Tutoring at University of Maryland in College Park, MD

Calculus I, II, II, and differential equations
College Park Tutors Blog


Mar 29, 2015

Tonight, on the World's Most Extreme Values. One 2-variable function. One closed region. One shot at glory. Don't miss it!

...sorry, had to get that out of my system. The problem we're going to look at today goes like this:

Find the absolute minimum(s) and maximum(s) of the function $f(x,y)=xe^y-x^2-e^y$ on the rectangle with vertices $(0,0)$, $(0,1)$, $(2,0)$, and $(2,1)$.

Ok, we've seen extreme value (i.e., maximum and minimum) problems like this in Calculus 1. If you don't remember the gist of this, please go back and check your notes/textbook first. Just to review, the basic idea is that we find the derivative of a function, set it equal to zero, and solve the resulting equation. Together with the points where the the function is non-differentiable, these solutions give us a set of critical points where the function might have a maximum, minimum, or inflection point.

Our example has two new issues we must confront. First of all, we have a function of two variables, so what does it mean to "set the de...

Continue Reading...

Mar 27, 2015

If you're reading this post, you've probably arrived at the point in your calculus course where they try to teach you how to do integration by partial fraction expansion. And if you're anything like I was when I first learned calculus, you're probably scratching your head and going whadafuhhhh? What is the point of all this?

Here's what you need to know. Partial fraction expansion is not an integration technique. It's an algebraic technique. That being said, it's useful for making certain algebraic expressions (i.e. rational expressions) easier to integrate by breaking them into smaller, simpler chunks. Sound familiar? Its that same mode of thinking that all your other calculus classes are pushing at - taking a difficult problem, decomposing it into several easier problems, and then putting the pieces back together.

Got it? Let's get down to business. Here's our problem:

Rewrite the following expression as the sum of three partial fraction terms: \[\frac{1}{(x-1)(x^2+x+1)}\]

Just to quickly review the step...

Continue Reading...

Mar 22, 2015

Or more specifically, a second-order linear homogeneous differential equation with complex roots. Yeesh, its always a mouthful with diff eq. Oh and, we'll throw in an initial condition just for sharks and goggles. The problem goes like this:

Find a real-valued solution to the initial value problem $y''+4y=0$, with $y(0)=0$ and $y'(0)=1$. Your solution must be real-valued or you will not receive full credit!

If you'll recall, the steps for solving a second-order homogeneous diff. eq. are as follows:

  1. Write down the characteristic equation.
  2. Find the roots of the characteristic equation.
  3. Use the roots to write down the two exponential basis solutions.
  4. Create a general solution using a linear combination of the two basis solutions.

For step 1, we simply take our differential equation and replace $y''$ with $r^2$, $y'$ with $r$, and $y$ with 1. Easy enough:

Finding the characteristic equation.

For step 2, we solve this quadratic equation to get two roots. The roots are going to be complex numbers, but that's ok:

Solving for the characteristic roots.

Step 3 tells us...

Continue Reading...

Mar 02, 2015

Howdy folks, Alex here! I thought I'd start this blog off right, with one of the most popular (and head-spinning) problems that get thrown at my calculus students. The problem goes something like this:

Find the area of the region enclosed by the lines $y=2x$, $y=3x$, and $y=2$. You must use calculus or you will not receive any credit!

Whoa. Strong words from the guy with the gradebook. Alright, well hopefully you've already seen problems that ask for the area between two functions. If not, break out your textbook ;-) The 2-function area problems are solved by integrating the difference between the "top" and "bottom" functions, like so:

Finding the area between two functions.

But wait a minute! Our problem is giving us THREE functions, not two. How in the name of Bieber can we apply the above formula to a region enclosed by three functions?

I'll tell you how. We're going to figure out a way to decompose, or break down, this difficult problem into several smaller, easier problems. Learning how to break down problems, by the way, is the reaso...

Continue Reading...