Bay Area Tutoring

Mar 02

When is torque useful in an equilibrium problem?

By Tutor GuyNo Comments

When you are trying to find a particular force for a system in equilibrium, you draw an FBD and set the sum of all the forces equal to zero. But sometimes, this doesn’t provide enough information for you to calculate the missing value. For example, check this problem out:

Example: A 6 m bench is supported by two posts at its two ends. The bench is a uniform piece of wood with a weight of 200 N. A dog with a weight of 100 N sits 2 m from one end of the bench. Find the magnitudes of the forces exerted on the bench by the two supports.

Solution: If we start by drawing a schematic of the bench, we can attach all the forces:

Obviously, the four forces sum to zero—the two support forces pushing up balance the weight of the bench and the dog. But that only gives us one equation and we have two unknowns—the forces upward by each leg of the bench.

How do we solve the problem? We recognize that the sum of all the torques also equals zero. But if the bench is not rotating, what’s the pivot point? We can use any point we want! Let’s pick the left end of the bench, because then the force from the left support does not exert a torque. There are then two clockwise torques (the center of mass of the bench and the dog) and one counterclockwise torque (the right leg of the bench). Now we have two equations:

$F_1+F_2-F_{bench}-F_{dog}=F_1+F_2-200-100=0$

$\tau_{bench}+\tau_{dog}-\tau_2=(200)(3)+(100)(4)-(F_2)(6)=0$

$\therefore F_2=167 \, N \text{ and } F_1=133 \, N$

Note that we could have chosen any other point along the bench as our pivot point, and we would have obtained the same result.

Physics

physics, torque

Feb 23

u substitutions with definite integrals

By Tutor GuyNo Comments

When a definite integral requires a u substitution to solve, be sure to substitute for the limits of integration as well. This way, you don’t need to substitute back in for the original function. Instead, you evaluate the integral using the new (u) limits. Here’s an example to show how this works.

Evaluate:

$\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta$

This is an obvious candidate for a u substitution. (See other posts on this website for more information on when to use u substitutions.)

Let $u= \tan \theta$ . Then $du=\sec^2 \theta \; d \theta$ .

But don’t stop there! Use your expression for u to determine the new limits as well.

$\theta =0 \rightarrow u=0; \; \theta= \dfrac{\pi}{4} \rightarrow u=1$

So the new integral becomes

$\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta =\int_0^1 u^2 \; du= \left. \dfrac{1}{3}u^3 \right |_0^1=\dfrac{1}{3}-0=\dfrac{1}{3}$

You have found the solution to the original integral without needing to put the integral back in terms of $\theta$ .

Calculus

calculus, integrals, math help, u substitution

Feb 16

Factoring ax² + bx + c, part 2

By Tutor GuyNo Comments

In my last post, I showed you a procedure for how to factor a quadratic when the leading coefficient is not equal to 1. Here I refine that method slightly with a short cut that makes your work even quicker. However, this method is a bit more difficult—if you are not really good at mental calculations, you will need to practice this a few times to get it down. To compare methods, I’ll use the same quadratic we factored in the last post.

Problem: Factor $12x^2+4x-5$

Our first step is the same as in the previous method—we need to find two numbers that multiply together to give $60$ (that’s $12*-5$ ) and add together to give $+4$ . Of course, those numbers are the same as they were in the last post: $+10$ and $-6$ .
……
Now here’s the shortcut. Divide both of these numbers by the leading coefficient, $12$ . That gives two fractions:
…..
….. $\dfrac{10}{12} \;\; \text{and } \dfrac{-6}{12}$
…..
Next, simplify the fractions if possible. In our example, we get
…..
….. $\dfrac{5}{6} \;\; \text{and } \dfrac{-1}{2}$
…..
In each fraction, the denominator is the coefficient of the x term in our factor and the numerator is the constant term. So our two factors must be $6x+5$ and $2x-1$ . See the diagram below:

Algebra 1, Algebra 2

factoring, math help, math homework

Feb 09

Factoring ax² + bx + c

By Tutor GuyNo Comments

You probably have a lot of experience factoring quadratics of the form $x^2+bx+c$ . (That is, where $a$ , the coefficient on the $x^2$ term is equal to $1$ .) But what do you do when $a$ isn’t $1$ ? For example, let’s factor $12x^2+4x-5$ . You may have been taught a trial-and-error technique where you look for all the numbers that multiply to give $12x^2$ and combine them with all the numbers that multiply to give $5$ and test each one to see which gives the correct middle term. But there’s a simple algorithm that takes all the guesswork out. I’ll show you the procedure with an example.

Problem: Factor $12x^2+4x-5$

Similar to the case where $a=1$ , look for two numbers that multiply together to give $a \cdot c$ (in our example, $12 \cdot -5=-60$ ) and add together to give $b$ (i.e., $+4$ in our example). [Actually, you can see that this is just like the $a=1$ case, because there you were looking for two numbers that multiply together to give $c$ , but when $a=1$ , $a \cdot c=c$ …] For our example, those numbers are $+10$ and $-6 \\$ .
…..
Bring down the first term without changing it:
…..
….. $12x^2$
…..
Write the linear term as a sum or difference using the two numbers you found in step A.:
…..
….. $12x^2$ $+10x-6x$
…..
(Don’t worry about which term comes first. As it turns out, you can write these two terms in either order.)
…..
Bring down the last term without changing it:
…..
….. $12x^2$ $+10x-6x$ $- \; 5$
…..
Now comes the magic. Look at the first two terms only from step D (cover up the last two terms if that helps you) and factor them:
…..
….. $2x$ $\!\!(6x+5)$
…..
Now look at the last two terms in step D (again, cover the first two terms if that helps you) and factor them. And this is very important: if the $x$ term has a minus sign on it, always factor out the minus sign:
…..
….. $2x$ $\!\! (6x+5)$ $- \; 1$ $\!\! (6x+5)$
…..
(Note that if nothing factors out of the last two terms, write a $1$ (you can always factor $1$ out of any term).
…..
Are the two expressions in the parentheses the same? They have to be. If they’re not, you’ve done something wrong. That’s one of the factors in your answer. The other factor is made up of the numbers outside the parentheses:
…..
….. $(2x-1)$ $\!\! (6x+5)$

Tada! I’ll leave it to you to show that if you had written the linear terms in step C in the reverse order, you’d get the same answer.

Algebra 1, Algebra 2

factoring, math help, math homework

Feb 02

The addition rule and multiplication rule in probability

By Tutor GuyNo Comments

Two of the many formulas you learn in statistics are referred to as the Multiplication Rule and the Addition Rule. How do you figure out when to use each one? This is actually a pretty easy one. The multiplication rule is used to determine the probability of two events both happening, one after the other. The key word to look for in a problem is ‘and’.

Example 1: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven first AND a heart second? Because you want the probability of both events occurring, you use the multiplication rule.

The addition rule is used to find the probability that either one of two events occurs. The key word here is “or’.

Example 2: You draw a single card from a standard deck, replace it and draw a second card. What is the probability you draw a seven on the first draw OR a heart on the second draw? Because you want the probability of either event occurring, you use the addition rule.

How do the rules work? The multiplication rule comes in two forms. If the two events are independent, that is, if the probability of the second event does not change when the first event occurs, then the formula is simple:

$P(A \cap B) = P(A) \cdot P(B)$

If the probability of the second event depends on the first event occurring, the formula is modified slightly to show this:

$P(A \cap B) = P(A) \cdot P(B|A)$

You can see from either formula why this is called the multiplication rule. The addition rule is just slightly more complicated.

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

You can see from this formula why it is called the addition rule. Let’s solve both problems.

Example 1: The two events are independent, so we use the first version of the rule:

$P(A \cap B) = P(A) \cdot P(B)=\dfrac{4}{52} \cdot \dfrac{13}{52}= \dfrac{1}{52}$

Example 2: (Note that we use the result from Example 1 in solving this problem.)

$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\dfrac{4}{52}+ \dfrac{13}{52}- \dfrac{1}{52}= \dfrac{16}{52}= \dfrac{4}{13}$

Algebra 2, Statistics

addition rule, math help, math homework, multiplication rule, probability, statistics

Jan 26

Finding the inverse of a 3X3 matrix

By Tutor GuyNo Comments

What’s the easiest way to find the inverse of a 3×3 matrix? Use your graphing calculator, of course! But if you need to find the inverse without a calculator, here’s a method that will give you the solution with the least amount of trouble. We’ll demonstrate with an example.

Find the inverse of the following matrix:

$\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{bmatrix}$

The first step in finding the inverse is to calculate the determinant of the matrix. The easiest way to calculate a 3×3 determinant is to write the matrix out, and append the first two columns at the end:

$\begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 10 \end{bmatrix} \begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6\\ \end{matrix}$

From here, you need to find the six different products along each of the diagonals shown:

Add the blue products together and add the red products together, then subtract the red total from the blue total: $(50 + 96 + 84) - (105 + 48 + 80) = -3$ . This is the determinant. By the way, if the determinant is $0$ , stop. Your matrix does not have an inverse.

Next, you need to find the elements of the inverse matrix. Here’s a clever trick that will help you do that. Start by writing the transpose of the original matrix. This is done by changing all the rows into columns:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix}$

Then write the first two columns over on the right and the first two rows over again on the bottom. Your array should look like this:

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix} \begin{matrix} 1 & 2 \\ 4 & 5 \\ 7 & 8 \end{matrix} \\ \begin{matrix} \text{ }1 & 2 & 3 & \,\, 1 & 2 \\ \text{ }4 & 5 & 6 & \,\, 4 & 5 \end{matrix}$

The next step is a little tricky to explain, though once you’ve done it, it’s pretty easy to figure out. For each of the nine positions in the matrix, you find the value of the determinant of the 2×2 array that is just below it and to the right. [One way to think of this is that each position in the matrix is the upper-left number of a 3×3 array. If you mentally delete the first row and column of that array, you have a 2×2 array left. This is the array for which you find a determinant. I demonstrate this below with the $1$ and the $6$ from the array above.

Do this process for all nine elements of the matrix and you should end up with this matrix:

$\begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}$

Finally, divide this matrix by the determinant you found above. The result will be the inverse of the original matrix:

$\dfrac{1}{-3} \begin{bmatrix} 2 & 2 & -3 \\ 4 & -11 & 6 \\ -3 & 6 & -3 \end{bmatrix}= \begin{bmatrix} ^{-2} \! /_3 & ^{-2} \! /_3 & 1 \\ ^{-4} \! /_3 & ^{11} \! /_3 & -2 \\ 1 & -2 & 1 \end{bmatrix}$

I’ll leave it to you to verify that this is indeed the inverse of the original matrix. For any 3×3 matrix that has an inverse, this method will calculate it for you correctly every time.

Algebra 2, Precalc/Trig

inverse of a matrix, math help, math homework, matrices

Jan 19

Completing the square

By Tutor GuyNo Comments

Are you an expert at completing the square? This is an important skill that is not nearly as difficult as a lot of students think it is. If you master just a couple of rules, you will be able to complete squares like a pro.

Let’s start with an explanation. Since we complete the square in algebra and not in geometry, exactly where is the square we are completing? This confuses many students and is probably the reason why they are frightened unnecessarily by the process. But it’s actually pretty simple. Let’s say we have the following expression:

$(x-3)^2$

This is obviously the square of $(x-3)$ , and so we call it a perfect square. If we expand the expression into $x^2-6x+9$ , we recognize that it is still a perfect square. Any trinomial of the form $ax^2+bx+c$ is a perfect square if it is the square of a binomial expression.

When we have an expression such as $x^2-6x$ , we look for a number we can add to make it a perfect square of a binomial. In this case, adding $9$ to $x^2-6x$ to get $x^2-6x+9$ gives us the square of $x-3$ . Therefore we have completed the square. That’s all the expression means!

Completing the square is actually quite easy in most cases. Because $(x+a)^2=x^2+2ax+a^2$ , anytime we have an expression of the form $x^2+2ax$ , all we need to do is add $a^2$ . You can do this in two quick steps:

Take the coefficient on the $x$ term and divide it by $2$ .
Square this number and add it to the end.

For example:

$x^2+8x$ : Divide the $8$ by $2$ (you get $4$ ) and square it (you get $16$ ), so the completed square is $x^2+8x \; +$ $16$
$x^2-12x$ : Divide the $-12$ by $2$ (you get $-6$ ) and square it (you get $36$ ), so the completed square is $x^2-12x \; +$ $36$
$x^2+9x$ : Divide the $9$ by $2$ (you get $\frac{9}{2}$ ) and square it (you get $\frac{81}{4}$ ), so the completed square is $x^2+9x \; +$ $\frac{81}{4}$
$x^2-13x$ : Divide the $-13$ by $2$ (you get $\frac{-13}{2}$ ) and square it (you get $\frac{169}{4}$ ), so the completed square is $x^2-13x \; +$ $\frac{169}{4}$

Next you write the factor as a square. The factor is always of the form $(x+a)^2$ or $(x-a)^2$ where the value of $a$ is the number that you squared. For the four examples above:

$x^2+8x \; +$ $16$ $= (x$ $+ \; 4$ $)^2$
$x^2-12x \; +$ $36$ $= (x$ $- \; 6$ $)^2$
$x^2+9x \; +$ $\frac{81}{4}$ $= (x$ $+ \; \frac{9}{2}$ $)^2$
$x^2-13x \; +$ $\frac{169}{4}$ $= (x$ $- \; \frac{13}{2}$ $)^2$

What if your expression starts with $ax^2$ where $a$ is not equal to $1$ ? Then you have to factor the $a$ out of both terms before you complete the steps above. It can appear very complicated, but the process is the same every time. Here are some examples:

$2x^2+8x=2(x^2+4x) \rightarrow 2(x^2+4x+4)=2(x+2)^2$
$4x^2-12x=4(x^2-3x) \rightarrow 4(x^2-3x+ \frac{9}{4}) = 4(x- \frac{3}{2})^2$
$2x^2+9x=2(x^2+ \frac{9}{2}x) \rightarrow 2(x^2+ \frac{9}{2}x+ \frac{81}{16}) = 2(x+ \frac{9}{4})^2$
$3x^2-13x=3(x^2- \frac{13}{3}x) \rightarrow 3(x^2- \frac{13}{3}x+ \frac{169}{36})=3(x- \frac{13}{6})^2$

Algebra 1, Algebra 2

completing the square, math help, math homework

Jan 12

How to approach integration by parts

By Tutor GuyNo Comments

In an earlier post, I described a strategy for approaching integrals and how to decide what technique to use. One question I suggested you ask yourself is “Can I use integration by parts?” In this post, I describe how you decide when integration by parts is the right approach, and how you decide which term will be u and which will be dv.

There is no product rule for integrals, but when you see an integrand that is the product of two functions [and u substitution has been ruled out], integration by parts is often the right approach. For example, each of the following integrals should be solved using integration by parts:

$\displaystyle \int x^2e^x \; dx \\ \displaystyle \int e^x \sin x \; dx \\ \displaystyle \int x \tan^{-1} x \; dx$

Once you decide that integration by parts is the correct technique, does it matter which factor you make u and which you make dv? Usually, yes. But don’t worry if you make the wrong choice—you’ll know that pretty quickly. Remember that the purpose of integration by parts is to take a complicated integrand and make it simpler. If you choose your u and dv incorrectly, your integrand will get more complicated. When you find your integral getting worse, just start again and switch your choices for u and dv. (If switching choices doesn’t make your integral look better, perhaps integration by parts isn’t the proper technique for the integral.)

Here’s a trick to help you decide which factor should be the u term: LIPET. This is a mnemonic to help you determine the priority for assigning factors to be u. The acronym stands for the following:

L: logarithmic functions
I: inverse trigonometric functions
P: polynomials
E: exponential functions
T: trigonometric functions

For example, the first integral above is the product of a polynomial and an exponential function. P precedes E in LIPET, so make x² the u term and e^x the dv term. The third integral above is the product of a polynomial and an inverse trig function, so let tan^-1x be the u term and x will be the dv term.

Finally, note that even though integration by parts creates a simpler integral, it won’t always create an integral you can evaluate by inspection. Often, you will need to use another technique to integrate the new integrand, or perhaps you will need to do integration by parts a second time.

Calculus

calculus, integration, integration by parts, math help, math homework

Jan 05

Strategies for finding integrals

By Tutor GuyNo Comments

How does one become an “expert integrator”? There are a lot of integration techniques, and recognizing which technique is the best is the first step to finding the integral. Many new calculus students look at an integral and don’t even know where to start. Here is a strategy you can adopt when faced with an integral. Carry on this internal dialog with yourself:

Is this something that I can integrate by inspection?
Is this an integral that can be turned into an elementary integral by simplifying?
Is there a trig identity that will simplify the integral?
Is it a candidate for a u substitution?
Can I use integration by parts?
Will decomposition by partial fractions be useful?
Does the integral involve a product of trig functions?
Would a trig substitution be effective?
Can I find it in a table of integrals?

Yes, this is a long list. But with practice, you can work your way through it pretty quickly. [If you are taking Calc AB this year, you will only learn the first four or five steps.] Think of each integration technique as a tool for your toolbox; with each integral you encounter, you decide which tool is the best for that problem.

Let’s look at a couple of examples to see how this works.

Example 1. Evaluate:

$\displaystyle \int {\sec}^2 x+x^2-2x \; dx$

This is an integral consisting of only elementary derivatives, so you can integrate it by inspection:

$\displaystyle \int {\sec}^2 x+x^2-2x \; dx = \tan x+ \dfrac{1}{3}x^3-x^2+C$

Note that you need to know the derivatives of the elementary functions in order to recognize when you can integrate by inspection. This means the derivatives of the trig functions, exponential and log functions and the inverse trig functions, not just polynomials.

Example 2. Evaluate:

$\displaystyle \int (x^2-1)(2x^2+3x) \; dx$

This is not an elementary derivative, so it cannot be integrated by inspection. But the integrand can by simplified by multiplying the terms out, turning it into an elementary derivative:

$\displaystyle \int (x^2-1)(2x^2+3x) \; dx \displaystyle \int 2x^4+3x^3-2x^2-3x \; dx=\dfrac{2}{5}x^5+\dfrac{3}{4}x^4-\dfrac{2}{3}x^3- \dfrac{3}{2}x^2+C$

Example 3. Evaluate:

$\displaystyle \int \tan^2 x \; dx$

This is not an elementary derivative. Is there a trig identity that will help? Yes, use the following Pythagorean identity to turn this into a simple integral:

$\displaystyle \int \tan^2 x \; dx = \displaystyle \int \sec^2 x-1 \; dx = \tan x-x+C$

At this point, you might be thinking “how am I supposed to see that identity?” Well, there’s no substitute for experience and practice. It’s important to know your basic trig identities, so you can rewrite a trig integral in a different form. If one identity doesn’t work, try another. Finding integrals is as much an art as it is a science. Be creative and be inventive.

(Look for other tips on integral problems on this website for strategies on u substitution, integration by parts, products of trig functions and trig substitutions.)

Calculus

calculus, integrals, math help, math homework

Dec 29

Sum( and Seq( commands on your calculator

By Tutor GuyNo Comments

Can you find the sum of the following series?

$\displaystyle \sum_{i=1}^{15}\dfrac{2n+1}{3n-2}$

This is neither an arithmetic nor a geometric series, so you don’t have a formula for it. This would be a tedious problem to do by hand. Fortunately, your graphing calculator can do these problems quickly and efficiently.

There are two functions you need to use on your calculator. The seq( command creates a sequence of terms based on a rule that you give. The sum( command adds together the terms in a sequence. Both functions are found on the LIST menu on your calculator. The seq( command is on the OPS submenu and the sum( command is on the MATH submenu.

To sum a series, you combine the two commands. If you have the new operating system on your calculator, it will prompt you for the entries when you select the seq( command. If you have the old operating system, you need to know the syntax for the command. The syntax for the series above is:

sum(seq((2x + 1)/(3x – 2),x,1,15))

Note that the seq( command has four parameters in the parentheses. From left to right, these are 1) the rule for the nth term of the sequence; 2) the variable name; 3) the first value of the variable; and 4) the final value of the variable. Now all you need to do is type this in to your calculator and let it do the crunching:

Algebra 2, Calculator, Calculus, Precalc/Trig

algebra 2, calculator, calculus, graphing calculator, math help, math homework, precalc, sequences, series, trig

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