Silly calculus problem I can't seem to do

Dr.Peterson said:
Actually, you weren't entirely wrong; you just wrote the integral too soon. I'll correct your work:


It's okay to move the radical when it isn't yet inside an integral.

Now finish:

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]​
​
[imath]\displaystyle \left.12t\right|_{0}^{1}=\left.2s^{1/2}\right|_{1}^{s}[/imath]​
​
[imath]\displaystyle 12=2s^{1/2}-2[/imath]​
​
[imath]\displaystyle 6=s^{1/2}-1[/imath]​
​
[imath]\displaystyle s=(6+1)^2=49[/imath]​

I hadn't noticed @khansaheb's small error.
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Thanks to you professor Dave. This means that my post #2 was correct! Why professor Steven did not take it?

🤔
 
MaxMath said:
[math]s=1+\int_0^t{v}dt[/math][math]v=12\sqrt{s}[/math][math]v=12\sqrt{1+\int_0^t{v}dt}[/math]Solve [imath]v=v(t)[/imath] from the last line (also using the initial conditions [imath]s=1[/imath] and [imath]v=12[/imath] when [imath]t=0[/imath]), then use the first line to get [imath]s[/imath] at [imath]t=1[/imath].
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Can someone help me to see if we can get the same result this way? Not doing this kind of things often, so my brain needs a bit lubricant.
 
MaxMath said:
Can someone help me to see if we can get the same result this way? Not doing this kind of things often, so my brain needs a bit lubricant.
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I think that you did one or two mistakes in setting the differential equation. Start over from the definition of velocity: \(\displaystyle v = \frac{ds}{dt}\).
 
mario99 said:
I think that you did one or two mistakes in setting the differential equation. Start over from the definition of velocity: \(\displaystyle v = \frac{ds}{dt}\).
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Thanks. That way already worked out, as it seems. I just want to tackle it in a different way so 1) to corroborate that method; 2) the better understand the problem; and potentially 3) to see if there are problems in that method.

And I think trying to pinpoint exactly what the problem is in a wrong answer (suppose this one is wrong) helps a lot in understanding the question and the tools in use.
 
Setting the equations like this is wrong:

[imath]\displaystyle s=1+\int_0^t{v}dt[/imath]
[imath]\displaystyle v=12\sqrt{s}[/imath]
[imath]\displaystyle v=12\sqrt{1+\int_0^t{v}dt}[/imath]

There are only two methods to solve the problem. When you set the integrals, you can either solve them with definite integral or indefinite integral.
 
mario99 said:
Setting the equations like this is wrong:

[imath]\displaystyle s=1+\int_0^t{v}dt[/imath]
[imath]\displaystyle v=12\sqrt{s}[/imath]
[imath]\displaystyle v=12\sqrt{1+\int_0^t{v}dt}[/imath]

There are only two methods to solve the problem. When you set the integrals, you can either solve them with definite integral or indefinite integral.
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MaxMath said:
Thanks. That way already worked out, as it seems. I just want to tackle it in a different way so 1) to corroborate that method; 2) the better understand the problem; and potentially 3) to see if there are problems in that method.
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The first lines physical meaning is clear. 1 is the position when t=0, the integral is a function of t, giving the distance travelled from time 0 to t.

The other two lines are self explanatory. I fail to see where exactly the problem is. Suppose we are not aiming at solving the original problem but are rather diagnosing these formulas.
 
MaxMath said:
The first lines physical meaning is clear. 1 is the position when t=0, the integral is a function of t, giving the distance travelled from time 0 to t.

The other two lines are self explanatory. I fail to see where exactly the problem is. Suppose we are not aiming at solving the original problem but are rather diagnosing these formulas.
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Show me the steps of how you got this:

[imath]\displaystyle s=1+\int_0^t{v}dt[/imath]
 
mario99 said:
Show me the steps of how you got this:

[imath]\displaystyle s=1+\int_0^t{v}dt[/imath]
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As I previously said, 1 is the initial condition when t=0. The integral part is the distance traveled starting from t=0 til any time t (as a variable). It in an indefinite form is clear,
[imath]\int{v dt}[/imath] means distance travelled of the particle with speed v (a function of t). Changing to definite, by adding 0 and t, makes it a function of time, distance travelled between 0 and t.

Does this makes sense?
 
MaxMath said:
As I previously said, 1 is the initial condition when t=0. The integral part is the distance traveled starting from t=0 til any time t (as a variable). It in an indefinite form is clear,
[imath]\int{v dt}[/imath] means distance travelled of the particle with speed v (a function of t). Changing to definite, by adding 0 and t, makes it a function of time, distance travelled between 0 and t.

Does this makes sense?
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Without writing clear steps, nothing will make sense. Therefore, let me write the steps which I think that you have used.

[imath]\displaystyle \frac{ds}{dt} = v[/imath]

[imath]\displaystyle ds = v \ dt[/imath]

[imath]\displaystyle \int ds = \int v \ dt[/imath]

[imath]\displaystyle s + C = \int v \ dt[/imath]

then you used the initial condition to find that [imath]\displaystyle C = -1[/imath]

[imath]\displaystyle s - 1 = \int v \ dt[/imath]

[imath]\displaystyle s = 1 + \int v \ dt[/imath]

[imath]\displaystyle s = 1 + \int_{0}^{t} v \ dt[/imath]

Is this what you did?
 
mario99 said:
Without writing clear steps, nothing will make sense. Therefore, let me write the steps which I think that you have used.

[imath]\displaystyle \frac{ds}{dt} = v[/imath]

[imath]\displaystyle ds = v \ dt[/imath]

[imath]\displaystyle \int ds = \int v \ dt[/imath]

[imath]\displaystyle s + C = \int v \ dt[/imath]

then you used the initial condition to find that [imath]\displaystyle C = -1[/imath]

[imath]\displaystyle s - 1 = \int v \ dt[/imath]

[imath]\displaystyle s = 1 + \int v \ dt[/imath]

[imath]\displaystyle s = 1 + \int_{0}^{t} v \ dt[/imath]

Is this what you did?
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I start right from the idea that distance travelled is the area under the curve (function) of speed [imath]v[/imath], i.e. [imath]\int_{t_1}^{t_2} v \ dt[/imath], between any given time [imath]t_1[/imath] and [imath]t_2[/imath]. So the distance travelled from [imath]t=0[/imath] to [imath]t[/imath] (as the variable itself) will be [imath] \int_{0}^{t} v \ dt[/imath]. Since the partical is at position 1 at the time of 0, then its position, relative to the origin, as a function of [imath]t[/imath], is

[imath]\displaystyle s = 1 + \int_{0}^{t} v \ dt[/imath]

Sorry I think I should try to more use mathematical language.

But anyway I think your steps correctly reflect my process.
 
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MaxMath said:
I start right from the idea that distance travelled is the area under the curve (function) of speed [imath]v[/imath], i.e. [imath]\int_{t_1}^{t_2} v \ dt[/imath], between any given time [imath]t_1[/imath] and [imath]t_2[/imath]. So the distance travelled from [imath]t=0[/imath] to [imath]t[/imath] (as the variable itself) will be [imath] \int_{0}^{t} v \ dt[/imath]. Since the partical is at position 1 at the time of 0, then its position, relative to the origin, as a function of [imath]t[/imath], is

[imath]\displaystyle s = 1 + \int_{0}^{t} v \ dt[/imath]

Sorry I think I should try to more use mathematical language.

But anyway I think your steps correctly reflect my process.
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The steps are completely wrong. Do you see why?
 
MaxMath said:
Please enlighten me.
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The steps are completely wrong because the differential element [imath]ds[/imath] depends on the function [imath]v(s)[/imath], but we ignored that fact completely by writing step number three like that.
 
If we have [imath]v(s)[/imath], it is wrong to write:

[imath]\displaystyle \int ds = \int v(s) \ dt[/imath]

But if we have [imath]v(t)[/imath], the steps are correct:

[imath]\displaystyle \frac{ds}{dt} = v[/imath]

[imath]\displaystyle ds = v \ dt[/imath]

[imath]\displaystyle \int ds = \int v(t) \ dt[/imath]

[imath]\displaystyle s + C = \int v(t) \ dt[/imath]

then you used the initial condition to find that [imath]\displaystyle C = -1[/imath]

[imath]\displaystyle s - 1 = \int v(t) \ dt[/imath]

[imath]\displaystyle s = 1 + \int v(t) \ dt[/imath]

[imath]\displaystyle s(t) = 1 + \int_{0}^{t} v(t) \ dt[/imath]

You will stop at this step and to solve the problem, [imath]v(t)[/imath] must be given.
 
mario99 said:
The steps are completely wrong because the differential element [imath]ds[/imath] depends on the function [imath]v(s)[/imath], but we ignored that fact completely by writing step number three like that.
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Thank you. But sorry I don’t get that. I still don’t understand why it’s wrong, particularly why my thought process is wrong.

Let park this here. So if we move on from this supposedly wrong formula, what can we get out of it?

Ps
Ok I see your post before me. This is very intriguing. I need to think about it a bit more.
 
MaxMath said:
Thank you. But sorry I don’t get that. I still don’t understand why it’s wrong, particularly why my thought process is wrong.

Let park this here. So if we move on from this supposedly wrong formula, what can we get out of it?

Ps
Ok I see your post before me. This is very intriguing. I need to think about it a bit more.
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It depends on the function [imath]v[/imath]. If it is [imath]v(s)[/imath] your process is wrong. If it is [imath]v(t)[/imath], your process is correct but you can't solve the problem if you don't know [imath]v(t)[/imath].

See post #35 to understand that.
 
Dr.Peterson said:
Actually, you weren't entirely wrong; you just wrote the integral too soon. I'll correct your work:


It's okay to move the radical when it isn't yet inside an integral.

Now finish:

[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]​
​
[imath]\displaystyle \left.12t\right|_{0}^{1}=\left.2s^{1/2}\right|_{1}^{s}[/imath]​
​
[imath]\displaystyle 12=2s^{1/2}-2[/imath]​
​
[imath]\displaystyle 6=s^{1/2}-1[/imath]​
​
[imath]\displaystyle s=(6+1)^2=49[/imath]​

I hadn't noticed @khansaheb's small error.
Click to expand...

This puzzled me. It looks like you edited the content of post #6 of @mario99 in the quote in your post #19? Not meant to be mean, but I don't think this is good practice. (Please correct me if I'm mistaken.)

Apart from that, a question on
[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]​

Why the lower/upper bounds of integrals, on the LHS and RHS, should be 0-1 and 1-s, respectively?

Or to ask in another way, why can we add definite integral this way on both sides to
[imath] 12 \ dt = \frac{1}{\sqrt{s}}\ ds[/imath] ?
 
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MaxMath said:
This puzzled me. It looks like you edited the content of post #6 of @mario99 in the quote in your post #19? Not meant to be mean, but I don't think this is good practice. (Please correct me if I'm mistaken.)
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It's done all the time. Why not, when I say explicitly what I am doing??
MaxMath said:
Apart from that, a question on
[imath]\displaystyle \int_{0}^{1} 12 \ dt = \int_{1}^{s} \frac{1}{\sqrt{s}} \ ds[/imath]​

Why the lower/upper bounds of integrals, on the LHS and RHS, should be 0-1 and 1-s, respectively?

Or to ask in another way, why can we add definite integral this way on both sides to
[imath] 12 \ dt = \frac{1}{\sqrt{s}}\ ds[/imath] ?
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Haven't you just had a long discussion with him (which I haven't read yet)? I didn't touch that aspect of his work, because that's not what I was correcting. When I have time, maybe I'll read your discussion and have something to add; but to be honest, the definite integral approach isn't one I use, so I don't have much to say about it to begin with.
 
MaxMath said:
I start right from the idea that distance travelled is the area under the curve (function) of speed [imath]v[/imath], i.e. [imath]\int_{t_1}^{t_2} v \ dt[/imath], between any given time [imath]t_1[/imath] and [imath]t_2[/imath]. So the distance travelled from [imath]t=0[/imath] to [imath]t[/imath] (as the variable itself) will be [imath] \int_{0}^{t} v \ dt[/imath]. Since the partical is at position 1 at the time of 0, then its position, relative to the origin, as a function of [imath]t[/imath], is

[imath]\displaystyle s = 1 + \int_{0}^{t} v \ dt[/imath]
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I think the main problem with your approach is that it doesn't help you actually solve anything. Since you don't know what function v is of t, you can't carry out the integral you wrote. It may be perfectly true, but useless. (This is what is said in #37).
MaxMath said:
Why the lower/upper bounds of integrals, on the LHS and RHS, should be 0-1 and 1-s, respectively?
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I have now read the discussion this morning, and see that you didn't cover the definite integral issue.

Besides asking the author of that solution, how about telling us your thoughts: What limits of integration would you consider using, and why?

Or, look it up. Here's a nice explanation I found for the general concept:

 
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