Business Calculus

Calculus Problem of the Week June 27, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week. Note: these are no longer updated, but you should find the old problems helpful!
[adsenseWide]

This week’s problem (taken from Calculus – Larson 4th ed.)

As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area. Show that the radius decreases at a constant rate.

Solution

To start out, let’s figure out exactly what this problem is asking for. We want to show that the radius of this raindrop decreases at a constant rate. The first thing you should notice, is that this is a way of describing a rate of change, i.e. a derivative. We want to show that $\dfrac{dr}{dt}$ is a constant where $r$ is the radius and $t$ is some unit of time.

So, we are being asked to show something about a rate of change, and if you read carefully, you will see we are also given a rate of change. We are told the rate at which the raindrop evaporates is proportional to its surface area. This would mean the raindrop is getting smaller at a rate that’s proportional to its surface area – i.e., it is decreasing in volume. The surface area of a sphere is:

$\text{surface area} = 4\pi r^2$

Mathematically, we can translate the fact that the rate is proportional to this into:

$\dfrac{dV}{dt}=c(4\pi r^2)$
where $V$ is volume and $c$ is a constant.

But, we could also find $\dfrac{dV}{dt}$ using implicit differentiation on the volume formula for a sphere. The formula for the volume of a sphere is:

$V=\dfrac{4}{3}\pi r^3$

Taking $\dfrac{d}{dt}$ of both sides:

$\begin{align} \dfrac{dV}{dt} &= \dfrac{4}{3}\pi (3r^2)\dfrac{dr}{dt}\\ &= 4\pi r^2\dfrac{dr}{dt}\end{align}$

Now we have two formulas for $\dfrac{dV}{dt}$, so they must equal:

$c(4\pi r^2)=4\pi r^2\dfrac{dr}{dt}$

Solving for $\dfrac{dr}{dt}$ now gives $c$, a constant, which is what we were trying to show.

$\begin{align} c(4\pi r^2)& = 4\pi r^2\dfrac{dr}{dt} \\ \dfrac{c(4\pi r^2)}{4\pi r^2} & = \dfrac{dr}{dt}\\ \dfrac{dr}{dt} &= c\end{align}$

Further study

If you are taking a calculus course, you should take a look at the free calculus lessons on MathBootCamps!

Calculus Problem of the Week June 20, 2011

This week’s problem
Find the derivative of $f(x)=x\sqrt{x-1}$

See the solution.

Looking at this function, the first thing you should notice is that it is a product. Yes, there is a composite function ( $\sqrt{x-1}$ ) but it is PART OF the product. So your first step should be to use the product rule.

$f'(x)=(x)'\sqrt{x-1}+x(\sqrt{x-1})'=\sqrt{x-1}+x(\sqrt{x-1})'$

Now the second part involves a derivative of the composite function, so use the chain rule.

$\sqrt{x-1}+x(\sqrt{x-1})'=\sqrt{x-1}+x(\dfrac{1}{2}(x-1)^{-\frac{1}{2}})$

At this point, the calculus part is done and it is time to simplify using algebra:

$=\sqrt{x-1}+x(\dfrac{1}{2\sqrt{x-1}})=\dfrac{2(x-1)}{2\sqrt{x-1}}+\dfrac{x}{2\sqrt{x-1}}$
$=\dfrac{3x-2}{2\sqrt{x-1}}$

Download: Differentiation Rules

Even though you can usually find all of the rules in the book, it helps to have them all on one page and handy. That’s why we wrote up this “cheat sheet” of the basic differentiation rules. Print this out to use as you are working through your calculus problems! (Note that if you combine this with the chain rule and you can find the derivative of just about anything.)

Basic Differentiation Rules (PDF) download

Calculus Problem of the Week: June 13, 2011

Starting this week, we will start posting a calculus problem with its solution (hidden of course!)! Sometimes easy and sometimes hard, these problems could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice!

This week’s problem
Find the integral: $\int\!\dfrac{\ln(x)}{x}\,\mathrm{d}x$

See the solution.

This problem can be done with a straightforward u substitution. If you let $u=\ln(x)$ then $\mathrm{d}u=\dfrac{1}{x}\mathrm{d}x$ . Solving for $\mathrm{d}x$ then gives $\mathrm{d}x = x\mathrm{d}u$ .

Therefore, our integral becomes:
$\int\!\dfrac{\ln(x)}{x}\,\mathrm{d}x=\int\!\dfrac{u}{x}\,x\mathrm{d}u=\int\!u\,\mathrm{d}u=\dfrac{u^2}{2}+C=\dfrac{(\ln(x))^2}{2}+C$

Hide this solution.

Finding Limits Using a Graph

What is a limit?

Calculus involves a major shift in perspective and one of the first shifts happens as you start learning limits. When I talk about the limit of a function $f(x)$ as $x$ approaches some value, I am not saying “what is $f(x)$ at this value” like I might in algebra! Instead, I am interested in what is happening to $f(x)$ when $x$ is close to this value.

[adsenseWide]

Notation

If we are interested in what is happening to the function $f(x)$ as $x$ gets close to some value $c$ from the right, we write:

$\lim_{x \to c^+} f(x)$
This is called the right handed limit.

If we are interested in what is happening to the function $f(x)$ as $x$ gets close to some value $c$ from the left, we write:

$\lim_{x \to c^-} f(x)$
This is called the left handed limit.

Example

Now for an actual example! I have graphed the function $f(x)=x^2+1$ below. We are going to find two limits: The limit of $f(x)$ as $x$ approaches 1 from the right and the limit as $x$ approaches 1 from the left. Remember: I don’t care what is happening when $x=1$, I only care about what is happening what $x$ is close to 1!

From the picture above, I can see that $\lim_{x \to 1^-} f(x)=2$ and $\lim_{x \to 1^+} f(x)=2$. When both the right hand and left hand limits exist (there will be a different discussion about when limits don’t exist) and equal, then we say the two sided limit equals that value (when people say “the limit” they usually mean the two sided limit). In this case $\lim_{x \to 1} f(x)=2$. If the left and right hand limits were different numbers, we would say the two sided limit doesn’t exist.

Did you notice that here it is also true that $f(1)=2$? This does not always have to be the case! Take a look at this piece-wise defined function (that means there is a different definition for the function for different parts of the domain).

Notice the closed and open circles. The open circle at a y-value means that is not a value of the function when you plug in $x$. For example, $f(-1) = -4$ since that is where the solid circle is. Additionally, $f(3)$ is undefined since there is no solid circle at $x=3$. But what about the limits?

From the picture above, we see that $\lim_{x \to 3^-} f(x)=2$ and $\lim_{x \to 3^+} f(x)=2$ therefore $\lim_{x \to 3} f(x)=2$ even though $f(3)$ is undefined! Again, it doesn’t matter what is happening when $x=3$ only what is happening near that value!

However, $\lim_{x \to -1^-} f(x)=-4$ and $\lim_{x \to -1^+} f(x)=2$. Therefore, $\lim_{x \to -1} f(x)$ does not exist, even though $f(-1)=-4$. I don’t have to say it again do I? (the whole thing about we are looking near x not at x… there I said it…)

As you continue to study limits, the plan is to develop ways to find limits without using the graph, but being able to find a limit this way can give you a much better understanding of exactly what a limit is, even if you aren’t using the formal definition.