this over a continuum of samples (n → ∞)?

 

[mimi.janson]

Maybe we could say "average slope" is the slope of a triangle's hypotenuse when that triangle has the same area as the area under the curve, and is on the same base? The result would be the same as finding the average height of the curve, i.e., area under the curve ÷ length of x-interval.

i don't think that it's the way my teacher want's me to do it.
We are having abour differentiating, so that's what i have to use.

 

[mimi.janson]

Suppose you measured the slope at n locations along the curve, equally spaced along the x-axis between 0 and 100 (where the portion of the curve under consideration ends). An average would be the sum of those slopes divided by n. Can you think of a way to do this over a continuum of samples (n → ∞)?

I have tried to measure.

I thought that:
f(x)=0,01x²-2x+100
f' (x)=0,02x-2
but then i began thinking..."this one i don't know on where exactly it is on the parable so mabe i have to first find the two points (see the first pic i added the blue one) on the parable without differentiating and after that differentiate them "

i tried to figure out how i could find what my f(x)=0,01x²-2x+100 is when my x is zero.

this one was easy because by simply looking at that pic i could see that it would be 100.
And i could also just put x=0 →
f(0)=0,01*0²-2*0+100
→
f(0)=100

and i cannot differentiate 100...that would just be zero again !
and i still am not able to put it in some kind of formula to that i can differentiate it. Besides i found what y is when x is zero, but what x is when y is zero that was very hard.

An average would be the sum of those slopes divided by n. Can you think of a way to do this over a continuum of samples (n → ∞)?

i have thought about finding the sum of the slopes and divided by n number of slopes...but i don't know how to find that sum ...its the exactly problem. it is for example when i have to differentiate that confuses me.

 

[gneill]

The function f'(x) gives you the slope at any point on your curve f(x), even when x=0 and at x=100. So finding f'(x) fro f(x) is the differentiation step.

Summing and averaging an infinite number of samples is something you can accomplish with an integration. Suppose you have some function g(x) and you want to find the average value of g(x) over some domain 0 ≤ x ≤ a. The area under the curve between x=0 and x=a is equal to the sum of g(x)*dx for all the individual dx's between 0 and a, in the limit as dx → 0. That's just the integral ∫g(x)dx. But the sum of all the dx's adds up to a. So if you divide the sum of all the g(x)*dx by a you'll have the average value of g(x). That is, average = (1/a)*∫g(x)dx.

You've got a function f'(x) which gives you the slope for any x, and you know the domain over which you want its average...

 

[mimi.janson]

The function f'(x) gives you the slope at any point on your curve f(x), even when x=0 and at x=100. So finding f'(x) fro f(x) is the differentiation step.

Summing and averaging an infinite number of samples is something you can accomplish with an integration. Suppose you have some function g(x) and you want to find the average value of g(x) over some domain 0 ≤ x ≤ a. The area under the curve between x=0 and x=a is equal to the sum of g(x)*dx for all the individual dx's between 0 and a, in the limit as dx → 0. That's just the integral ∫g(x)dx. But the sum of all the dx's adds up to a. So if you divide the sum of all the g(x)*dx by a you'll have the average value of g(x). That is, average = (1/a)*∫g(x)dx.

You've got a function f'(x) which gives you the slope for any x, and you know the domain over which you want its average...

But why are you speaking about the area under the curve ? I am doing differential and not integral math ?

and could you show me an example by calculating of what you are trying to make me understand couse my english is a bit bad. and the numbers are the same all around the world ...

 

[gneill]

But why are you speaking about the area under the curve ? I am doing differential and not integral math ?

The problem needs both differentiation and integration. Differentiation is required in order to find a function that gives the slope for any value of x. Integration is needed to find the average of an infinite number of samples along that curve. The integral of curve yields the area under that curve.

and could you show me an example by calculating of what you are trying to make me understand couse my english is a bit bad. and the numbers are the same all around the world ...

The average of the function sin(x) for 0 ≤ x ≤ $\pi$/2 is:

$$avg = \frac{1}{\pi/2} \int_0^{\pi/2} sin(x) dx $$

 

[mimi.janson]

The problem needs both differentiation and integration. Differentiation is required in order to find a function that gives the slope for any value of x. Integration is needed to find the average of an infinite number of samples along that curve. The integral of curve yields the area under that curve.


The average of the function sin(x) for 0 ≤ x ≤ $\pi$/2 is:

$$avg = \frac{1}{\pi/2} \int_0^{\pi/2} sin(x) dx $$

I really don't think it is that way. Because it has to be more simple and we didnt even learn ingetration.

 

[gneill]

I suppose you could always just draw a straight line between the end points and use its slope as the average. In fact, there's probably a theorem that states that the average slope of a curve between two points is the same as the slope of the straight line drawn between those points. That's something I'd have to investigate.

EDIT: Well, I remembered correctly. See the page at this link.

 

[NascentOxygen]

The problem needs both differentiation and integration.

True, but ... https://www.physicsforums.com/images/icons/icon6.gif

If we are going to integrate what we have just differentiated, then we arrive back at the quadratic. So if the integration is over the region between the specified points, then all we'll end up having to do is evaluate the parabola's quadratic at both these points, then subtract and average.

That sounds like what your cite probably shows, though I haven't yet clicked on it.

In essence, everyone here is right! https://www.physicsforums.com/images/icons/icon10.gif

 

[NascentOxygen]

Suppose you measured the slope at n locations along the curve, equally spaced along the x-axis between 0 and 100 (where the portion of the curve under consideration ends). An average would be the sum of those slopes divided by n.

Differentiating, then using integration over 0→100 gives an average of the slope as -1

This result is identical with a line joining (0,100) to (100,0), this line has slope = -1

So mimi.janson your "shortcut" method is valid and gives the correct answer, though for reasons which we didn't appreciate at the time you wrote them.

I wrote:

In essence, everyone here is right! https://www.physicsforums.com/images/icons/icon10.gif

That is, everyone but me.