Determining rate order from initial reaction rate

By Tutor GuyNo Comments

When you are studying chemical kinetics, a typical problem will look something like this:

Compounds A and B react together according to the following reaction:

A + 2 B → C + D

The following initial reaction rate data was collected. Determine the rate equation for this reaction.

Rxn [A] [B] Initial rxn rate
1 0.0500 M 0.0500 M 1.2 x 10-4 M/s
2 0.100 M 0.0500 M 2.4 x 10-4 M/s
3 0.100 M 0.100 M 9.6 x 10-4 M/s


These problems are pretty easy to solve if you have learned exponential equations in algebra 2. The trick is to compare two different reactions where only one compound changes its concentration. For example, if you compare reaction 1 with reaction 2, you will see that [A] changes and [B] is the same. If you compare reactions 2 and 3, you see that [B] changes and [A] is held constant. However, when you compare reactions 1 and 3, both [A] and [B] change.  So do not compare reactions 1 and 3.

Next, remember that your answer will be in the form

Rate = k [A]m [B]n

Therefore, your job is to determine the values of m, n, and k. You do this by taking the values from the table and plugging them into this equation.

Let’s find the value of m first. You need two equations where [A] changes and [B] does not—that is, equations 1 and 2.

\begin{matrix} (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \\ (1) \quad & 1.2 \cdot 10^{-4} = k(0.0500)^m(0.0500)^n \end{matrix}

Now divide the first equation by the second. The magic here is that k and the term with the n exponent both cancel out, leaving you with a simple expression:

 2.0 = 2^m

It should be pretty obvious that m = 1. What if the numbers don’t divide evenly and you end up with an equation such as

2.03 = 1.982^m?

Unless told otherwise, you can assume that the exponent is supposed to be an integer, and therefore m= 1 in this problem too.

Now, let’s solve for n. We need two equations where [B] changes and [A] does not. That would be equations 2 and 3. Set it up the same way as before:

\begin{matrix} (3) \quad & 9.6 \cdot 10^{-4} = k(0.100)^m(0.100)^n \\ (2) \quad & 2.4 \cdot 10^{-4} = k(0.100)^m(0.0500)^n \end{matrix}

And divide as before:

4.0 = 2^n

Thus n = 2. Now we can solve for k by taking any of the three original equations and plugging in all the given values. Let’s use equation 3.

9.6 \cdot 10^{-4} = k(0.100)^1(0.100)^2

Solving this equation gives k = 0.96. Therefore, our final answer is

Rate = 0.96 [A] [B]2

Now here’s an additional trick I give you at no extra charge: If you are pretty good with exponents and scientific notation, you should see that you can often solve for the values of m and n in your head. In our first step, the rate doubled when [A] doubled, so the rate with respect to A is first order (m = 1). In our second step, the rate quadrupled when [B] doubled, so the rate with respect to B is second order (n = 2). (That was certainly a lot easier than the work we did above!) Then plug the known values into the rate equation to solve for k.


Differential rate law vs. integrated rate law

By Tutor GuyNo Comments


When we study rates of chemical reactions, we look for an equation that describes the reaction rate. The equation can be in one of two forms, known as the differential rate law and the integrated rate law.

If you have taken calculus, you’ll recognize these names and you even know how to derive the integrated law from the differential law. If you haven’t taken calculus, don’t worry! You probably won’t be asked to derive the equations. However, you should recognize each form of the law for zeroth, first and second order reactions.

The differential rate law gives the reaction rate as a function of the concentration of the reactants. The integrated rate law gives the concentration as a function of time. Both equations are useful, and you will probably use both in your chemistry class.

The table below shows the two rate laws for each reaction order in terms of a single reactant A.

Order  Differential rate law  Integrated rate law 
 Zeroth  Rate = k  [A] = -kt + [A]_0
 First  Rate = k[A]  \ln{[A]} = -kt + \ln{[A]_0}
 Second  Rate = k[A]2  \dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0}


You might have noticed that each of the integrated law equations is in the form y = mx + b, the equation for a line. In the old days (before graphing calculators), this is the method that chemists used to determine the reaction order. If the graph of concentration vs. time is a straight line, the reaction is zeroth order. If the graph of ln(concentration) vs. time is a straight line, the reaction is first order. And if the graph of 1/concentration vs. time is a straight line, the reaction is second order.



Solving integrals of the form tanm x secn x

By Tutor GuyNo Comments


In a recent post, I showed you how to approach integrals of the form

\displaystyle \int \sin^m{x} \; \cos^n{x} \; dx

In this post, we look at the similar integral of the form

\displaystyle \int \tan^m{x} \; \sec^n{x} \; dx

As before, m and n are integers. The technique we use requires m to be odd or n to be even and we factor out terms of the associated function. We demonstrate with an example. Evaluate:

\displaystyle \int \tan^5{x} \; \sec^6{x} \; dx

Here, m is odd and n is even, so we can factor either the tan x or the sec x function. We’ll do it both ways here to show both techniques. First, when m is odd, we factor out a single power of both tan x and sec x and the rest of the integral follows the method we saw in the sinm x cosnx integral example:

\displaystyle \int \tan^5{x} \; \sec^6{x} \; dx = \int \sec{x} \; \tan{x} \; \tan^4{x} \; \sec^5{x} \; dx =

\displaystyle \int \sec{x} \; \tan{x} \;  (\tan^2{x})^2 \; \sec^5{x} \; dx = \int \sec{x} \; \tan{x} \; (\sec^2{x} - 1)^2 \; \sec^5{x} \; dx =

\displaystyle \int \sec{x} \; \tan{x} \; (\sec^4{x} - 2 \sec^2{x} + 1) \; \sec^5{x} \; dx = 

\displaystyle  \int \sec{x} \; \tan{x} \; (\sec^9{x} -2 \sec^7{x} + \sec^5{x}) \; dx =

Then letting u = sec x leads to

\displaystyle f(x) = \frac{1}{10} \sec^{10}{x} - \frac{1}{4} \sec^8{x} + \frac{1}{6} \sec^6{x} + C

Now let’s do the same integral the other way. When n is even, we factor out a sec2 x, and the rest of the process is similar to the first method:

\displaystyle \int \tan^5{x} \; \sec^6{x} \; dx = \int \sec^2{x} \; \tan^5{x} \; \sec^4{x} \; dx =

\displaystyle \int \sec^2{x} \; \tan^5{x} \; (\sec^2{x})^2 \; dx = \int \sec^2{x} \; \tan^5{x}\; (\tan^2{x} + 1)^2 \; dx =

\displaystyle \int \sec^2{x} \; \tan^5{x} \; (\tan^4{x} +2\tan^2{x} + 1) \; dx = 

\displaystyle  \int \sec^2{x} (\tan^9{x} + 2 \tan^7{x} + \tan^5{x}) \; dx

Then letting u=tan x,

\displaystyle f(x) = \frac{1}{10} \tan^{10}{x} + \frac{1}{4} \tan^8{x} + \frac{1}{6} \tan^6{x} + C

Wait! That’s not the same integral we got in the method above. Well, actually, it is! After a trig substitution (1 + tan2 x = sec2 x) and a lot of messy algebra, you can show they are the same function. I’ll leave you to work out the details. Better yet, graph them both on your graphing calculator and you will see they are the same.


Solving integrals of the form sinm x cosn x when m and n are even

By Tutor GuyNo Comments


In my last post, I showed you the method for integrating functions of the form

\int \sin^m{x} \; \cos^n{x} \; dx

when either m or n is odd. If they are both even, the process is more complicated. In this post, I’ll show you how to attack this problem. There are two different methods you can use; if one doesn’t work so well, try the other. We will demonstrate both methods here with an example. Let’s integrate

\int \sin^2{x} \; \cos^4{x} \; dx

Method 1

First, rewrite the integral so that the factors are powers of sinx and cosx.

\int \sin^2{x} \; (\cos^2{x})^2 \; dx

Use the power reducing formulas

\sin^2{x} = \frac{1}{2}(1- \cos{2x}) \text{ and  } \cos^2{x} = \frac{1}{2}(1+\cos{2x})

to replace both the sinx and the cosx terms.

\displaystyle \int \sin^2{x} \; (\cos^2{x})^2 \; dx = \int \left(\frac{1}{2}(1- \cos{2x})\right) \left(\frac{1}{2} (1+\cos{2x}) \right)^2 \; dx

Then expand each term and multiply it all out (hey, I said it was more complicated…)

\displaystyle \int \left(\frac{1}{2}(1- \cos{2x})\right) \left(\frac{1}{2} (1+\cos{2x}) \right)^2 \; dx =

\displaystyle \frac{1}{8} \int (1- \cos{2x})(1 + 2 \cos{2x} + \cos^2{2x}) \; dx =

\displaystyle \frac{1}{8} \int 1 + 2 \cos{2x} + \cos^2{2x} - \cos{2x} - 2 \cos^2{2x} - \cos^3{2x} \; dx

\displaystyle \frac{1}{8} \int 1 + \cos{2x} - \cos^2{2x} - \cos^3{2x} \; dx

Now we are finally ready to integrate. The first two terms can be integrated by inspection. The third term is attacked by using the power reducing formula. In the last term, the exponent is odd, so apply the technique we used in the last blog post:

\displaystyle \frac{1}{8} \int 1 + \cos{2x} - \frac{1}{2} (1+\cos{4x}) - \cos{2x}\cos^2{2x} \; dx =

\displaystyle \frac{1}{8} \int 1 + \cos{2x} - \frac{1}{2} (1+\cos{4x}) - \cos{2x}(1- \sin^2{2x}) \; dx =

\displaystyle \frac{1}{8} \int \frac{1}{2} - \frac{1}{2} \cos{4x} + \cos{2x} \sin^2{2x} \; dx =

\displaystyle \frac{1}{8} \left( \frac{1}{2}x - \frac{1}{8} \sin{4x} + \frac{1}{6} \sin^3{2x} \right) + C = \frac{1}{16}x - \frac{1}{64} \sin{4x} + \frac{1}{48} \sin^3{2x} + C


Method 2:

First, rearrange terms so that there is a power of (sin x cos x).

\displaystyle \int \sin^2{x} \cos^4{x} \; dx = \int \sin^2{x} \cos^2{x} \cos^2{x} \; dx =

\displaystyle \int (\sin{x} \cos{x})^2 \cos^2{x} \; dx

Then use the double angle formula \sin{x} \cos{x} = \frac{1}{2} \sin{2x} and the power reducing formulas as necessary:

\displaystyle \int (\sin{x} \cos{x})^2 \cos^2{x} \; dx = \int \left(\frac{1}{2} \sin{2x} \right)^2 \left(\frac{1}{2} (1+ \cos{2x}) \right) \; dx

Expand the terms and multiply out:

\displaystyle \int \left(\frac{1}{2} \sin{2x} \right)^2 \left(\frac{1}{2} (1+ \cos{2x}) \right) \; dx = \frac{1}{8} \int \sin^2{2x} (1+ \cos{2x}) \; dx =

\displaystyle \frac{1}{8} \int \sin^2{2x} + \sin^2{2x} \cos{2x} \; dx = \frac{1}{8} \int \frac{1}{2} (1- \cos{4x}) + \sin^2{2x} \cos{2x} \; dx =

\displaystyle \frac{1}{8} \left( \frac{1}{2}x - \frac{1}{8} \sin{4x} + \frac{1}{6} \sin^3{2x} \right) + C = \frac{1}{16}x - \frac{1}{64} \sin{4x} + \frac{1}{48} \sin^3{2x} + C

Note that this is the same result as in Method 1 above.


Solving integrals of the form sinm (x) cosn (x)

By Tutor GuyNo Comments


An important class of integrals is of the form:

\int \sin^m{x} \; \cos^n{x} \; dx

where m and n are integers. If either m or n is odd, factor out a single power of that function and rewrite the integral to solve with a u substitution. This is best demonstrated with an example:

\int \sin^5{x} \; \cos^7{x} \; dx

Here, both m and n are odd, so we can select either function to factor. The process is easier when you pick the smaller of the two exponents, so let’s choose the sin x and we factor out one power as follows:

\int \sin{x} \; \sin^4{x} \; \cos^7{x} \; dx

We then rewrite the even power as a power of sin2 x so that we can apply a trig identity:

\int \sin{x} \; \sin^4{x} \; \cos^7{x} \; dx = \int \sin{x} \; (\sin^2{x})^2 \; \cos^7{x} \; dx =

\int \sin{x} \; (1-\cos^2{x})^2 \; \cos^7{x} \; dx = \int \sin{x} \; (1-2 \cos^2{x} + \cos^4{x}) \; \cos^7{x} \; dx =

\int \sin{x} \; (\cos^7{x} -2 \cos^9{x} + \cos^{11}{x}) \; dx

This is easily integrated with a u substitution (let u = cos x):

\dfrac{1}{8} \cos^8{x} - \dfrac{1}{5} \cos^{10}{x} + \dfrac{1}{12} \cos^{12}{x} + C

So what do you do if both m and n are even? Well, most students just skip the problem and go on to the next one. ☺ But if you want to see the technique, look for my next blog post.


Making sense of math induction

By Tutor GuyNo Comments


Do you know the sum of the first five positive odd numbers? How about the sum of the first 20 positive odd numbers? The first 100?  Wouldn’t it be great if there were a formula for the sum of the first n positive odd numbers so we could easily answer this question? Let’s see if we can find a pattern:

n Expression Sum
1 1 1
2 1 +3 4
3 1 + 3 + 5 9
4 1 + 3 + 5 + 7 16
5 1 + 3 + 5 + 7 + 9 25


It certainly appears that the sum of the first n positive odd numbers is n2. Or, if we write this symbolically:

\displaystyle \sum_{i=1}^n (2i - 1) = n^2

We’ve just shown that this formula is true for the first five values of n, but how do we know it is true for every value of n? We could test our theorem for n = 6, n = 7, n = 8, and so forth. But you can see that this process would take us a very long time to show it for every value of n. Like, forever! Fortunately, there is a better method for proving conjectures of this sort. It is called mathematical induction. Math induction is a very powerful tool in analysis. If you think about it, it allows us to prove an infinite number of theorems. And we can do this in only three steps!

How does math induction work? It depends on a subtle idea: if the theorem is true for one particular value of n (say n= k) and we can use this fact to show that it’s true for the next value of n (that is, n=k+1), then it will be true for every value of n ≥ k. That is, one proof turns into an infinite number of proofs! How is this possible? Because each value of n leads to the next value of n. If it’s true for n=3 then we’ve shown it’s true for n=4. But since it’s true for n=4, we’ve also shown it’s true for n=5. And therefore it’s also true for n=6, and then n=7, etc. Most textbooks that cover math induction use a dominoes analogy—if you knock down the first domino, it knocks down the second, the second knocks down the third and so on.


Here are the three steps in math induction, and you will see that they are quite simple to describe:

Step 1: prove the theorem is true for n=1. This is a really easy step. You just plug in n=1 into the theorem and show that it is true. (Note: on occasion, you will start at a value of n greater than 1. This does not change the validity of the process.)

Step 2: Assume that the theorem is true for n = k. (Note that you aren’t even proving anything in this step. You just write out what the theorem would look like if it were true for n=k.)

Step 3: Use the expression in Step 2 to show that the theorem is true for n=k+1. (You usually do this by applying a little bit of algebra.)

That’s all there is to it! Let’s see how it works on the theorem we proposed above. We restate the theorem here for convenience:

\displaystyle \sum_{i=1}^n (2i - 1) = n^2

Step 1: Prove for n = 1.

1 = 1^2     (See? This is a really easy step.)

Step 2: Assume for n=k

1+3+5+ \dots + (2k-1) = k^2    (Just write down the formula for n = k.)

Step 3: Prove for n = k+1

1+3+5+ \dots + (2k-1) = k^2  \; \; \text{[line 1]}

1+3+5+ \dots + (2k-1)+(2k+1) = k^2 + (2k+1)  \; \; \text{[line 2]}

k^2 + (2k+1) = (k+1)^2 \; \; \text{[line 3]}

[This step is a little harder to follow, so I’ve numbered the lines to explain the process.

  • Line 1 is simply a repeat of Step 2. I wrote it here for clarity; you don’t need to repeat your work.
  • In line 2, I’ve added (2k + 1) to both sides. That’s a valid algebra operation, so if Line 1 is true, Line 2 is also true. Why did I add (2k +1)? Because it’s the next term in the sum on the left side of the equation.
  • In Line 3, I’ve simplified the right side of the equation.]

But (k+1)2 is exactly what our theorem says the right side of the equation will be when n = k+1. So the assumption that the theorem is true for n=k led to the conclusion that it is also true for n=k+1. And our proof is complete.


Managing exponential equations

By Tutor GuyNo Comments


Many algebra 2 students get intimidated by exponential equations because the answers are rarely nice simple integers. But keep in mind that like all the equations you solved in algebra 1, you are merely trying to find the value of x that satisfies the equation. The process is a little more complicated because your variable is now in an exponent, but just follow these steps and you’ll soon be an exponential expert!

There are two types of exponential equations and they each have a preferred strategy for solving them. The first type has an exponential expression on both sides of the equal sign, and the two bases are both powers of the same number. The second type of equation has exponential expression with bases that aren’t powers of the same number (or has an exponential expression on only one side of the equals sign).

Type 1. Bases that are powers of the same number. Check out these two problems.

\text{a.  } 4^{x+4} = 8^{x-1}

Solution: note that the two bases (4 and 8) are both powers of 2. We rewrite both bases as powers of 2 and simplify:

\text{a.  } 4^{x+4} = 8^{x-1} \rightarrow (2^2)^{x+4} = (2^3)^{x-1} \rightarrow (2)^{2x+8} = (2)^{3x-3}

Now we take advantage of a simple property that says if a^x = a^y \text{ then } x = y.  Since both bases are the same, we just set the exponents equal to each other:

(2)^{2x+8} = (2)^{3x-3} \rightarrow 2x + 8 = 3x - 3. Solving this gives x = 11.


\text{b.  } 5^{4x} = 125^{x+1}

Solution: note that both bases are powers of 5. Proceed exactly as in the last problem:

\text{b.  } 5^{4x} = 125^{x+1} \rightarrow 5^{4x} = (5^3)^{x+1} \rightarrow 5^{4x} = 5^{3x+3} \rightarrow 4x = 3x + 3 \rightarrow x = 3


Type 2: Bases are not powers of the same number. To solve these types of problems, you will need to use logarithms. Here are two examples.

\text{a.  } 2^{x+3} = 12

Solution: The bases (2 and 12) are not powers of the same base. (Always check this first, because the method shown in the Type 1 examples above is almost always simpler.) So we solve by taking logs of each side. In the old days (before graphing calculators) you would need to use a common log (base 10) or a natural log (base e). I’ll show you that method first:

\text{a.  } 2^{x+3} = 12 \rightarrow \ln{(2^{x+3})} = \ln{12} \rightarrow (x+3) \ln{2} = \ln{12}

\therefore x = \dfrac{\ln{12}}{\ln{2}} - 3 \approx 0.5840

If your graphing calculator has the logbase command on it, you can solve this problem even more easily by taking a log base 2 of each side:

\text{a.  } 2^{x+3} = 12 \rightarrow \log_2{(2^{x+3})} = \log_2{12} \rightarrow x+3 = \log_2{12}

\therefore x = \log_2{12} - 3 \approx 0.5840


\text{b.  } 5^{x+3} = 7^{x-2}

Solution: This looks a lot uglier than the previous example, but the solution starts the same way. Take the log of both sides:

\text{b.  } 5^{x+3} = 7^{x-2} \rightarrow \log_5{(5^{x+3})} = \log_5{(7^{x-2})} \rightarrow (x+3) = (x-2) \log_5{7}

We need to collect the x terms to solve for x, so distribute on the right side and solve:

(x+3) = (x-2) \log_5{7} \rightarrow (x+3) = x \; \log_5{(7)} - 2 \; \log_5{(7)} \rightarrow

3 + 2 \log_5{(7)} = x \; \log_5{(7)} - x \rightarrow 3 + 2 \; \log_5{7} = x (\log_5{(7)} - 1) \rightarrow

x = \dfrac{3 + 2 \; \log_5{(7)}}{\log_5{(7)} - 1} \approx 25.92


Algebra 2, Precalc/Trig

Right-hand rules

By Tutor GuyNo Comments


The “right-hand rule” is a valuable tool in physics to help you determine the direction of vectors and fields. When you use the right-hand rule properly, your work is a lot easier. In this post, I describe a number of right-hand rules and show you how to apply them. But first, a warning:

Never, ever use a left-hand rule!

You may already know that one use of the right-hand rule is to find the direction of the force on a positively charged particle. If you want to know the direction of the force on a negatively charged particle, you use the right-hand rule and then reverse the direction you obtain. Almost every year, I get a student who comes up with the clever idea to use a left-hand rule on negative particles. Does this work? Absolutely! So why is it a bad idea? Because once you’ve used the left-hand rule, you’ve created a muscle memory that says it is okay. And the next time you need to use a right-hand rule, you may pick up your left hand without even thinking about it and you will get the wrong result. (This is particularly true if you are right handed—you already have a pencil in your right hand, so it seems perfectly natural to use your free hand.) You won’t even realize you got the wrong answer. When I teach students the right-hand rule for the first time, I train them to sit on their left hands so they won’t be tempted to use them by mistake. After 10 or 20 times using the right-hand rule, you’ll develop a muscle memory that says “only the right hand will do” and you will be less likely to make this mistake.

  • Cross products: When you find the cross product of two vectors, the result is perpendicular to each of the original vectors. But does the resultant vector point “above” the plane of the two vectors or “below” the plane? Use the right-hand rule to determine the direction of this vector. There are a number of ways to implement this right-hand rule. I’ve seen textbooks that teach you to imagine an arrow coming out of your palm. My dad learned to contort his fingers like this, with his thumb up, his index finger pointing out and his middle finger pointing perpendicular to his palm:

He taught this method to me, but I think it’s hard to remember which finger goes with which vector.






Here’s the technique I think works best: Let’s say you want to find the direction of \vec{u} \times \vec{v}. Point your fingers in the direction of \vec{u}.  Curl your fingers so that they point in the direction of \vec{v}.  (Find the smallest angle between \vec{u} and \vec{v}. ) Your outstretched thumb points in the direction of the cross product. Note that you may have to rotate your wrist before you curl your fingers so that they will point in the correct direction.








Here are some vector products where you can use the right-hand rule to determine the direction of the vector product:

\bullet \quad \vec{\tau} = \vec{r} \times \vec{F} (torque)

\bullet \quad \vec{L} = \vec{r} \times \vec{p} (angular momentum)

\bullet \quad \vec{F}_B = q\vec{v} \times \vec{B} (magnetic force on a moving charge)

\bullet \quad \vec{F}_{wire} = I\vec{\ell} \times \vec{B} (magnetic force on a wire due to a current)

Use this same rule when you are constructing coordinate axes in space. Use the rule to point the three positive axes in the correct direction given that \vec{x} \times \vec{y} = \vec{z}.

  • Magnetic field due to the current in a wire: When a current travels through a wire, it generates magnetic field lines that form concentric circles around the wire. But does the magnetic field point clockwise or counterclockwise? Use the right-hand rule to determine the direction of the field. Grip the wire with your right hand so that your thumb points in the direction of the (conventional) current. Then your fingers curl around the wire in the direction of the field. If the wire is bent into a loop, this same method tells you which direction through the loop the field points.

In this photo, the current is moving to the left. Point your thumb to the left, and you see the field lines are moving down behind the wire and are moving up in front of the wire.





  • Lenz’ law: When a loop of wire is placed in a location where the magnetic flux is changing, a current is induced in the wire. But in which direction is the induced current? Use a right-hand rule to determine the direction of the current. First determine the direction of the induced magnetic field predicted by Lenz’ law. If the flux is increasing through the loop the induced magnetic field has to point in the direction opposite the flux. If the flux is decreasing, the induced field points in the same direction as the flux. Now wrap your fingers around the wire so that they are pointing in the direction of this induced flux. Your thumb points in the direction of the induced current. Note that this right-hand rule is essentially the reverse of the previous rule.

In this photo, we have determined the induced flux must point out of the page towards our point of view. We curl our fingers to show this direction and we see the induced emf and induced current will be counterclockwise.







  • Ampere’s law: The loop integral \oint \vec{B} \cdot d \vec{s} around an Amperian loop is proportional to the net current through the loop. But in which direction through the loop is the current positive? Use a right-hand rule to determine the direction of positive current. Curl your fingers around the loop in the direction of integration. Your thumb points in the direction of positive current.

Our Amperian loop is shown in orange in this photo. The two red circles represent current-carrying wires. The top wire has current coming out of the page and the bottom wire has current flowing into the page.  We choose to integrate in a counterclockwise direction. Curl your fingers in that direction and your thumb points out of the page (towards your point of view). Therefore the top wire will be assigned a positive current and the bottom wire will be assigned a negative current in order to apply Ampere’s law.


Solving differential equations by separation of variables

By Tutor GuyNo Comments


A common differential equation is of the form y' = f(x) \cdot g(y) . In this situation, the equation can be solved by a technique called “separation of variables”. It involves putting all the y terms on one side of the equation and all the x terms on the other side. Then integration on both sides leads to a solution. Let’s look at a couple of examples.

Example 1

Solve: \dfrac{dy}{dx} = \dfrac{2x}{y^2} with the initial condition y(0) = 2.

First, multiply both sides by y2 and by dx to separate the variables (all the y terms on one side and all the x terms on the other):

y^2 \; dy = 2x \; dx

Note that dy/dx isn’t really a fraction, but you can treat it as a fraction to separate the variables. Now you have an exact differential on each side, so you can integrate both sides:

\int y^2 \; dy = \int 2x \; dx

\dfrac{1}{3} y^3 = x^2 + C

Why isn’t there a constant of integration on the left side? Well, there was, but we subtracted it and combined it with the constant of integration on the right so there is only one constant. Always do it this way, and put the constant on the independent variable side. Now solve for y:

y = \sqrt[3]{3x^2 + C}

Why didn’t we multiply the constant by 3? Well, we did, but a constant times three is still a constant, so we simplified it. Always keep your constant as simple as possible. Now solve for the constant, using the initial condition given:

2 = \sqrt[3]{3(0)^2 + C} \rightarrow C = 8

\therefore y = \sqrt[3]{3x^2 + 8}

Example 2:

Solve: \dfrac{dy}{dx} = \dfrac{y}{\sqrt{x}}  with the initial condition y(0) = 2.

As before, separate the variables:

\dfrac{dy}{y} = \dfrac{dx}{\sqrt{x}}

Note that the dy and dx terms must always be in the numerator. Now integrate and solve for y:

\displaystyle \int \dfrac{dy}{y} = \int \dfrac{dx}{\sqrt{x}}

\ln |y| = 2 \sqrt{x} + C

y = e^{2 \sqrt{x} + C} = Ce^{2 \sqrt{x}}

Wait, how did the constant move from the exponent to the coefficient? It’s a simplification trick. Proceed as follows:

y = e^{2 \sqrt{x} + C} = e^{2 \sqrt{x}} \cdot e^C 

But eC is a constant too, so just call it C. Where did the absolute value sign go? Because C can be either positive or negative, we can drop the absolute value sign around the y. Let’s finish the problem by solving for C:

2 = Ce^{2 \sqrt{0}} \rightarrow C = 2

\therefore y = 2e^{2 \sqrt{x}}


Tabular integration

By Tutor GuyNo Comments


Occasionally, you will be faced with a complicated integral such as

\int x^4 \cos 2x \; dx

If you’ve read some of my other posts, you know that this requires integration by parts. Let’s get started on it and see where it leads.

\begin{matrix} u = x^4 & dv = \cos 2x \; dx \\ du = 4x^3 \; dx & v = \dfrac{\sin 2x}{2} \end{matrix}

Then (after a bit of simplifying),

\int x^4 \cos 2x \; dx = \dfrac{1}{2} x^4 \sin 2x - 2 \int x^3 \sin 2x \; dx

Now what? Well, the new integral requires integration by parts too! So you go through all the steps again. Then you have to do integration by parts on this result. And again on the next result! It’s tedious, but eventually you get to the correct answer. However, if one of the two functions in the original integral is a polynomial, there is a faster way to do this process. It’s called tabular integration, because all the parts of the integral are found by filling in a table.

Make a two-column table, with the columns labeled u and dv. Put the polynomial function (in our example, x4) under the u. Then take the derivative of this function and write it below the original function. Continue taking the derivative until you reach 0. In our example, your first column now looks like this.

\begin{matrix} \underline{u} & \underline{dv} \\ x^4 & \text{ } \\ 4x^3 & \text{ } \\ 12x^2 & \text{ } \\ 24x & \text { } \\ 24 & \text{ } \\ 0 & \text{ } \end{matrix}

Next, place the other function in the dv column. Integrate this function repeatedly until there are entries in every row. In our example, your completed table will look like this:

\begin{matrix} \underline{u} & \underline{dv} \\ x^4 & \cos 2x \\ \text{ } \\4x^3 & \frac{1}{2} \sin 2x \\ \text{ } \\12x^2 & \frac{-1}{4} \cos 2x \\ \text{ } \\24x & \frac{-1}{8} \sin 2x \\ \text{ } \\24 & \frac{1}{16} \cos 2x \\ \text{ } \\0 & \frac{1}{32} \sin 2x \end{matrix}

You’ve done all the hard work! Now you can write out the integral from the values in this table. First draw an arrow from each u value (except 0) to the dv value on the next line below, as shown here:







Next, place alternating plus and minus signs on each arrow:







Now multiply each u value by the dv value at the other end of the arrow. The plus or minus sign tells you whether to add or subtract this term. For this problem, the integral would be:

(x^4)(\frac{1}{2} \sin 2x) - (4x^3)(\frac{-1}{4} \cos 2x) + (12x^2)(\frac{-1}{8} \sin 2x) - (24x)(\frac{1}{16} \cos 2x) + (24)(\frac{1}{32} \sin 2x)

Simplifying gives the final answer (don’t forget the C!):

\frac{1}{2} x^4 \sin 2x \; + \; x^3 \cos 2x \; - \; \frac{3}{2} x^2 \sin 2x \; - \; \frac{3}{2} \cos 2x \; + \;  \frac{3}{4} \sin 2x \; + \; C 


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