Post by: a weaponized ikea chair on July 03, 2020, 03:26:01 pm
See attached. Here's what I understand so far:
A) I understand a, except... well see b..
B) The answer states that it is [0,2.5). How can the instantaneous velocity be positive at t=0 if 0 is not positive or negative. The question states that the particle starts at rest so at t=0 there is no velocity, and thus is not positive.
C) Answer says 6m; if you look at graph, it goes slightly higher than 6; i put 6.5.
D) I understand this question.
E) Here it gets wishy-washy. So the answer is 3, apparently, but since the equation of the parabola is not known, I can't take the derivative and are stuck how they inferred this.
Any help is gladly appreciated.
Post by: Evolio on July 03, 2020, 03:56:01 pm
For B) you need to understand that velocity is the change in displacement over change in time. So when they ask for the values of t where the instantaneous velocity is position, they're essentially asking you for the domain where the displacement is positive, that is where the particle is moving forward but not backward. It is including 0 as that is the start of the domain and 2.5 here is non-inclusive because there is no slope or gradient at the top of the parabola since it is a turning point (stationary point).
For C) Yeah, I believe you are correct. In the actual Methods exam, the reading of the graph for the values won't be this ambiguous and unclear.
For E) You don't need the equation of the parabola. Remember: velocity(speed)= change in displacement/change in time. / denotes divided by.
Here, change in time is 1-0 seconds= 1 second so this would be the denominator. For change in displacement, well, at 0, the particle has 0 displacement but at 1 it has a displacement of 3 m. So, 3-0 m=3 m. This is our numerator. Since, 3/1=3, that is our instantaneous velocity.
Hope this helps!
Post by: 1729 on July 03, 2020, 04:36:36 pm
Hi,Okay I am going to give you two ways to approach this problem
See attached. Here's what I understand so far:
A) I understand a, except... well see b..
B) The answer states that it is [0,2.5). How can the instantaneous velocity be positive at t=0 if 0 is not positive or negative. The question states that the particle starts at rest so at t=0 there is no velocity, and thus is not positive.
C) Answer says 6m; if you look at graph, it goes slightly higher than 6; i put 6.5.
D) I understand this question.
E) Here it gets wishy-washy. So the answer is 3, apparently, but since the equation of the parabola is not known, I can't take the derivative and are stuck how they inferred this.
Any help is gladly appreciated.
So first of all you have the following conditions:
s(0)=0
s(5)=0
s’(2.5)=0
So you can make this differential equation
Integrating both sides will give you this:
Now we know that s’(2.5)=0, so we can plug that in:
So we have:
We can integrate both sides of the differential equation again to get this:
So since we know that c is -2.5a we can make a system of equations.
However this is a very much complicated method of doing it, defintely go with Evolios method, I'm only saying this assuming it's not an exam. However if this was an exam I wouldn't think about building a differential equation. Also (b) is (0, 2.5) since it starts at rest as you said, (c) is anything at least 6 is reasonable. Just eyeball it! It's reasonable to guess anything between 2 and 4. I'd pick 3 because it looks very 3ish to me. If you really wanted to get a precise answer, you'd guess that it's a graph of the form ax(x-5), guess a, then take derivatives. I think guessing a=-1 turns out to be nice so you have f'(1)=3. You can do differential equations if you want but you wouldn't bring a bulldozer to knock down a lego house If your end goal is to do well on an exam, setting up and solving a differential equation would waste at least 5 minutes vs just saying, "looks like 3".
Post by: a weaponized ikea chair on July 03, 2020, 04:50:34 pm
Hello.
For B) you need to understand that velocity is the change in displacement over change in time. So when they ask for the values of t where the instantaneous velocity is position, they're essentially asking you for the domain where the displacement is positive, that is where the particle is moving forward but not backward. It is including 0 as that is the start of the domain and 2.5 here is non-inclusive because there is no slope or gradient at the top of the parabola since it is a turning point (stationary point).
For C) Yeah, I believe you are correct. In the actual Methods exam, the reading of the graph for the values won't be this ambiguous and unclear.
For E) You don't need the equation of the parabola. Remember: velocity(speed)= change in displacement/change in time. / denotes divided by.
Here, change in time is 1-0 seconds= 1 second so this would be the denominator. For change in displacement, well, at 0, the particle has 0 displacement but at 1 it has a displacement of 3 m. So, 3-0 m=3 m. This is our numerator. Since, 3/1=3, that is our instantaneous velocity.
Hope this helps!
Thank you for your answer,
I do not understand how the displacement is 3; when t = 1, x looks like 4 to me.
Post by: Evolio on July 03, 2020, 04:55:01 pm
Also just a heads up that differential equations aren't on the Math Methods Study Design, although they are on the Specialist Maths one.
Post by: a weaponized ikea chair on July 03, 2020, 04:58:24 pm
Ah, okay. Well, that's probably just like Part C then. On the Methods exam, it won't be this ambiguous. As long as you understand the concept behind it, then you're good.I think 1729 mentioned something bout this, but:
Also, 1729, I'm pretty sure differential equations aren't on the Methods Study Design, although they are on the Specialist one.
assume equation of graph is ax(x-5)
a= -1
-x(x-5)
-x^2 +5x
d/dx = -2x + 5
x = 1
= 3
Coincidence or?
Post by: Evolio on July 03, 2020, 05:05:40 pm
Post by: a weaponized ikea chair on July 03, 2020, 05:07:50 pm
Post by: Evolio on July 03, 2020, 05:19:47 pm
The answer would have to be the same either way, whether you were using the derivative method or the other one.
I think they put 3 because they measured the x value as 3 so depending on how accurately you measure the x value, people would get different answers which is what happened here.
Post by: keltingmeith on July 03, 2020, 08:41:56 pm
Well, if it's taught before differentiation, then the method they were intending for you to use is through looking at the graph and that's probably the method they used as well (velocity=change in displacement/time) as it's simpler.
The answer would have to be the same either way, whether you were using the derivative method or the other one.
I think they put 3 because they measured the x value as 3 so depending on how accurately you measure the x value, people would get different answers which is what happened here.
Just going to chime in here and say this seems like a stupid way of doing things, that's because it is. (not to say Evolio is - they're right on the mark on how the textbook would want you to answer this, y'all should give an upvote for good advice)
You will never be asked for this in a 3/4 exam, and it is highly unlikely your teachers will ask you about this in any SAC or test. Because it's stupid and highly subjective - and literally immediately after this chapter, you're probably going to learn about differentiation -anyway- so why ask you to calculate the gradient like this when they can get two questions for the price of 1 by also asking you to calculate the derivative? So basically, if you're struggling with that part of the question, just ignore it and move on - it's not worth your time, your textbook is probably the only place you'll see it.
Post by: a weaponized ikea chair on July 03, 2020, 11:46:09 pm
Just going to chime in here and say this seems like a stupid way of doing things, that's because it is. (not to say Evolio is - they're right on the mark on how the textbook would want you to answer this, y'all should give an upvote for good advice)Agreed. They expect you to determine the gradient at a point of a crudely drawn graph which is extremely vague and of course, the answer is never what you estimated. Yeah, I will just take the derivative.
You will never be asked for this in a 3/4 exam, and it is highly unlikely your teachers will ask you about this in any SAC or test. Because it's stupid and highly subjective - and literally immediately after this chapter, you're probably going to learn about differentiation -anyway- so why ask you to calculate the gradient like this when they can get two questions for the price of 1 by also asking you to calculate the derivative? So basically, if you're struggling with that part of the question, just ignore it and move on - it's not worth your time, your textbook is probably the only place you'll see it.
Thanks
Post by: S_R_K on July 04, 2020, 02:31:07 pm
Hello.
For B) you need to understand that velocity is the change in displacement over change in time. So when they ask for the values of t where the instantaneous velocity is position, they're essentially asking you for the domain where the displacement is positive, that is where the particle is moving forward but not backward. It is including 0 as that is the start of the domain and 2.5 here is non-inclusive because there is no slope or gradient at the top of the parabola since it is a turning point (stationary point).
Hmmm... I think that OP is actually pointing out that the answer contradicts the question. The question states that the object begins from rest. What does this mean, if not that velocity = 0 at t = 0? Hence t = 0 should not be included in the set of times at which the instantaneous velocity is positive.
One could also argue that the left-sided time-derivative of position does not exist at t = 0, so the velocity is undefined there, but this is looking ahead to later chapters.
Post by: 1729 on July 04, 2020, 03:31:25 pm
What does this mean, if not that velocity = 0 at t = 0? Hence t = 0 should not be included in the set of times at which the instantaneous velocity is positive.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.
Post by: a weaponized ikea chair on July 05, 2020, 11:09:06 am
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.
I'm still not convinced.
At t = 0, velocity is zero.
Zero is not positive or negative. The particle is not moving at all.
The question asks interval for which velocity is positive. Still doesn't make sense.
Remember that this is taught before calculus so it seems unreasonable to expect students to use derivatives.
Post by: 1729 on July 05, 2020, 01:24:03 pm
I'm still not convinced.Ahh, okay. If you don't know what derivatives are disregard what I said.
At t = 0, velocity is zero.
Zero is not positive or negative. The particle is not moving at all.
The question asks interval for which velocity is positive. Still doesn't make sense.
Remember that this is taught before calculus so it seems unreasonable to expect students to use derivatives.
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.
Post by: a weaponized ikea chair on July 05, 2020, 08:20:09 pm
Ahh, okay. If you don't know what derivatives are disregard what I said.Ah, alright, makes sense now.
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.
Post by: S_R_K on July 05, 2020, 09:49:21 pm
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).
Post by: 1729 on July 06, 2020, 10:49:33 am
Increasing does not imply positive derivative.You are wrong.
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).
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:
(https://lh3.googleusercontent.com/-tGEk-IY2HN0/XwJ00d1VCPI/AAAAAAAABl8/2yOC9yKefZgospwzeXXy5KxNzVBi6vUegCK8BGAsYHg/s0/2020-07-05.jpg)
Post by: keltingmeith on July 06, 2020, 11:20:45 am
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:
(https://lh3.googleusercontent.com/-tGEk-IY2HN0/XwJ00d1VCPI/AAAAAAAABl8/2yOC9yKefZgospwzeXXy5KxNzVBi6vUegCK8BGAsYHg/s0/2020-07-05.jpg)
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
Post by: S_R_K on July 06, 2020, 12:02:21 pm
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:
(https://lh3.googleusercontent.com/-tGEk-IY2HN0/XwJ00d1VCPI/AAAAAAAABl8/2yOC9yKefZgospwzeXXy5KxNzVBi6vUegCK8BGAsYHg/s0/2020-07-05.jpg)
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.
Post by: keltingmeith on July 06, 2020, 01:14:13 pm
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:
If that limit does not exist, then the derivative does not exist. And in this case:
Since f(0+h) is NOT defined for h<0, then the limit CANNOT exist.
Post by: a weaponized ikea chair on July 06, 2020, 03:05:22 pm
thanks
Post by: 1729 on July 06, 2020, 03:08:55 pm
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.It's on the domain ]0,2.5[ thats where v is positive. Otherwise it is either negative or undefined.
thanks
Post by: keltingmeith on July 06, 2020, 03:10:58 pm
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