MHB How do you calculate current in a circuit using integrals and limits?

[Nemo1]

Hi Community,

I have this question:
View attachment 5007

If I take the information provided $$\frac{dI}{dt}=\frac{-100}{t^2}$$

Setup my Integral $$\int\frac{-100}{t^2}\,dt$$

Take the constant out $$-100\int\frac{-1}{t^2}\,dt$$

Simplify $$\frac{-1}{t^2}= t^{-2}$$

To get $$-100\int{t^{-2}}\,dt$$

Apply the power rule $$-100\frac{t^{-2+1}}{-2+1}$$

Simplify and add the constant $$\frac{100}{t}+c$$

We know that when $$t=2$$ the current is $$150amps$$

So we now can setup an equation $$150=\frac{100}{2}+c$$

Solving for $$c$$ I can refine $$150=50+c$$

Then take $$50$$ from both sides to get $$c=100$$

Now I know what $$c$$ is I can plug it into my equation to get $$I(t)=\frac{100}{t}+c$$ becomes $$I(t)=\frac{100}{t}+100$$

$$I(t)=\frac{100}{t}+100$$ when I plugin $$t=2$$ I get $$I(t)=\frac{100}{2}+100$$ $$=$$$$150$$ which is correct as per the initial statement "When $$t=2$$, the current is $$150amps$$."

For the second part when I plugin $$t=20$$ I get $$I(t)=\frac{100}{20}+100$$ $$=$$$$105$$ So I can see that the current is decreasing as $$t$$ increases.

For the third part of what happens to the current as $$t \to \infty$$

Setting this up I get $$\lim_{{t}\to{\infty}}\left(100+\frac{100}{t}\right)$$

Separate into two easier limits $$\lim_{{t}\to{\infty}}100$$ $$+$$ $$100\left(\lim_{{t}\to{\infty}}\frac{1}{t}\right)$$

When $$\lim_{{t}\to{\infty}}\frac{1}{t}=0$$

My limit becomes $$\lim_{{t}\to{\infty}}100+100\cdot0=100$$

I would really appreciate it if my working out could be checked to see if I am making any mistakes with clear explanations of incorrect terminology and so forth. I want to be able to learn how to solve any similar questions with confidence.

Many thanks for your time in advance.

 

[MarkFL]

Hello, Nemo! :D

I will show you how I would work this problem. We are given an IVP, and so to solve part (a) I would write:

$$\int_{150}^{I(t)}\,du=-100\int_{2}^{t} v^{-2}\,dv$$

I have replace the dummy variables of integration, and used the given boundaries as the limits of integration. Applying the FTOC, we obtain:

$$I(t)-150=100\left[\frac{1}{v}\right]_2^t=100\left(\frac{1}{t}-\frac{1}{2}\right)=\frac{100}{t}-50$$

Hence:

$$I(t)=\frac{100}{t}+100$$

This agrees with your result. :)

Now, for part (b), we may write:

$$I(20)=\frac{100}{20}+100=105$$

This also agrees with your work. :)

And finally for part (c), we may write:

$$I(\infty)=\lim_{t\to\infty}\left(\frac{100}{t}+100\right)=100$$

So, I agree with your results in all 3 parts of the problem. (Yes)

The only issue I really see is in your $\LaTeX$ and exponents. User the caret character, and if an exponent is more than 1 character, then wrap that exponent in curly braces. For example:

t^2 produces $t^2$

t^{x+y} produces $t^{x+y}$

 

[Nemo1]

Hello Mark,

With your definite integral

$$\int_{150}^{I(t)}\,du=-100\int_{2}^{t} v^{-2}\,dv$$

I would like to understand the process in how you created it, the way I am reading it is that:

$$\int_{150}^{I(t)}\,du$$ is a place holder to put the limit from 150 to $$I(t)$$ which directly corresponds to the next step of applying the FTOC which gives us $$I(t)-150$$

And that the $$u$$ is the dummy variable you mentioned.

Next is the $$-100\int_{2}^{t} v^{-2}\,dv$$ the constant $$-100$$ has been taken from $$\frac{-100}{t^2}$$ moved to the front.

The remaining $$\frac{1}{t^2}$$ has had the exponent property of $$\frac{1}{a^n}={a^{-n}}$$ applied to get $${t^{-2}}$$ and the dummy variable $$v$$ has been applied.

Put together we get $$\int_{150}^{I(t)}\,du=-100\int_{2}^{t} v^{-2}\,dv$$

I am unsure why the lower and upper limit $$\int_{2}^{t}$$ is set to from $$2$$ $$\to$$ $$t$$ at this point?

The remaining section after applying the FTOC I understand to get $$100\left[\frac{1}{v}\right]_2^t$$

Put together we get $$I(t)-150=100\left[\frac{1}{v}\right]_2^t$$

Also has the negative sign has been removed when applying the FTOC to $$-100\int_{2}^{t} v^{-2}\,dv$$ when making the section $$100\left[\frac{1}{v}\right]_2^t$$ because the negative on $$-100$$ would cancel when times by $$\left[-\frac{1}{v}\right]$$ from the $$\int v^{-2}dv$$?

In the next step:

$$\frac{100}{t}-50$$

How does it become:

$$\frac{100}{t}+100$$

Graphing the function I get the below which appears to correspond with my answer that the current approaches $$100amps$$ as $$t\to\infty$$

View attachment 5013

Many thanks, Nemo :)

 

[MarkFL]

The type of problem you were givien is what's called an Initial Value Problem (IVP). This consists (in this case) of an Ordinary Differential Equation (ODE) relating 2 variables and an Initial Value, which is a point on the curve representing the solution. We are given:

$$\d{I}{t}=-\frac{100}{t^2}$$ where $$I(2)=150$$

So, we are given the point $(2,150)$, and we wish to be able to define any point $(t,I(t))$...these are the boundaries, that is the initial and final points. In this view, $t$ is initially 2 and $I(t)$ is initially 150. The initial values become the lower limits of integration and the final values are the upper limits.

When we solve the ODE, we may use these limits and thereby eliminate the need to solve for a constant of integration inherent in using indefinite integrals. Because the variables used within a definite integral are "integrated out" they are called "dummy variables" and may be replaced to avoid the bad form of having the same variables in the limits of integration as we have within the integrals themselves.

...Also has the negative sign has been removed when applying the FTOC to $$-100\int_{2}^{t} v^{-2}\,dv$$ when making the section $$100\left[\frac{1}{v}\right]_2^t$$ because the negative on $$-100$$ would cancel when times by $$\left[-\frac{1}{v}\right]$$ from the $$\int v^{-2}dv$$?

Yes, that's correct. :)

In the next step:

$$\frac{100}{t}-50$$

How does it become:

$$\frac{100}{t}+100$$

We have at this point:

$$I(t)-150=\frac{100}{t}-50$$

Since we wish to solve for $I(t)$ we add 150 to both sides to get:

$$I(t)=\frac{100}{t}+100$$

And this is the solution to the given IVP.

 

[Nemo1]

Hi Mark,

Thanks for the detailed explanation.

Of course:

$$I(t)-150=\frac{100}{t}-50$$

Solving for $I(t)$ we add 150 to both sides to get:

$$I(t)=\frac{100}{t}+100$$

I was over cooking it in my head and looking for a complex answer when there wasn't one. (something I tend to do often :( )

Thanks for your help, I will continue to work thru it all to better my understanding.

Cheers Nemo :)