Sunday, February 5, 2017

High School Math Solutions – Derivative Applications Calculator, Tangent Line

We learned in previous posts how to take the derivative of a function. Now, it’s time to see the applications of derivatives.

We use derivatives when we find the equation of a tangent line. A tangent line is a straight line that just touches the curve at a point (on the curve). Tangent lines can help us find the length of the curve and their slopes tell us what the curve looks like and where we can find maximum and minimums. When we take the derivative of the function of the curve at a particular point, we get the slope of the tangent line.



Let’s see how we can use taking the derivative to find the equation of a tangent line.

Steps to find the equation of a tangent line at a point:

1.  Find the tangent point

  • Plug in the value for x into the function to find the y coordinate 

2.  Compute the slope of the function

  • Take the derivative of the function

3.  Compute the slope of the function at the given x coordinate

  • Plug in the value for x into the derivative 

4.  Use the point-slope formula to find the equation of the tangent line

  • y-y_1=m(x-x_1) 
  • Get (x_1, y_1) from Step 1 and get m from Step 3

We’ll now go over some examples.

First example (click here):

                                      Find the tangent line of \f(x)=\sqrt{x^2+1}  at  x=-1

1. Find the tangent point

                                                  \f(-1)=\sqrt{(-1)^2+1}=\sqrt{2}

                                                                 (-1,\sqrt{2})

2. Compute the slope of the function

                                                  \f(x)=(x^2+1)^(\frac{1}{2})

                                        \f^' (x)=\frac{1}{2} (x^2+1)^(\frac{-1}{2})∙2x

                                                    \f^' (x)=\frac{x}{\sqrt{x^2+1}

3. Compute the slope of the function at the given x coordinate

                                         \f^' (-1)=\frac{-1}{\sqrt{(-1)^2+1}}=\frac{-1}{\sqrt{2}}

                                                             m=\frac{-1}{\sqrt{2}}

4. Use the point-slope formula to find the equation of the tangent line

                                                  (-1,\sqrt{2})            m=\frac{-1}{\sqrt{2}}

                                                       y-y_1=m(x-x_1)

                                                      y-\sqrt{2}=\frac{-1}{\sqrt{2}}(x-(-1))

                                                      y-\sqrt{2}=\frac{-1}{\sqrt{2}} x-\frac{1}{\sqrt{2}}

                                                        y=\frac{-1}{\sqrt{2}} x+\frac{1}{\sqrt{2}}

Next example (click here):

                                                 Find the tangent line of \f(x)=\frac{1}{x^2}   at (-1,1)
1. Find the tangent point

             We can skip this step because the tangent point is given

2. Compute the slope of the function

                                                                        \f(x)=x^(-2)

                                                                     \f^' (x)=-2x^(-3)

                                                                      \f^' (x)=\frac{-2}{x^3}

3. Compute the slope of the function at the x coordinate

                                                                  \f^' (-1)=\frac{-2}{(-1)^3} =2

                                                                               m=2

4. Use the point-slope formula to find the equation of the tangent line

                                                                   (-1,1)          m=2

                                                                      y-1=2(x-(-1))

                                                                           y=2x+3

Last example (click here):

                                             Find the tangent line of \f(x)=x^2+2x+3 at  x=2
1. Find the tangent point

                                                              \f(2)=2^2+2(2)+3=11

2. Compute the slope of the function

                                                                     \f^' (x)=2x+2

3. Compute the slope of the function at the x coordinate

                                                                 \f^' (2)=2(2)+2=6

                                                                           m=6

4. Use the point-slope formula to find the equation of the tangent line

                                                                  (2,11)        m=6

                                                                     y-11=6(x-2)

                                                                         y=6x-1

As you can see, finding the equation of a tangent line of a point on a curve is not too hard. As long as you’ve mastered computing derivatives and the steps to finding the equation of the tangent line, you will be able to solve these problems quick and easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Tuesday, January 24, 2017

Advanced Math Solutions – Vector Calculator, Advanced Vectors

In the last blog, we covered some of the simpler vector topics. This week, we will go into some of the heavier vector topics. This includes dot product, cross product, and projection. So let’s hop into it.

























Let’s see some examples using the dot product . . .
First example (click here):



Next example (click here):




Now, we will learn about cross product. In multiplication, we often see 1×1 or 1∙1, which both equal the same thing. However, these symbols are very different when talking about vectors. It is important to not interchange the symbols.
















While finding the cross product of two vectors, it is important to know what direction the cross product of the two vectors will point. This is where the right hand rule comes into play.








We’ll go over some properties of cross products.














Now that you have a brief overview on cross products, here’s an example on how to find the cross product of two vectors (click here):




When finding the cross product of two vectors, I find it easier to make a matrix and derive the answer from the matrix, rather than memorizing the formula for the answer.


Onto the last topic, projection . . .












Simple enough, let’s see an example (click here):




In this blog post, we learned about three heavy vector topics. The key to being successful in these topics is to memorize the formulas. If you do that, this will be a piece of cake. Use the definitions, properties, and facts to familiarize yourself with how to find if two vectors are orthogonal or parallel; these tend to be teachers’ favorite questions.

Until next time,

Leah

Thursday, December 22, 2016

Advanced Math Solutions – Vector Calculator, Simple Vector Arithmetic

Vectors are used to represent anything that has a direction and magnitude, length. The most popular example of a vector is velocity. Given a car’s velocity of 50 miles per hour going north from the origin, we can draw a vector. In this blog post, we will focus on the simpler aspects of vectors. We won’t talk about how to graph vectors.

Position vectors are vectors that give the position of a point from the origin. The vector is denoted as \vec{v}=<a_1,\:a_2,\:a_3> and starts at point A=(0, 0, 0) and ends at point B=(a_1,\:a_2,\:a_3).

This brings us to how to find a vector given an initial and final point. Given two points A=(a_1,\:a_2,\:a_3) and B=(b_1,\:b_2,\:b_3), the vector \vec{AB}, which goes from point A to B, is \vec{v}=<b_1-a_1,\:b_2-a_2,\:b_3-a_3>.


Adding vectors is very simple

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}+\vec{b}=<a_1+b_1,\:a_2+b_2,\:a_3+b_3>

Here’s an example of adding vectors (click here):




Subtracting vectors is just as simple

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}-\vec{b}=<a_1-b_1,\:a_2-b_2,\:a_3-b_3>

Here’s an example subtracting vectors (click here):





Scalar multiplication is used to lengthen or shorten a vector
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and any number c, c\vec{a}=<ca_1,\:ca_2\:ca_3>


Here’s an example of scalar multiplication (click here):



Every vector has a magnitude and a direction. The direction is where its arrow is pointed and the magnitude is the length of the vector. If the magnitude of a vector is 1, then we call that vector a unit vector.


Magnitude is denoted as |\vec{a}| or ||\vec{a}||.
We will use ||\vec{a}||, so we don't get confused with absolute values.
||\vec{a}||=\sqrt{a_x^2+a_y^2}


Here’s an example of finding a vector’s magnitude (click here):




A unit vector \hat{u\:}, is a vector with length 1.
\hat{u\:}=\frac{u}{||u||}


Here’s an example of finding the unit vector of a vector (click here):






Here are some properties to memorize about basic vector arithmetic:











The topics we covered in this blog are simple. I recommend practicing a few examples and memorizing the formulas, and you should be good to go. We are going to cover some of the heavier vector topics in next blog.

Until next time,

Leah

Tuesday, December 13, 2016

High School Math Solutions – Inequalities Calculator, Exponential Inequalities

Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential inequalities. The method of solving exponential inequalities is very similar to solving logarithmic inequalities. In these problems, the variable is in the exponent and our goal is to isolate the variable, which means getting it out of the exponent. We will see some these problems in the forms:

                                                                   e^x>a

                                                                   a^x<b

Let’s see how to solve these problems step by step.

1. Use algebraic manipulation to move everything that is not in the exponential expression to one side
2. Isolate the variable by getting rid of the exponential expression

Example:
e^x>a
e^x>e^{\ln(⁡a)}
x>\ln(a)

Example:
a^x<b
a^x<a^{\log_a(b)}
x<\log_a⁡(b)

3. Solve the inequality

These problems are a little simpler than solving logarithmic inequalities. Let’s see how to solve an example step by step.

First example (click here):

                                                     3^{x+1}+1<2

Step 1: Use algebraic manipulation to move everything that is not in the exponential expression to one side

                                                    3^{2x+1}+1<2

                                                       3^{2x+1}<1

Step 2: Isolate the variable by getting rid of the exponential expression

                                                       3^{2x+1}<1

                                                  3^{2x+1}<3^{\log_3(1)}

                                                       2x+1<\log_3⁡(1)

Step 3: Solve the inequality

                                                       2x+1<\log_3(1)

                                                           2x+1<0

                                                            x<\frac{-1}{2}

That was pretty simple. Let’s see some more examples.

Second example (click here):



Last example (click here):



Solving exponential inequalities are simpler than solving logarithmic inequalities. However, it can still get a little tricky when solving these inequalities with more parts. For more help and practice on solving exponential inequalities, visit Symbolab’s practice.

Until next time,

Leah

Tuesday, November 29, 2016

High School Math Solutions – Inequalities Calculator, Logarithmic Inequalities

Last post, we talked about radical inequalities. In this post, we will talk about how to solve logarithmic inequalities. We’ll see logarithmic inequalities in forms such as \log_b(f(x))<a or \ln(⁡f(x))<a. In order to solve these inequalities, the goal will be to isolate the variable, just as in any inequality, and we will do this by getting rid of the log function. Let’s dive in and see how to solve logarithmic inequalities.

Steps to solve logarithmic inequalities:
1. Use algebraic manipulation to move anything that is not in the logarithmic expression to one side
2. Combine logarithmic expressions
3. Isolate the variable by getting rid of the logarithmic expression
             Ex:    \log_b⁡(f(x))<a

b^(\log_b⁡(f(x))) <b^a

f(x)<b^a

Ex: \ln(f(x))<a

\ln⁡(f(x))<\ln⁡(e^a )

f(x)<e^a
4. Solve inequality
5. Get the range for the expression in the original log function
             Ex:  \log_b⁡(f(x))
Range: f(x)>0
6. Combine ranges

Let’s do an example step by step now.

First example (click here):

                                            \log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

Step 1: Use algebraic manipulation to move anything that is not in the logarithmic expression to one side

There’s nothing to move, so we can skip this step.

Step 2: Combine logarithmic expressions

                                            \log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

                                                      \log_4⁡(\frac{x+3}{x+2})≥\frac{3}{2}


Step 3: Isolate the variable by getting rid of the logarithmic expression

                                                 \log_4⁡(\frac{x+3}{x+2})≥\frac{3}{2}

                                                4^(\log_4⁡(\frac{x+3}{x+2})) ≥4^(\frac{3}{2})

                                                         \frac{x+3}{x+2}≥8

Step 4: Solve inequality

                                                          \frac{x+3}{x+2}≥8

We can see that this is now a rational inequality. We won’t solve this step by step; I will show the answer after solving this inequality. If you are struggling with solving this inequality, visit the blog post on rational inequalities.

                                                             -2<x≤\frac{-13}{7}

Step 5: Get the range for the expression in the original log function

                                                   \log_4⁡(x+3)-\log_4⁡(x+2)\ge\frac{3}{2}

                                                       x+3>0              x+2>0

                                                       x>-3              x>-2

Step 6: Combine ranges

                                                    -2<x≤\frac{-13}{7},    x>-2,    x>-3

                                                              -2<x≤\frac{-13}{7}

That wasn’t too bad! Let’s see some more examples.


Second example (click here):



Last example (click here):




Solving logarithmic inequalities is not too difficult. Just remember to get the ranges inside the logarithmic expressions and to double check your work. For more help and practice on this topic visit Symbolab’s  practice.

Until next time,

Leah


Tuesday, November 22, 2016

High School Math Solutions – Inequalities Calculator, Radical Inequalities

Last post, we went over how to solve absolute value inequalities. For today’s post, we will talk about how to solve radical inequalities. Solving radical inequalities is easier than solving absolute value inequalities and require fewer steps. Let’s see the steps on how to solve these inequalities.

Steps:
  1. Isolate the square root
  2. Check that the inequality is true (i.e. not less than 0)
  3. Find the real region for the square root, (i.e. see when the expression inside square root igreater than or equal to 0)
  4. Simplify and compute the inequality
  5. Combine the ranges


Let’s see how to do one example step by step.

First example (click here):

                                                                    \sqrt{5+x}-1<3

Step 1: Isolate the square root

                                                                    \sqrt{5+x}<4

Step 2: Check that the inequality is true

Yes, this inequality is true the radical is not less than 0.

Step 3: Find the real region for the square root

                                                                       5+x≥0

                                                                         x≥-5

Step 4: Simplify and compute the inequality

                                                                    \sqrt{5+x}<4

                                                                (\sqrt{5+x})^2<4^2

                                                                     5+x<16

                                                                       x<11
Step 5: Combine the ranges

                                                                  x≥-5 and x<11

                                                                     -5≤x<11

That wasn’t too difficult. Let’s see some more examples.

Second example (click here):



Last example (click here):



Solving radical inequalities isn’t too difficult, however, they require practice. For more practice examples check out Symbolab’s practice.

Until next time,

Leah

Thursday, November 3, 2016

High School Math Solutions – Polynomials Calculator, Dividing Polynomials (Long Division)

Last post, we talked dividing polynomials using factoring and splitting up the fraction. In this post, we will talk about another method for dividing polynomials, long division. Long division with polynomials is similar to the basic numerical long division, except we are dividing variables. This is where it gets tricky. I will talk about the steps to dividing polynomials using long division to help make the process easier and go into detail.

Steps for polynomial long division:

1.  Organize each polynomial by higher order
  • We want to make sure that each polynomial is written in order of the variable with the highest exponent to the variable with the lowest exponent
  • you can skip this step if they are already in high order
2.  Set up in long division form
  • The denominator becomes the divisor and the numerator becomes the dividend

3.  Write 0 as the coefficient for missing terms in the dividend
  • Since we’ve put in order the terms based on their exponent, we can see which terms are missing (i.e. x^4+x^2 we can see we are missing an x^3 term so we will add that in to its proper spot and make the coefficient 0) 
  • This will help you with step 6, so you don’t subtract the wrong terms 
  • You can skip this step if there are no missing terms
4.   Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
  • This is allows us to see what we need to multiply the divisor by to get rid of the first term of the dividend
5.   Multiply the divisor by that term
  •  Write the term down on top of line where the term that is getting eliminated is 
6.  Subtract this from the dividend 
  • This gives you a new polynomial to work with
7.  Repeat steps 4-6 until you get a remainder 
  • When you repeat step 4, move onto the newest first term from step 6
8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial

This may seem a bit confusing, so we will go through two examples step by step to understand better how to solve these problems.

First example (click here):
                                                         \frac{(x^4+6x^2+2)}{(x^2+5)}

1.  Organize each polynomial by high order
     We can skip this step because the polynomials are already in high order
2.  Set up in long division form
3.  Write 0 as the coefficient for missing terms in the dividend

4.  Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
                                                   \frac{x^4}{x^2}= x^2

5.  Multiply the divisor by that term
     x^2∙(x^2+5)=x^4+5x^2

6.  Subtract this from the dividend

7.  Repeat step 4-6 until you get a remainder
     \frac{x^2}{x^2} =1

8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial
                                                              x^2+1+\frac{(-3)}{(x^2+5)}

 Last example (click here):
                                                             \frac{(2x^2-18+5x)}{(x+4)}

1.  Organize each polynomial by high order
                                                              \frac{(2x^2+5x-18)}{(x+4)}

2.  Set up in long division form

3.  Write 0 as the coefficient for missing terms in the dividend
     We can skip this step because there are no missing terms.

4.  Divide the first term of the dividend (numerator) by the first term of the divisor (denominator)
                                                                \frac{2x^2}{x}=2x

5.  Multiply the divisor by that term
     2x(x+4)=2x^2+8x

6.  Subtract this from the dividend

7.  Repeat steps 4-6 until you get the remainder
     \frac{(-3x)}{x}=-3

8.  Put the remainder over the divisor to create a fraction and add it to the new polynomial
                                                                  2x-3+\frac{(-6)}{(x+4)}

Dividing polynomials using long division is very tricky. It is so easy to skip an exponent, have an algebraic error, and forget a step. This is why practicing this type of problem is so important. The only way to get better at it is to keep practicing it. Check out Symbolab’s Practice for practice problems and quizzes.

Until next time,

Leah.