At what point do you need Calculus?

[BeautifulLight]

I was under the impression that you needed Calculus to find the instantaneous velocity at a point when dealing with non-linear functions. I'm near positive that statement is false or at least only partially correct. f(x)=√x is a non-linear function, but the "line" (or curve?) is accelerating uniformly, so you can figure this out without the use of any Calculus.

Pick a point, B.

Pick another point to the left of point B and call it point A.

Pick yet another point on the curve, but this time move to the RIGHT of point B and call it point C.

You might have to modify the location of one of your points, but just make sure point A and point C are equidistant to your midpoint B.

Find the slope of AB.

Now find the slope of BC

(you probably see where this is going)

If you take the mean of the two, you should have figured out Vinstant at point B. This method should work for any non-linear function with a "line" that is accelerating uniformly, and you don't need any Calculus. So how should I edit my statement from my intro? You need Calculus to find the instantaneous velocity when dealing with a non-linear function that isn't accelerating uniformly?

 

[sk1105]

This method gives you an estimate for the velocity at B, but it is not perfectly accurate because you are approximating a smooth function with a series of straight lines. You should be able to reason that the approximation becomes more accurate with shorter lines (and therefore more of them), so we ask what would the velocity at B be if we had an infinite number of infinitesimally short lines. The value in this case is obtained by differentiation i.e. through calculus!

However, the method you have described is very similar to methods used to approximate functions in computer models, where we work numerically rather than algebraically.

 

[BeautifulLight]

This method gives you an estimate for the velocity at B, but it is not perfectly accurate because you are approximating a smooth function with a series of straight lines.

In this case, it is perfectly accurate because f(x)=√x is a smooth function.

Consider f(x)=x^2

I want to find the instantaneous velocity when x=7

So, f(7)=7^2

f(7)=49

(7,49) This is B, which will be my midpoint.

I went 2 units to both the left and the right of my midpoint to get my two other points A & C.

Point A = (5,25)
Point C = (9,81)

Slope of AB = 12
Slope of BC = 16

Take the mean of the two and you get 14. 14 is the instantaneous velocity at B.

Using Calculus, you end up with f'(x)=2x

When x=7, you get 14. 14 is the instantaneous velocity at B.

 

[Ishaan S]

Yes, this is correct. With parabolas, the acceleration would be constant; however, how would one possibly be able to know that this curve has a uniform acceleration without the use of calculus to prove it? Someone would have probably already told you that the curve had a constant acceleration. If not, then I would be more than eager to hear your explanation of a proof showing that the particle following this graph is accelerating uniformly without calculus.

 

[BeautifulLight]

Someone would have probably already told you that the curve had a constant acceleration. If not, then I would be more than eager to hear your explanation of a proof showing that the particle following this graph is accelerating uniformly without calculus.

Simply by observing the particle's displacement over time along the curve. Differentiate and measure the slope. If it's zero, then you know the particle is accelerating uniformly.

There are a few things that bother me though. What if I only differentiate the particle's displacement over time graph from time 5 to time 25. How do I know the particle didn't all of a sudden change it's acceleration at time 150? And just imagine what that curve may look like from time 5 to time 25. Maybe it looks like a sinusoid. How do I know that what really looks like a sinusoid is actually a sinusoid built of many many sinusoids. So now it looks like I'm going to have to observe the particle's displacement over time at really really really small increments.

This is what I don't understand about nonlinear equations with acceleration that isn't uniform. Can you have an acceleration of an acceleration? In other words, can you have a fundamental acceleration with other acceleration's superimposed on it?I probably should have asked this question back in Algebra, but if a line is defined by an equation, then you assume some kind of symmetry. Take 2x for example. That is a straight line, but not just any straight line, it's a straight line that continues in both directions forever. Same with trigonometric functions. Now how about we graph a particle's displacement over time. Let's not make this equation a straight line nor a curve who's acceleration is uniform. Get as random as possible, but still make sure it'll pass the vertical line test so we can define it. Okay, so now we have graphed the particle's position in space from time zero to time oh, 25. If you were to differentiate this graph, what happens after time 25? Am I suppose to start back at the particle's position at time zero? So it would oscillate (in both directions forever) from time 0 to time 25, correct?

 

[RaulTheUCSCSlug]

Keep in mind that certain functions will be piece wise functions in which there is a point that is seemingly "random" where the derivative at that point is something different. This is why you may not know whether the slope of the graph is the same through out, and that is why the fundamental theorem of calculus has certain guidelines that must be met.
Also, the function : ƒ(x) = x^½ does not "accelerate uniformly." Take a look at the derivative and you will see what I mean. Or maybe I am confused by what you mean? And differentiation will be helpful in the long run making lots of functions painless and easy!