MHB Mastering Complex Math Problems: Solving for Volume, Surface Area, and Time

[Nemo1]

Hi Community,

I have the following problem and I am completely stuck. I really struggle to get my head around how to break down these questions into chunks that I can then apply the math to.
View attachment 5625

From what I can see so far, I have a to be able calculate the surface area at any height to get the volume and therefore somehow link this to time and that the volume of the water is decreasing proportionally with the surface area.

Many thanks for your time in advance.

Cheers Nemo.

 

[MarkFL]

For part a), take the given hint:

$$V=\int_0^h A(x)\,dx$$

and differentiate both sides w.r.t $h$...what do you get?

 

[Nemo1]

For part a), take the given hint:

$$V=\int_0^h A(x)\,dx$$

and differentiate both sides w.r.t $h$...what do you get?

Hi Mark,

When I work it out I get:

$$V=\int_0^h A(x)\,dx$$ = $$\frac{1}{2}Ah^2$$

By using the Power rule and the F.T.O.C, this seems very abstract for me as I am struggling to understand what the dx truly means in this context. At a basic level, I know that dx is the same as $$\Delta x$$ and visually I see it as the increment along the $x$ $axis$. Then $"time"$ needs to be put in and that's where I get completely confused.

I feel like as much as I do these problems over and over, I am only plugging numbers and missing the underlying concepts.

Cheers Nemo.

 

[MarkFL]

Let's begin with:

$$V=\int_0^h A(x)\,dx$$

Now, if we differentiate both sides w.r.t $t$ (not $h$ as I originally said, although that would work too), we get:

$$\d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t}$$

Now, use $$\d{V}{t}=-kA(h)$$ and solve for $$\d{h}{t}$$. :D

 

[Nemo1]

Let's begin with:

$$V=\int_0^h A(x)\,dx$$

Now, if we differentiate both sides w.r.t $t$ (not $h$ as I originally said, although that would work too), we get:

$$\d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t}$$

Now, use $$\d{V}{t}=-kA(h)$$ and solve for $$\d{h}{t}$$. :D

Hi Mark,

I am completely lost now,

When I read: $$\d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t}$$

I am thinking that the derivative of the volume with respect to time is equal to the derivative of time from $0$ to $h$ of $A(x)dx$

If $A(x)dx$ is the area of the surface, the volume decreases as $h$ decreases because the volume is directly linked to the area time the height $h$ which is proportional.

Can you please layout the steps in between, as I said in my previous post, I get confused where time comes into the equations and derivatives.

I feel like I am not grasping some key concepts and therefore in perpetual confusion. I do thank you so much for all of your help you have given me over the past year.

Cheers Nemo

 

[MarkFL]

Okay, let's look at what I wrote here:

$$\d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t}$$

Now, there is a very important theorem in calculus, so important it is called the Fundamental Theorem Of Calculus (FTOC). It forms a powerful conduit between the two main branches of the calculus: integral and differential. It comes in two forms, one of which is the derivative form, which states:

[box=blue]

Fundamental Theorem Of Calculus -- Derivative Form​

---

Let $f$ be continuous on $[a,b]$ and let $x$ be any number in the interval. If $G(x)$ is a function defined by:

$$G(x)=\int_a^x f(t)\,dt$$ then $$G'(x)=f(x)$$​

[/box]

Now, when I utilized this theorem for this problem, I had to apply the chain rule as well, because we are differentiating w.r.t $t$ instead of $h$ and $h$ is a function of $t$.

Now, you are likely more familiar with the anti-derivative form of the FTOC, which states:

[box=green]

Fundamental Theorem Of Calculus -- Antiderivative Form​

---

Let $f$ be continuous on $[a,b]$ and let $F$ be any function for which $F'(x)=f(x)$. Then:

$$\int_a^b f(x)\,dx=F(b)-F(a)$$​

[/box]

So, if we define $B'(x)=A(x)$, then we could write:

$$V=\int_0^h A(x)\,dx=B(h)-B(0)$$

Now if we differentiate w.r.t $t$, we get (using the chain rule as well and noting that $B(0)$ is a constant):

$$\d{V}{t}=A(h)\d{h}{t}$$

which is the same thing we obtained using the derivative form of the FTOC, although in my opinion, the first method is more direct. I recommend you become comfortable with both forms of the FTOC, as they will come up often. :D

 

[Nemo1]

Okay, let's look at what I wrote here:

$$\d{V}{t}=\frac{d}{dt}\int_0^h A(x)\,dx=A(h)\d{h}{t}$$

Now, there is a very important theorem in calculus, so important it is called the Fundamental Theorem Of Calculus (FTOC). It forms a powerful conduit between the two main branches of the calculus: integral and differential. It comes in two forms, one of which is the derivative form, which states:

[box=blue]

Fundamental Theorem Of Calculus -- Derivative Form​

---

Let $f$ be continuous on $[a,b]$ and let $x$ be any number in the interval. If $G(x)$ is a function defined by:

$$G(x)=\int_a^x f(t)\,dt$$ then $$G'(x)=f(x)$$​

[/box]

Now, when I utilized this theorem for this problem, I had to apply the chain rule as well, because we are differentiating w.r.t $t$ instead of $h$ and $h$ is a function of $t$.

Now, you are likely more familiar with the anti-derivative form of the FTOC, which states:

[box=green]

Fundamental Theorem Of Calculus -- Antiderivative Form​

---

Let $f$ be continuous on $[a,b]$ and let $F$ be any function for which $F'(x)=f(x)$. Then:

$$\int_a^b f(x)\,dx=F(b)-F(a)$$​

[/box]

So, if we define $B'(x)=A(x)$, then we could write:

$$V=\int_0^h A(x)\,dx=B(h)-B(0)$$

Now if we differentiate w.r.t $t$, we get (using the chain rule as well and noting that $B(0)$ is a constant):

$$\d{V}{t}=A(h)\d{h}{t}$$

which is the same thing we obtained using the derivative form of the FTOC, although in my opinion, the first method is more direct. I recommend you become comfortable with both forms of the FTOC, as they will come up often. :D

I am still confused even with your explanation, sorry.

How do you differentiate a function or apply the chain rule when it is only $A(x)$.

Totally lost.

Cheers Nemo

 

[MarkFL]

Well, the main thing you want to observe here is that the definite integral:

$$V=\int_0^h A(x)\,dx$$

is a function of $h$...$x$ is just a "dummy" variable in the sense that it simply gets integrated out...we could just as easily write:

$$V=\int_0^h A(\varphi)\,d\varphi$$

This has the same meaning as the original, we have just used a different dummy variable.

Since this definite integral is a function of $h$, and $h$ is a function of $t$, we can then differentiate by $t$, bearing in mind that the chain rule must be applied.

 

[Nemo1]

Hi Mark, Ok, so I have been working on this one for a while now, and think I have solved part a from your help.
I feel like it is simple algebraic manipulation but wish I fully understood the concepts.

So if we have:

$$V=\int_{0}^{h}A(x) \,dx$$

Then setup the derivative with respect to $t$:

$$\d{V}{t}=\d{}{t}=\int_{0}^{h}A(x) \,dx = A(h)\d{h}{t}$$

Knowing that:

$$\d{V}{t}=-kA(h)$$

As stated in the question we can setup the equation:

$$A(h)\cdot\d{h}{t}=-k \cdot A(h)$$

We can divide both side by $A(h)$:

$$\frac{A(h)}{A(h)}\cdot \d{h}{t}=-k\cdot \frac{A(h)}{A(h)}$$

To then arrive at:

$$\d{h}{t}=-k$$

For part b:

I don't know where to start.(Doh) Each time I think I have a clue it just adds to the confusion.

Cheers Nemo

 

[MarkFL]

Okay, for part b) we are being asked to solve the IVP:

$$\frac{dh}{dt}=-k$$ where $$h(0)=h_0$$

Solving this IVP will give us $h$ as a function of $t$, which we can solve for $t$ and then set $h=0$ to determine the time at which the container is empty. So, what I would do to solve the ODE is integrate both sides with respect to $t$ (replace the dummy variables of integration), using the initial and final conditions as the limits of integration:

$$\int_{h_0}^{h(t)}\,du=-k\int_0^t\,dv$$

What does this give us after applying the FTOC?

 

[Nemo1]

Okay, for part b) we are being asked to solve the IVP:

$$\frac{dh}{dt}=-k$$ where $$h(0)=h_0$$

Solving this IVP will give us $h$ as a function of $t$, which we can solve for $t$ and then set $h=0$ to determine the time at which the container is empty. So, what I would do to solve the ODE is integrate both sides with respect to $t$ (replace the dummy variables of integration), using the initial and final conditions as the limits of integration:

$$\int_{h_0}^{h(t)}\,du=-k\int_0^t\,dv$$

What does this give us after applying the FTOC?

Hi Mark,

So after applying the FTOC:

$$\d{}{x}\int_{a}^{x} f(t) \,dt = f(x)$$$$h(t)=-k\cdot t+c$$

Which set to zero:

$$h(0)=-k\cdot 0+c$$

We get:

$$c=0$$Which makes sense as when the height is zero the surface area is also zero.

Does my explanation make sense?

Cheers Nemo

 

[MarkFL]

Here is what I had in mind:

If we continue from here...

$$\int_{h_0}^{h(t)}\,du=-k\int_0^t\,dv$$

We then obtain (by using the anti-derivative form of the FTOC):

$$h(t)-h_0=-k(t-0)$$

Now, solving for $t$, we obtain:

$$t(h)=\frac{h_0-h}{k}$$

When all of the water has evaporated, we must have $h=0$, and so the time it takes for the water to evaporate is:

$$t(0)=\frac{h_0}{k}$$

 

[Nemo1]

Thanks Mark for the extra explanation.

What happens to the $"+c"$ after the anti-derivative?

Is it a case of $+c$ on the first integral and $-c$ on the other?

Cheers Nemo

 

[MarkFL]

Thanks Mark for the extra explanation.

What happens to the $"+c"$ after the anti-derivative?

Is it a case of $+c$ on the first integral and $-c$ on the other?

Cheers Nemo

Definite integrals don't have a "constant of integration"...that's just indefinite integrals. :)

You could solve the problem using indefinite integrals though (I find it is simpler to use the boundaries as the imits of integration in definite integrals.

We could go back to:

$$\d{h}{t}=-k$$

Integrate both sides w.r.t $t$ to get:

$$h(t)=-kt+C$$

We are told to let $h(0)=h_0$ and so we may write:

$$h(0)=-k(0)+C=C=h_0$$

And so we find:

$$h(t)=-kt+h_0$$

This is equivalent to what we found using definite integrals, which I will demonstrate:

Suppose you have the initial value problem (IVP):

$$\frac{dy}{dx}=f(x)$$ where $$y\left(x_0\right)=y_0$$

Now, separating variables and using indefinite integrals, we may write:

$$\int\,dy=\int f(x)\,dx$$

And upon integrating, we find

$$y(x)=F(x)+C$$ where $$\frac{d}{dx}\left(F(x) \right)=f(x)$$

Using the initial condition, we get

$$y\left(x_0 \right)=F\left(x_0 \right)+C$$

Solving for $C$ and using $$y\left(x_0\right)=y_0$$, we obtain:

$$C=y_0-F\left(x_0 \right)$$ thus:

$$y(x)=F(x)+y_0-F\left(x_0 \right)$$

which we may rewrite as:

$$y(x)-y_0=F(x)-F\left(x_0 \right)$$

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

$$\int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx$$

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

$$\int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv$$

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.

 

[Nemo1]

Hmm, your explanation is great! I am yet to confidently say I get it tho!

My teacher has given me some sample problems from Schaum's Outlines to go through which start easy and get harder.
I am hoping to get through them before the exam and also some practice at $related rates$ word problems.

I don't think I have a firm grasp of applying $time$ into these formulas / equations yet.
In the practice exam I can see L'Hopital limits, vectors (dot, cross and addition), definite and indefinite integrals, related rates, areas from vector points, parametric curves and derivative of a natural log. (Doh)

Will have to work hard over the next month to get confident.

Many thanks for your help! I really appreciate your time and effort.

Cheers Nemo