#criticalthinkingcc – Julia’s Blog

Desmos Art Functions Card 2022

This desmos project took several hours to complete, yet I am happy with the result. I did this project in grade 10, so I was familiar with the constant, linear, quadratic, cubic, square root, cube root, absolute value, exponential and rational functions. This year I learned how to graph and manipulate sin, cos, and tan graphs, as well as non-functions (relations) such as sideways parabolas and circles. When I needed to graph a new line, I would sketch the basic shape and then compared the shape to the arsenal of functions I knew, which is how I chose which equation to utilize. The main challenge I encountered was trying to shade with the relations (sideways parabolas and circles), because I am not familiar with this; however, I overcame this challenge when I rewrote the non-functions as functions (for example, when I wanted to shade a sideways parabola, I made two square root functions derived from the original equation, and then utilized them to shade the sideways parabola). Another challenge was when I realized I could not graph SpongeBob’s face with accuracy in the time I was given. So, I made a decision to modify the face so that it was easier to graph yet still held the integrity and youthfulness of SpongeBob. The aha moments of this project were when I learned how much easier it was to manipulate a basic function with function notation and transformations rather than making a new equation/function for each line, which I did in grade 10. I completed this project alone, with no help, but I did assist others. Some people were making new equations for each line rather than transforming one function over and over, and so I showed them how to transform their equations to be transformations of the same function. I told them to make a folder with all of their basic functions, and then refer to those functions when they wanted to make a new one. The main strategy I used was to make the outline of the image with functions and transformations before beginning to shade. This was beneficial, because when if I wanted to change the shape of something, I would not have to change all of the shading as well. This strategy was efficient, and I was able to complete the shading with ease. This assignment helped me understand more about transformations of functions and relations, because I spent so much time manipulating and transforming them. I was able to witness how changing the domain, range, vertical/horizontal stretch, vertical/horizontal translation could have such a drastic impact on the shape of the line.

Link to my desmos project: https://student.desmos.com/activitybuilder/instance/63b47d10b9d7d3fd3f789875/student/63c3ab49e647cadd6bde721d#screenId=6fe1849c-43b9-4a5d-b03b-31bc3cfd24d8

Original photo

My desmos project

Desmos Art Functions Card 2020

For my pre-calc 10 math class, I was challenged to replicate a photo using functions and shading. At the beginning I was mainly utilizing the functions I was the most familiar with, such as the linear function, constant function, quadratic function, and exponential function. However, as I started to slowly incorporate the functions that were still fairly new to me, the square root function, absolute value function, and the cubic function, I began using them more frequently. My knowledge on these functions and how various components of an equation can manipulate the function it produces definitely developed with this project. I figured out what functions to use by analyzing my original photo and searching for where specific functions could be utilized. If I needed a slanted, straight line, I would use the linear function or absolute function. Since vertical lines are not functions, we weren’t allowed to use them, so often when my photo required a vertical line I would use the exponential function. The constant function would be used for horizontal lines. The cubic function, quadratic function, and square root function were used for any curved lines.

When working on this project, I started by replicating myself first, since I was in the centre of the photograph. Then I tackled the road, and eventually completed the houses from left to right, attempting to adjust the shading in the process to depict the brightness from the sun and how much darker the surrounding area is that the sun doesn’t reach.

There were definitely challenges along the way that I had to overcome. The shading is a significant aspect to consider when replicating a photo. There were only six colours available on Desmos, so I had to adapt and figure out how to properly overlap colours, and adjust their transparency and weight to create colours that were similar to my photo. The sun beaming down in my photo is something I really wanted to capture in my project, so I overlapped various colours and transparency levels to attempt to replicate the beautiful orange and yellow tones. Another challenge I encountered was the construction of the large tree. At first, I thought it would be easier to create an abundance of lines to form the tree rather than create an outline and shade it in. However, that was extremely difficult and unrealistic to accomplish, so I decided to challenge myself in another way and utilize square root functions instead, and so the left portion of the tree is created using square root functions as an outline.

As I was completing this assignment I had a realization; although I was tempted to utilize many small lines to achieve the perfect outline of an object or part, I learned that it is much more visually appealing when you attempt to use fewer lines since you can create this smooth, elegant shape rather than having multiple angles from trying to connect numerous little lines.

A strategy I utilized for this project was outlining a specific part or object, and then immediately shading it in after. This technique helped me tremendously since I was able to fully complete an outline and then shade the section in while still having the equations fresh in my mind, and then just entirely move on. Another strategy I used was just ordering my functions in each folder in a specific order, and this was extremely beneficial since I wouldn’t have to continuously check which functions I need to shade because I knew what order they were in.

Overall I enjoyed this project, and persistently utilizing these functions strengthened my ability to manipulate and maneuver previously learned and new functions.

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Math Study Matrix – Core Competency Reflection

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Studio Arts 3D Final Assignment – Pop Up Self Reflection

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The Basic Box Pop-Up

Bridge Layers Pop-Up

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Linear Relations Desmos Project and Core Competency Reflection

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Everything I know about exponents

Diagram of a power:

2. Describe how powers represent repeated multiplication

Powers represent repeated multiplication since repeated multiplication is the same as multiplying a number by itself a certain amount of times. Powers are a short way of writing out repeated multiplication for an individual number.

If you had $3^4$ , the exponent (4) is representing the number of times the base (3) is written out in a repeated multiplication expression.

When written out in the form of repeated multiplication, $3^4$ = 3 $\times$ 3 $\times$ 3 $\times$ 3 = 81.

For example, $2^5$ = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 = 32.

4. Demonstrate the difference between two given powers in which the exponent and the base are interchanged by using repeated multiplication, such as $2^3$ and $3^2$ .

There is a difference between two given powers when the exponent and base of the powers are interchanged. Interchanging the base and exponent is not the same as switching the order of numbers in a multiplication question, and getting the same answer either way. The exponent is the amount of factors of the base you have.

For example $2^3$ represents 2 $\times$ 2 $\times$ 2 which equals 8, and $3^2$ represents 3 $\times$ 3 which equals 9.

6. Explain the role of parentheses in powers by evaluating a given set of powers such as $(-2)^4$ , $(-2^4)$ , and $-2^4$ .

Parentheses have a significant role in determining whether a power will have a positive or negative outcome.

For $(-2)^4$ , when the negative sign is within the brackets, and the exponent is outside the brackets, it implies that the base is negative. You would apply the exponent to the negative base.

However, whenever there is an exponent inside the brackets, such as $(-2^4)$ , or there isn’t any brackets at all, such as $-2^4$ , the negative sign stands for a coefficient (-1) and the base is positive, so you apply the exponent to the positive base, and than the negative coefficient afterwards, since Exponents is before Multiplication in BEDMAS.

For example, if you have $(-2)^4$ , it would be written out as:

(-2) $\times$ (-2) $\times$ (-2) $\times$ (-2), which would equal +16.

If you had $(-2^4)$ , it would be written out as:

(-1 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2), which would equal (-16).

If you had $-2^4$ , it would be written out as:

-1 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2, which would equal -16.

8. Explain the exponent laws for raising a product and quotient to an exponent.

When given the expression $(4\times 2)^2$ , people tend to use BEDMAS instead of using the raising a product to a power exponent law.

For example, they would multiply the numbers within the brackets first (4 x 2) = 8

Then they would apply the exponent, so they would have the expression $8^2$ which equals 64.

However, if you are multiplying two or more numbers inside of brackets, and on the outside of the brackets there is an exponent, you can apply that exponent to every individual number instead of applying that exponent to the product of the numbers inside the brackets. This is called raising a product to an exponent, which is an exponent law.

For example, if you have the expression $(4\times 2)^2$ you could write it as:

$4^2$ $\times$ $2^2$ which is equivalent to 16 $\times$ 4 = 64.

In certain expressions, such as $(4\times 2)^2$ , it would be easier to apply BEDMAS, however in other expressions using the raising a product to an exponent law is more efficient.

For example, if you have the expression $(3a\times 4)^3$ instead of using BEDMAS and evaluating the expression within the brackets first, $(3a\times 4)^3$ = $12a^3$ = $1728a^3$ , you could apply the raising a product to an exponent law,

$(3a\times 4)^3$ = $(3^3 a^3\times 4^3)$ = 27 $\times$ $a^3$ $\times$ 64 = $1728a^3$ .

27 $\times$ $a^3$ $\times$ 64 is easier to evaluate than $12^3$ .

When given the expression $(\frac{4}{2})^2$ , people also tend to apply BEDMAS instead of using the raising a quotient to a power exponent law.

For example, they would divide the numbers within the brackets first $(\frac{4}{2})$ = 2

Then they would apply the exponent, so they would have the expression $2^2$ which equals 4.

However, if you are dividing one number by another inside of brackets, and on the outside of the brackets there is an exponent, you can apply that exponent to each number individually rather than applying the exponent to the quotient of the two numbers inside the brackets. This is called raising a quotient to an exponent, which is also an exponent law.

For example, if you have the expression $(\frac{4}{2})^2$ you could write it as:

$(\frac{4^2}{2^2})$ which is equivalent to $(\frac{16}{4})$ = 4.

In certain expressions, such as $(\frac{4}{2})^2$ , it would be easier to apply BEDMAS, however in other expressions using the raising a quotient to an exponent law is more efficient.

For example, if you have the expression $(\frac{4}{3})^2$ , instead of using BEDMAS and evaluating the expression within the brackets first, $(\frac{4}{3})^2$ = $1.\overline{33}^2$ ≈ 1.78, you could apply the raising a quotient to an exponent law.

$(\frac{4}{3})^2$ = $4^2$ $\div$ $3^2$ = 16 $\div$ 9 = $1.\overline{7}$ ≈ 1.78.

64 $\div$ 9 is easier to evaluate than $1.\overline{33}^2$ .

10. Use patterns to show that a power with an exponent of zero is equal to one.

$2^4$ =16, $2^3$ = 8, $2^2$ = 4, $2^1$ = 2, so $2^0$ has to equal 1.

When you write out all the powers of a certain base, and find the answers for all of them, you will notice a pattern. As the exponent decreases by 1, the answer is divided once by the base. As shown above, as the exponent decreases by 1 from 5 to 4, the answer for $2^5$ which is 32 is divided by 2 which is 16 since $2^4$ = 16.

So if $2^1$ = 2, than if you decrease the exponent by 1, which would equal 0, you would have to divide the previous answer by 2. $2^1$ = 2, and $\frac{2}{2}$ = 1.

12. Use patterns to explain the negative exponent law.

$4^4$ = 256, $4^3$ = 64, $4^2$ = 26, $4^1$ = 4, $4^0$ =1, $4^{-1}$ = $\frac{1}{4}$ , $4^{-2}$ = $\frac{1}{16}$ .

As said above, when you write out all the powers of a certain base, as you subtract 1 from the exponent, the answer will be equivalent to dividing the previous answer by the base number.

If you take $4^0$ = 1, and you subtract 1 from the exponent, which would make it $4^{-1}$ , to find the answer you would divide the 1 (the previous answer) by 4 (the base number) where you would get 1 $\div$ 4, which is the same as $\frac{1}{4}$ .

As you keep dividing by the base (which is 4) and you have negative exponents, your answer is going to be the reciprocal of what the answer would have been if the exponent was positive.

For example, $4^2$ = 16 while $4^{-2}$ = $\frac{1}{16}$ . They are the reciprocal of each other.

14. I can identify the error in a simplification of an expression involving powers.

A common error in the simplification of an expression using powers is when people apply the product and quotient law when trying to find the sum and difference of powers. The quotient law applies to powers when you are dividing two powers with the same base. The product law applies to powers when you are multiplying two powers with the same base. You have to apply BEDMAS when adding and subtracting powers.

A typical mistake made with the quotient law is that it is applied to a subtraction question, such as

$2^6$ – $2^2$ = $2^4$ which would lead them to assume the answer is 16.

However, you cannot apply the quotient law to a subtraction question, so the solution should be written out as:

$2^6$ – $2^2$ = 64 – 4 which is 60.

The product law applies to powers when you are multiplying two numbers with the same base. People often apply the product law in an addition question, such as $2^2$ + $2^3$ = $2^5$ which would lead them to believe it equals 32.

Although, you can’t apply the product law to an addition question, so the solution should be written out as:

$2^2$ + $2^3$ = 4 + 8 which is 12.

In addition, there is sometimes confusion with the power law, since the power law is when you multiply the exponent within the brackets with the exponent on the outside of the brackets. This is used so you only have to multiply the base by one exponent.

A common error is that people confuse the power law with the product law, and add the two exponents together, such as:

$(2^6)^2$ = $2^8$ which would guide them to think the answer is 256. However, the solution should be written out as $(2^6)^2$ which is $2^{12}$ which is 4096.

16. Determine the sum and difference of two powers.

You can easily and quickly determine the sum and difference of two powers by evalutating the answer for each individual power, and either use addition or subtraction to answer the problem. Do not apply the product or quotient exponent law since you are adding and subtracting, and these laws only apply when you’re multiplying or dividing two powers with the same base.

For example, $3^3$ + $5^3$ you would evaluate $3^3$ , which is 27, and then $5^3$ which is 125, and then you would add the two numbers together. 27 + 125 is 152.

18. Use powers to solve problems (measurement problems)

You can use powers to solve a variety of measurement problems, one of them being to find the surface area and volume of a cube. To find the surface area of a cube you need to find the area of one side of the cube, and multiply that number by 6 (since there is 6 sides on a cube). You can find the surface area of a 5cm x 5cm x 5cm cube by using the expression

$6(5^2)$ since $5^2$ is equal to 5cm x 5cm = the area of one side of the cube. $6(5^2)$ = $150cm^2$ . To find the volume of the cube, you would need to multiply the length, width, and the height of the cube together. You can find the volume of the cube using the expression $(5^3)$ since the length, width, and height of a 5cm x 5cm x 5cm cube is the same. $(5^3)$ = $125cm^3$ .

You can also use powers to solve Pythagorean theorem questions, when trying to find the length of one side of a right triangle.
For example the formula for finding the hypotenuse of a right triangle is $(a^2)$ + $(b^2)$ = $(c^2)$ .

Step 1. If we were finding the hypotenuse of a right triangle where side a = 3mm and side b = 4mm we would start out by replacing the variables in the formula with the numbers assigned to them.The equation would look like:

$(3^2)$ + $(4^2)$ = $(c^2)$

Step 2. Then you would evaluate each individual power. $(3mm)^2$ = $9mm^2$ , $(4mm)^2$ = $16mm^2$ , so the equation would look like:

9 + 16 = $(c^2)$ .

Step 3. Next we would add the two side lengths together which would equal:

25 = $(c^2)$ .

Step 4. The next step would be to find the square root of each number to isolate the c.

5mm x 5mm = 25mm so 5 is the square root of 25.

5mm= c so the missing side length (the hypotenuse) equals 5mm.

The answer is: 5mm

Another way you could use powers to solve problems is by using them to find the total volume of two shapes. As said above, you find the volume of a cube by multiplying the length, width, and height the cube together. You can find the volume a 3mm x 3mm x 3mm cube by using the power $(3mm)^3$ . Then you would find the volume of the other cube, which is 2mm x 2mm x 2mm, by using the power $(2mm)^3$ . Then you would add the two volumes to evaluate the volume of the whole shape altogether. $3^3$ = 27mm and $2^3$ =8mm. $27mm^3$ + $8mm^3$ = $35mm^3$

20. Applying the order of operations on expressions with powers involving negative exponents and variable bases.

When simplifying a question that includes negative exponents and variable bases, you have to take it step by step.

I’ll be applying the steps to the expression $\frac{(a^2b^3) (b^7a^3)}{a^6b^8}$ . Always follow BEDMAS, and make sure you evaluate powers before multiplying, dividing, adding, or subtracting, unless you are using the product or quotient law. If using the product law, it allows you to add the exponents when multiplying powers with the same base. If using the quotient law, it allows you too subtract exponents when dividing powers with the same base.

Step 1. Apply the product law. You add the exponent of the same base together when multiplying.

$\frac{a^5 b^{10}}{a^6b^8}$

Step 2. Apply the quotient law. You subtract the exponent of the same base when dividing.

$a^{-1} b^2$

Step 3. Get rid of any negative exponents. If there is a negative exponent, you put the reciprocal of the power to make the exponent

$\frac{b^2}{a}$

The answer is: $\frac{b^2}{a}$

In addition, I’ll be applying the steps to the expression $(2\times a^2\times b^3)^{-2}$ $(4\times a^{-3})^{-3}$ . Always follow BEDMAS, and make sure you evaluate powers before multiplying, dividing, adding, or subtracting, unless you are using the product or quotient law. If using the product law, it allows you to add the exponents when multiplying powers with the same base. If using the quotient law, it allows you too subtract exponents when dividing powers with the same base.

Step 1. Apply the exponent. For this expression, we can apply the raising a product to a power law, which as said above means to apply the exponent individually to each number and variable.

$(2\times a^2\times b^3)^{-2}$ = $2^{-2}$ $a^{-4}$ $b^{-6}$