Surface area of sphere increase at a constant rate

[Nemo1]

Hi Community,

I have this tutorial question.
View attachment 5517

When I look at the first question (a) I think it is FALSE as the surface area would not increase at the same rate as the radius.

For the second question I am not sure if I am interpreting it correctly.

If \(\displaystyle r=\sqrt{\frac{Ct}{4\varPi}+2}\) where \(\displaystyle Ct\) is the derivative of \(\displaystyle s=4\varPi r^2\) which I have calculated to be \(\displaystyle 8\varPi r\)

How does this all fit together in the bigger picture of the Modelling of the Math that is trying to be taught, I feel like I am missing a few steps.

Cheers

Nemo

 

[MarkFL]

a) The term:

\(\displaystyle \d{S}{t}\)

represents the rate at which the surface area $S$ changes with time $t$. So, if we are given:

\(\displaystyle \d{S}{t}=C\)

where $C$ is a fixed constant, then what do we know about how the surface area is changing?

 

[Nemo1]

Hmm, I think I get it, if \(\displaystyle C\) is an integer say for example \(\displaystyle 5\) then the surface area would be increasing at a constant rate.

How does this fit in with part b, as $C$ is the numerator of a squared function?

 

[MarkFL]

If $S=4\pi r^2$ and $r$ is a function of time $t$, then what is another expression for \(\displaystyle \d{S}{t}\) that we can derive?

 

[Nemo1]

If $S=4\pi r^2$ and $r$ is a function of time $t$, then what is another expression for \(\displaystyle \d{S}{t}\) that we can derive?

If \(\displaystyle r=\sqrt{\frac{Ct}{4\varPi}+2}\) where \(\displaystyle Ct\) is the derivative of \(\displaystyle s=4\varPi r^2\) which I have calculated to be \(\displaystyle 8\varPi r\)

Then

\(\displaystyle r=\sqrt{\frac{8\varPi r}{4\varPi}+2}\)

Or should I rearrange \(\displaystyle s=4\varPi r^2\) to be \(\displaystyle r=\frac{\sqrt{s}}{\sqrt{\varPi}}\) in terms of $r$ and plug that in for $Ct$ to get

\(\displaystyle r=\sqrt{\frac{\frac{\sqrt{s}}{\sqrt{\varPi}}}{4\varPi}+2}\)

 

[MarkFL]

What you want to do here is implicitly differentiate the following w.r.t $t$:

\(\displaystyle S=4\pi r^2\)

This will give you an expression equal to \(\displaystyle \d{S}{t}\), and since we know \(\displaystyle \d{S}{t}=C\), we can equate the two, and the result is a separable first order ODE. We can then integrate, using the boundaries:

\(\displaystyle r(0)=\sqrt{2},\,r(t)=r\)

and the result will be the radius $r$ as a function of time $t$.

So, what do you get when you do the implicit differentiation?

 

[MarkFL]

I fear I may have lost you here, and I can tell you are truly wanting to understand, and not just have your work done, so I am going to explain things more fully and show you how to work this problem. :)

We know:

\(\displaystyle \d{S}{t}=C\)

That is, the rate at which the surface area of the sun is changing with respect to time is a constant.

Now, because the sun is reasonably being modeled as a sphere, we know the surface area will be given by:

\(\displaystyle S=4\pi r^2\)

where $r$ is the radius of the sun. Because this sun will be expanding, we know that $r$ is a function of time $t$. And so, if we implicitly differentiate the above w.r.t $t$, we find:

\(\displaystyle \d{S}{t}=4\pi\left(2r\d{r}{t}\right)=8\pi r\d{r}{t}\)

Now, since \(\displaystyle \d{S}{t}=C\), we may now write:

\(\displaystyle 8\pi r\d{r}{t}=C\)

Now, if we let the initial time be $t=0$ and the final time be $t$, and we observe that the game developer intends for $r(0)=\sqrt{2}$, then upon integrating both sides w.r.t $t$ and switching out the dummy variables of integration (the variables in definite integrals get "integrated out" and are thus referred to as "dummy variables"), we may now write:

\(\displaystyle 8\pi\int_{\sqrt{2}}^{r} u\,du=C\int_0^t\,dv\)

Applying the FTOC, we get:

\(\displaystyle 4\pi\left(r^2-2\right)=Ct\)

And, once we solve for $r$, we obtain:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+2}\)

 

[Nemo1]

Thanks Mark,

I am going to need some time to process this, I am sure I will have more questions.

Cheers Nemo

 

[Nemo1]

I fear I may have lost you here, and I can tell you are truly wanting to understand, and not just have your work done, so I am going to explain things more fully and show you how to work this problem. :)

We know:

\(\displaystyle \d{S}{t}=C\)

That is, the rate at which the surface area of the sun is changing with respect to time is a constant.

Now, because the sun is reasonably being modeled as a sphere, we know the surface area will be given by:

\(\displaystyle S=4\pi r^2\)

where $r$ is the radius of the sun. Because this sun will be expanding, we know that $r$ is a function of time $t$. And so, if we implicitly differentiate the above w.r.t $t$, we find:

\(\displaystyle \d{S}{t}=4\pi\left(2r\d{r}{t}\right)=8\pi r\d{r}{t}\)

Now, since \(\displaystyle \d{S}{t}=C\), we may now write:

\(\displaystyle 8\pi r\d{r}{t}=C\)

Now, if we let the initial time be $t=0$ and the final time be $t$, and we observe that the game developer intends for $r(0)=\sqrt{2}$, then upon integrating both sides w.r.t $t$ and switching out the dummy variables of integration (the variables in definite integrals get "integrated out" and are thus referred to as "dummy variables"), we may now write:

\(\displaystyle 8\pi\int_{\sqrt{2}}^{r} u\,du=C\int_0^t\,dv\)

Applying the FTOC, we get:

\(\displaystyle 4\pi\left(r^2-2\right)=Ct\)

And, once we solve for $r$, we obtain:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+2}\)

Hi Mark,

So far I am following:

\(\displaystyle \d{S}{t}=C\)

That is, the rate at which the surface area of the sun is changing with respect to time is a constant.

Now, because the sun is reasonably being modeled as a sphere, we know the surface area will be given by:

\(\displaystyle S=4\pi r^2\)

where $r$ is the radius of the sun. Because this sun will be expanding, we know that $r$ is a function of time $t$. And so, if we implicitly differentiate the above w.r.t $t$, we find:

\(\displaystyle \d{S}{t}=4\pi\left(2r\d{r}{t}\right)=8\pi r\d{r}{t}\)

Now, since \(\displaystyle \d{S}{t}=C\), we may now write:

\(\displaystyle 8\pi r\d{r}{t}=C\)

Where I am getting confused now is in the next step, I am unsure of where $r(0)=\sqrt{2}$ is observed and also how integrating \(\displaystyle 8\pi\int_{\sqrt{2}}^{r} u\,du=C\int_0^t\,dv\) fits into the bigger picture.

Could it be derived instead or is integration a simpler approach? My understanding is that a derivative is the slope at a point whereas integration is the area under the curve. So my working theory is that you have integrated it because we are looking at the surface area of the sun expanding at a constant rate $C$.

Maybe I am looking for a complex answer where none exists?

Cheers Nemo.

 

[MarkFL]

...Where I am getting confused now is in the next step, I am unsure of where $r(0)=\sqrt{2}$ is observed and also how integrating \(\displaystyle 8\pi\int_{\sqrt{2}}^{r} u\,du=C\int_0^t\,dv\) fits into the bigger picture.

Could it be derived instead or is integration a simpler approach? My understanding is that a derivative is the slope at a point whereas integration is the area under the curve. So my working theory is that you have integrated it because we are looking at the surface area of the sun expanding at a constant rate $C$.

Maybe I am looking for a complex answer where none exists?

Cheers Nemo.

Let's pick up at this point:

\(\displaystyle 8\pi r\d{r}{t}=C\)

What we have here is a first order (the highest order derivative in the equation is a first derivative) separable (the variable $r$ and $t$ may be separated to their own side of the equation), ordinary differential equation, or ODE. We use integration to get $r$ as a function of $t$, because integration is the inverse operation to differentiation.

If we look at the function we were given to verify:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+2}\)

this is where we find:

\(\displaystyle r(0)=\sqrt{2}\)

Perhaps the concept of using boundaries as the limits of integration is new to you (I didn't start using it until I had experience solving differential equations). So, let's use indefinite integrals instead. We begin with:

\(\displaystyle 8\pi r\d{r}{t}=C\)

And the integrate both sides w.r.t $t$:

\(\displaystyle \int 8\pi r\d{r}{t}\,dt=\int C\,dt\)

Now, we can move the constant factors in our integrands outside the integrals, and simplify the differential on the LHS:

\(\displaystyle 8\pi\int r\,dr=C\int \,dt\)

Next, compute the anti-derivatives:

\(\displaystyle 8\pi\cdot\frac{r^2}{2}=Ct+ k\) where $k$ is the constant of integration.

Simplify:

\(\displaystyle 4\pi r^2=Ct+k\)

Now, when $t=0$ (and letting $r(0)=r_0$), we have:

\(\displaystyle 4\pi r_0^2=k\)

And so we obtain:

\(\displaystyle 4\pi r^2=Ct+4\pi r_0^2\)

Dividing through by $4\pi$, we get:

\(\displaystyle r^2=\frac{Ct}{4\pi}+r_0^2\)

Since $r$ is a linear physical measure, we take the positive root:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+r_0^2}\)

This would be the general solution, for any valid initial radius.

Now, if we observe that the programmer intends for $r(0)=\sqrt{2}$, we then get:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+2}\)

 

[Nemo1]

Thanks Mark, it marks more sense now.

I have had limited exposure to integrals and have it in next semester. I am currently working my way thru Calculus for Dummies as the explanations don't skip over the algebra that is assumed. Also I have not done much with differential equations which would also not help. My math learning is on an exponential curve at the moment.

I'll need to keep practicing for the exam in 5 weeks time.

Cheers Nemo

 

[MarkFL]

Thanks Mark, it marks more sense now.

I have had limited exposure to integrals and have it in next semester. I am currently working my way thru Calculus for Dummies as the explanations don't skip over the algebra that is assumed. Also I have not done much with differential equations which would also not help. My math learning is on an exponential curve at the moment.

I'll need to keep practicing for the exam in 5 weeks time.

Cheers Nemo

I had no idea you haven't been exposed officially to integration yet.

We could work part b) without integration...after all we are only asked to verify that the equation the programmer derives satisfies the condition that the radius increases at a constant rate.

So, let's begin with it:

\(\displaystyle r=\sqrt{\frac{Ct}{4\pi}+2}\)

And differentiate both sides w.r.t $t$:

\(\displaystyle \d{r}{t}=\frac{\dfrac{C}{4\pi}}{2\sqrt{\dfrac{Ct}{4\pi}+2}}=\frac{C}{8\pi r}\)

Now, if we observe that:

\(\displaystyle \d{S}{r}=8\pi r\)

then we have:

\(\displaystyle \d{r}{t}=\frac{C}{\dfrac{dS}{dr}}\implies \d{S}{t}=C\)

This is likely what the author had in mind if you are not expected to use integration to derive the formula yourself. :)