August 14, 2020, 05:28:05 pm

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Re: Need help with rates of change question
« Reply #15 on: July 05, 2020, 08:20:09 pm »
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Ahh, okay. If you don't know what derivatives are disregard what I said.
However if you are aware of what derivatives are briefly, when thinking about the derivative of this function, which is velocity in this case, it's important to understand that when t=0, that doesn't mean the velocity is 0. In short that's equivalent to saying f(0)=f'(0) which isn't the case so when t=0, the function is still increasing.
Ah, alright, makes sense now.
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S_R_K

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Re: Need help with rates of change question
« Reply #16 on: July 05, 2020, 09:49:21 pm »
+1
t=0 is still part of the interval at which the function is increasing. You shouldn’t think of it as positive or negative, you should think of it as increasing or decreasing. Because positive derivative just = increasing = tangent at that point is positive. Because that’s what we’re relatively looking for when applying calculus to motion and velocity is just the derivative.

Increasing does not imply positive derivative.

The displacement is increasing at t=0, but the velocity is undefined (here assuming that both left- and right-hand derivatives must be defined for the velocity to be defined).

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Re: Need help with rates of change question
« Reply #17 on: July 06, 2020, 10:49:33 am »
+2
Increasing does not imply positive derivative.

The displacement is increasing at t=0, but the velocity is undefined (here assuming that both left- and right-hand derivatives must be defined for the velocity to be defined).
You are wrong.
At t=0, v does not equal 0. The derivative of this function is in the form y’=ax+b. So at t=0 the velocity is equal to b. A non zero value. This function is very obviously continuous so there is no way that the derivative could be undefined anywhere on it.
Here is the graph of the velocity:
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keltingmeith

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Re: Need help with rates of change question
« Reply #18 on: July 06, 2020, 11:20:45 am »
+2
You are wrong.
At t=0, v does not equal 0. The derivative of this function is in the form y’=ax+b. So at t=0 the velocity is equal to b. A non zero value. This function is very obviously continuous so there is no way that the derivative could be undefined anywhere on it.
Here is the graph of the velocity:
(Image removed from quote.)

Having not looked at the question in... Well, question.... I would like to chime in and add:

The definition of "increasing" and "decreasing" has nothing to do with the derivative. A function is defined as increasing over an interval I if for every a and b in I, a<b ==> f(a)<=f(b). A function is instead defined as decreasing if for a<b, f(a)>=f(b)

This is confusing, I'm sure, but what I want to point out is that under this definition, the graph of f(x)=|x+2|-|x-4| is always increasing.

At this link, I've drawn the function in red and its derivative in blue. We know the function is increasing if every bigger x-value returns a y-value that is the same or bigger. That's a check for every real number. But notice that the derivative isn't even defined for every real number, and in fact is 0 everywhere that isn't between -2 and 4. So, you'd think that this function ISN'T increasing much at all based on its derivative, even though it's CONSTANTLY increasing

Looking at the question your interested in, this gives a very important note - it doesn't make sense for something to be increasing at a single point. Something can only be increasing over an interval.

Also also, that graph you drew is very circular, and but very parabolic, soz
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S_R_K

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Re: Need help with rates of change question
« Reply #19 on: July 06, 2020, 12:02:21 pm »
0
You are wrong.
At t=0, v does not equal 0. The derivative of this function is in the form y’=ax+b. So at t=0 the velocity is equal to b. A non zero value. This function is very obviously continuous so there is no way that the derivative could be undefined anywhere on it.
Here is the graph of the velocity:
(Image removed from quote.)

Sorry, I am not confused at all about this. I think you are reading too close a connection between continuity, differentiability, and increasing / decreasing. These notions can all come apart in various ways, although the ways they do so is generally considered pathological for VCE maths, and is more typically studied in a university level analysis course.

1) Continuity does not imply differentiability, so quoting the continuity of the function does not prove anything.

2) Even though the function f(x) = ax + b is differentiable for all x in its maximal domain (all reals), the domain of the function given in the question is not all reals, so that point is moot. Questions about differentiability need to be considered in the context of the given domain. Here the domain of the displacement function is t ≥ 0.

3) In VCE maths, the convention is the derivatives are undefined at endpoints of closed (or half-closed) intervals, since both left- and right-hand limits of the difference quotient are undefined. Since t = 0 is an endpoint of the domain, we should consider the derivative undefined there.

4) As KeltingMeith points out above, increasing does not imply positive derivative (although the converse is certainly true). I note that KeltingMeith's example illustrates only that non-decreasing does not imply positive derivative, but it's also true that strictly increasing (ie. replace the partial order in the definition with a strict order) does not imply positive derivative. The standard counterexample is f(x) = x^3, where the function is strictly increasing for all real numbers x, but the derivative is not positive at x = 0.

Hence, I think the best thing to say about the question in the OP is that the answer contradicts the information in the question. If the object is initially at rest, then v = 0 at t = 0, but because it is projected with positive velocity, the velocity function is not continuous at t = 0, and hence we have positive velocity for 0 < t < 2.5.

keltingmeith

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Re: Need help with rates of change question
« Reply #20 on: July 06, 2020, 01:14:13 pm »
+2
It is a proven and taught fact that a function increasing or decreasing is based on the sign of its derivative

Here's proof that you're incorrect. The FACT is that the derivative is not present in the definition of an increasing function whatsoever. I've even provided a counter-example of a function with a discontinuous derivative which is always increasing. The Wolfram article even says this, giving another example of a function that is always increasing despite having a function that doesn't have a defined derivative at all points.

I can prove it right now in this example:
-snip-
This is an example function. And I’m finding where it increases/decreases. First you have to solve for when the derivative of the function is 0 or undefined. I plot these values on a number line and I use the principle of continuity to find out whether the derivative is positive or negative. Near those values, the + means increasing, the - means decreasing
Let’s see if I’m correct:
-snip-
There’s the actual function.

And now we see that your method is wrong - the domain for which your function is increasing INCLUDES the point x=+1. Firstly, f(+1)=1/(1-2-3)=-1/4. -1/4 is BIGGER THAN OR EQUAL TO every other point on the interval x in (-1,1) - so, we would include it in the interval, so the function is ACTUALLY increasing on the interval x in (-1,1]. You are correct that the derivative can be used (and is often easier to do so) to find when a function is increasing, but because the definition IS NOT f'(x)>0, you need to be careful when applying it.

That’s because it’s a point of inflexion. Which is where the second derivative is equal to 0.

What's that gotta do with the price of fish in China? It's not even clear which point of S_R_K's you're trying to argue with here. What you've said is correct, but that doesn't mean ANYTHING that they've said is incorrect.

You have missed my point completely. It doesn’t even matter the domain. The instantaneous velocity will still be positive at t=0
Do you realize that the velocity of an objects displacement is equal to the first derivative of the function of said displacement?

No, you are missing the point here. The domain DOES matter, because it determines where a derivative is defined. Again, as S_R_K told you before, this function is not defined for t<0 - as a consequence, this means that the function's derivative CANNOT be defined for t=0, because the left-hand limit (or the limit from the negative side) cannot be defined. Remember, even though we have a bunch of fancy rules for derivatives, the fundamental truth of the derivative is:

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

If that limit does not exist, then the derivative does not exist. And in this case:

$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{f(0+h)}{h} - \lim_{h\to 0}\frac{f(0)}{h}$

Since f(0+h) is NOT defined for h<0, then the limit CANNOT exist.
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Re: Need help with rates of change question
« Reply #21 on: July 06, 2020, 03:05:22 pm »
0
Without any math mumbo jumbo, what is the verdict? Is the answer 0<t<2.5 (or whatever number it is) or is it 0<=t<2.5.

thanks
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Re: Need help with rates of change question
« Reply #22 on: July 06, 2020, 03:08:55 pm »
0
Without any math mumbo jumbo, what is the verdict? Is the answer 0<t<2.5 (or whatever number it is) or is it 0<=t<2.5.

thanks
It's on the domain ]0,2.5[ thats where v is positive. Otherwise it is either negative or undefined.
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keltingmeith

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Re: Need help with rates of change question
« Reply #23 on: July 06, 2020, 03:10:58 pm »
0
Without any math mumbo jumbo, what is the verdict? Is the answer 0<t<2.5 (or whatever number it is) or is it 0<=t<2.5.

thanks

Maths mumbo jumbo you should understand 😏 the answer is probably the first one, but there is a chance the writer had something in mind when they said it was the other one. My response is the whole question is dumb and ambiguous and you should just ignore it
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