Weekly Math Problem #62

Improper Integral. Ever so often, I find my way back to calculus, and this week it's via an improper integral. By my recollection, the technique is okay. The type of function you're dealing with can definitely be the cause that the problem is annoying to do. Let's just hope this week's problem isn't annoying to you. 😬 

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     How to solve an improper integral
        ✔️     Methods of integration

     

WMP! #62 says to...

Happy solving!

Check back on Saturday, November 6th for the solution, which will be posted below ⬇️.

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Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Alright...let's get into this problem! If you looked up integral tables and used a calculator, that's good. I did too! I wasn't expecting anyone to have the information needed to solve this problem memorized. I didn't have it memorized, but I felt confident that I knew how to use the assistance.

I started this problem by using an integral table. Why? Well, I saw that the function was rational and knew that most, if not all, integral tables have some rational functions on them. When I saw what I needed, I rewrote the function in a way that help me begin solving the problem.

With the problem rewritten, I could see that I needed to invoke the substitution method to keep going. Since the substitution method uses a new variable, I also adjusted the lower bound of integration to match the new variable.

Once that was done, I could now move forward and integrate.

Since this is a definite integral, we need to evaluate! Evaluating the second term, which comes from the lower bound was simple. If you're familiar with this special value, then you can simple write the value. You can also use a calculator like I did. 😁 It's the first term, that requires a bit more work. Infinity isn't a value to plug in, so I used my calculator, again, to see if the function converges as my variable approaches infinity. It does!!!

Now that I have the values I need, I can plug them in and simplify.

I don't want to leave my answer like that, so I did a little "clean-up" by rationalizing the denominator.

Here is an image of the function. Keep in mind that the variable is approaching infinity. Even though, the shape doesn't end, the area under the curve converges to value above. 

**This plot was created using Geogebra's Graphing Calculator. Click image to enlarge.

▪️ Were you able to solve the improper integral??

▪️ Let me know what you thought about this week's problem in the comments section. 

Thanks for solving with me this week!

WMP! #63. is up next. 👩🏿‍🏫

Cheers!

The Younge Lady

Weekly Math Problem #60

Logarithmic Equation. Honestly, I don't want to lose momentum this week. I want to say more but...I...am...tired! 😩😩

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     Properties of Logarithms
        ✔️     Methods to solve a quadratic equation

     

WMP! #60 want us to...

Happy solving!

Check back on Saturday, October 23rd for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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If you've solved the problem above, then you would've seen that, indeed, solving a quadratic equation was required. There are options to solving a quadratic equation. How many options you have available to you depends on the type of quadratic equation you have. Below, you will see that I chose to use the completing the square method to solve the resulting quadratic equation. 

Did you forget that I like visuals? (I don't blame you if you did. 😅) The graph below shows the intersection where the LHS of the equation meets the RHS of the equation. As you can see, the intersection occurs when x=2...our solution! 👏🏿

**This plot was created using Geogebra's Graphing Calculator. Click image to enlarge.

▪️ Were you able to solve this week's equation??

▪️ Let me know what you thought about this week's problem in the comments section. 

Thanks for solving with me this week!

Up next...WMP! #61. 👍🏿

Cheers!

The Younge Lady

Weekly Math Problem! #41

Arc Length. Nothing strange going on here this week. I'm just getting back to a calculus problem I wanted to do last week, this week. A little application situation in the form of finding the arc length. 👀 There are times when it is needed to find the length of a curve and there aren't any regular geometric techniques that'll help with that, so calculus steps in to save the day. 😁

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     How to find the first derivative
        ✔️     How to work with hyperbolic functions
        ✔️     How to do integration

WMP! #41 says to:

Happy solving!

Check back on Friday, March 12th for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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And we're back to get this problem done. So, let's see... The problem is asking us to find the arc length of a function on a closed interval. There is a formula for such a situation.

In regards, to our problem, we need to identify the function, take its first derivative, and use the interval. Starting with the function, we're dealing with hyperbolic cosine who has a first derivative that is hyperbolic sine.

Now that we have the first derivative of the function and the interval, we can plug the information in the formula and perform the integration. Remember, we will come out with a numerical answer that represents the length of the function (a curve) for the specified endpoints; it makes that we are working on a definite integral.

As you can see, the arc length of f(x) = cosh(x) on [0, ln 2] is 3/4 units. Let me point out that a scientific or graphing calculator can easily evaluate hyperbolic functions for you. However, I still evaluated it by hand so you can see exactly where the final answer comes from. To assist in evaluating by hand, the exponential representation of hyperbolic cosine is used.

Here is an image of the given function and the part of the function for which we found the arc length.

 

**This plot was created using Geogebra's Graphing Calculator. Click image to enlarge.

 

▪️ We're you able to find the arc length?

▪️ Leave your response down below and let me know what you thought about this week's problem.

Thanks for solving with me this week!

Moving right along to WMP! #42. 👩🏿‍🏫

Cheers!

The Younge Lady

Weekly Math Problem! #34

Inverse of a Function. I have different ways of choosing problems to solve on this blog. Sometimes I go through books and notes I have. Other times, I use Google. 😁 This week, Google won as my method due to a lack of time. 🤷🏿‍♀️ What can I say? So, I decided to check out some precal topics and found a lighter problem to use. 

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     How to find the inverse of a function

That's it? Yup...that's it. WMP! #34 wants us to...

Happy solving!

Check back on Friday, January 22nd for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Let's go right on ahead and jump in this thing! 

Here comes the fun 🤪 part! (At least I think it's fun. 🤷🏿 ) The equation right above here is a literal equation. We are going to solve it but, the answer we get will not be a number. By placing x over 1, it can be more clearly seen that the above equation is a proportion.

We're just about done. Just one more thing...a mere formality:

That's it! We've found the inverse function, f -1. 🤸🏿‍♀️ Yay! Like the original function, the inverse is also a rational function, with a variable present in the denominator. That means we cannot use, or plug-in, any value for x that we like. To continue with the second part of the question, we will find the domain for f -1 by identifying which value of x will cause the function to be undefined. I did something slightly unconventional to find the value. Instead of setting the denominator equal to zero, I set the denominator not equal zero and proceeded normally:

Now, I can write the domain formally. I don't really have a preference for notation, so I did it two ways:

 

You know I wouldn't end this blog post without graphing the functions, right? Okay, great! (I tried something a little different with the color scheme. What do you think?) When a function and its inverse are plotted on the same set of axes, they will be mirror images of each other. The mirror is the line y = x.

**This plot was created using Geogebra's Graphing Gaclulator.

▪️ Let me know what you thought about this week's problem down below.

Thanks for solving with me this week!

Up next...WMP! #35

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Cheers!

The Younge Lady

Weekly Math Problem! #15

Quadrilaterals. Listen, geometry is not necessarily a favorite topic in mathematics for a lot of people. So, now you're thinking, "What about you, Lori? How do you feel about geometry?"  Well, it's not my favorite either 😅; I'm not in love with geometry. 🤷🏿 I will say, I am very visual, so I love the fact images, and the need to draw pictures, are a crucial part of geometry. I have lots of respect for this topic. Geometry is everywhere.

This week's problem was taken from the last administered NYS Regents High School Examination in Geometry before quarantine--January 2020, Part III, question 32. To solve this week's problem in completion, you need to recall the following math skills:

    ✔️  Plotting points on the coordinate plane

    ✔️  Slope, distance, midpoint formulas

    ✔️  Properties and theorems for quadrilaterals

    

WMP! #15 says...

Happy solving! 

Check back on Friday, August 14th for the solution, which will be posted below ⬇️.

Shameless plug: Follow me on Instagram @TheYoungeLady

✏️📓 Solution Time! 📓✏️

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In high school I learned the two-column proof method for geometry proofs. The proofs I remember doing the most were triangle proofs. I rarely, if ever, used the paragraph-style for writing geometric proofs. So, for this problem I'll complete the necessary computations and accompany them with statements that support those computations. 

There are various way to prove that a quadrilateral is a rhombus. This is because there are various ways to prove that a quadrilateral is a parallelogram, of which a rhombus is a special case. I will prove that quadrilateral NATS is a rhombus with the following method:

            I.   Prove that quadrilateral NATS is a parallelogram

            II.  Prove that the parallelogram is a rhombus.

 

But first, here is an image of quadrilateral NATS (the rhombus in question):

**This plot was created using GeoGebra's Geometry App.

I. Prove that quadrilateral NATS is a parallelogram. I will do this by showing that the diagonals of NATS bisect each other. The midpoint formula was used.

Diagonal NT and diagonal AS share a midpoint. NM = TM and AM = SM. Therefore, diagonals NT and AS bisect each other. Therefore, quadrilateral NATS is a parallelogram. ✔️

II. Prove that parallelogram NATS is a rhombus. I will do this by showing that the diagonals are perpendicular. 

The slopes of diagonal NT and diagonal AS are negative reciprocals of one another. Therefore, diagonal NT and diagonal AS are perpendicular. Therefore, parallelogram NATS is a rhombus. ∎

It's been quite a while since I've done one of these. This has me thinking 🤔 back to when I took my high school math Regents. I remember doing logic proofs. Maybe I'll explore that topic in a future WMP!

◾️ How do you feel about geometry?? 

◾️ What method did you use to complete the proof?? 

◾️ Comment below with your responses and let me know what you thought about this week's problem.

Without further ado, on to WMP! #16 

Cheers!

The Younge Lady

Weekly Math Problem! #13

Integration. Are you familiar with CLEP exams? I've heard the acronym before but didn't know what it stood for or what kind of exams they were. Now, I know. I got to find out a little more about CLEP by working with a student who will be taking a CLEP Calculus exam. CLEP stands for College-Level Examination Program. CLEP exams allow students to "demonstrate their mastery of introductory college-level material and earn college credit". There are 34 CLEP exams one can take. To get more information about CLEP exams, click here.

In helping a student prepare for the CLEP Calculus exam, the question below showed up in some practice questions.

To solve this week's problem in completion, you need to recall the following math skills:

    ✔️  How to solve a definite integral

    ✔️  Substitution method in integration

    

I present to you WMP! #13...

Happy solving! 

Check back on Friday, July 31st for the solution, which will be posted below ⬇️.

Shameless plug: Follow me on Instagram @TheYoungeLady

✏️📓 Solution Time! 📓✏️

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Here is plot of the function on the domain:

**This plot was created using Geogebra's Graphing Gaclulator.

As you can see, it's a semicircle. I can find the area of this via geometry with a straightforward calculation.

Then, there is the calculus way to do this problem. In the world of calculus, trigonometric substitution is the method to use. I decided to use the cosine substitution, because I already used the sine substitution.

As you can see, I arrived at the same answer. For me, this trig sub problem wasn't so bad to complete. However, I feel like the ones that involve tangent/cotangent, won't be so easy. 🤷🏿‍♀️ I'll present one in a future WMP.

◾️ Do you understand how to use trigonometric substitution?? 

◾️ If you've taken calculus before, what is your favorite topic from calculus?? 

◾️ Comment below with your responses and let me know what you thought about this week's problem.

Tired yet? No? GREAT! 😉

Up next...WMP! #14 😁

Cheers!

The Younge Lady