August 04, 2020, 02:52:40 pm

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#### a weaponized ikea chair

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##### Need help with rates of change question
« on: July 03, 2020, 03:26:01 pm »
+3
Hi,

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.

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#### Evolio

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##### Re: Need help with rates of change question
« Reply #1 on: July 03, 2020, 03:56:01 pm »
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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!
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#### 1729

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##### Re: Need help with rates of change question
« Reply #2 on: July 03, 2020, 04:36:36 pm »
+8
Hi,

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.

Okay I am going to give you two ways to approach this problem
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
$\frac{\text{d}^2s}{\text{d}t^2}$
Integrating both sides will give you this:
$\frac{ds}{dt}=ax+c$
Now we know that s’(2.5)=0, so we can plug that in:
$0=2.5a+c$
So we have:
$c=-2.5a$
We can integrate both sides of the differential equation again to get this:
$s(t)=\frac{a}{2}x^2+cx+d$
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".
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#### a weaponized ikea chair

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##### Re: Need help with rates of change question
« Reply #3 on: July 03, 2020, 04:50:34 pm »
+3
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!

I do not understand how the displacement is 3; when t = 1, x looks like 4 to me.
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#### Evolio

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##### Re: Need help with rates of change question
« Reply #4 on: July 03, 2020, 04:55:01 pm »
+4
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.

Also just a heads up that differential equations aren't on the Math Methods Study Design, although they are on the Specialist Maths one.
« Last Edit: July 03, 2020, 09:47:28 pm by Evolio »
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#### a weaponized ikea chair

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##### Re: Need help with rates of change question
« Reply #5 on: July 03, 2020, 04:58:24 pm »
+4
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.

Also, 1729, I'm pretty sure differential equations aren't on the Methods Study Design, although they are on the Specialist one.
I think 1729 mentioned something bout this, but:

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?
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#### Evolio

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##### Re: Need help with rates of change question
« Reply #6 on: July 03, 2020, 05:05:40 pm »
+3
No, that is correct. That is another way to do it.
« Last Edit: July 03, 2020, 05:07:41 pm by Evolio »
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#### a weaponized ikea chair

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##### Re: Need help with rates of change question
« Reply #7 on: July 03, 2020, 05:07:50 pm »
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Gosh this is confusing. So it's not four but three because of the derivative. Sorry for asking so many questions, but this chapter is before differentiation is taught, so....
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#### Evolio

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##### Re: Need help with rates of change question
« Reply #8 on: July 03, 2020, 05:19:47 pm »
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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.
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#### keltingmeith

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##### Re: Need help with rates of change question
« Reply #9 on: July 03, 2020, 08:41:56 pm »
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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)

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#### a weaponized ikea chair

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##### Re: Need help with rates of change question
« Reply #10 on: July 03, 2020, 11:46:09 pm »
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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.

Thanks
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#### S_R_K

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##### Re: Need help with rates of change question
« Reply #11 on: July 04, 2020, 02:31:07 pm »
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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.

#### 1729

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##### Re: Need help with rates of change question
« Reply #12 on: July 04, 2020, 03:31:25 pm »
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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.
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#### a weaponized ikea chair

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##### Re: Need help with rates of change question
« Reply #13 on: July 05, 2020, 11:09:06 am »
0
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.
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#### 1729

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##### Re: Need help with rates of change question
« Reply #14 on: July 05, 2020, 01:24:03 pm »
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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.
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.
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