( or it could have been Awesome that stuff is so cool! If it was something like that then you do not have to read this but you still can ) I mean when would you EVER have to use that in regular life ?
My article today strives to answer at least part of this question " why calculus " and how it is awesome!
Calculus , like any topic, can be made boring and dissatisfying through mental prejudice and unenthusiasm. When you put it into your brain ( quite unfairly ) that this is not going to be fun and it will never be fun , in a nutshell that you dislike it. This approach must not be the way you look at calculus if you ever want to come to love it and have fun doing it, but rather you at least have to have a open approach to the subject. Preferably you should have a positive, optimistic approach to the subject having a hope that it will be fun and exciting.
What makes Calculus fun is that with knowledge you can solve problems, thereby thereby using your brainpower to get answers. Calculus is really not fun when you do not know how to do it, so learning it is an essential part of the process. Learning can be hard at times and so you need to press forward continue trying and rely on the infinite grace and mercy of our Heavenly Father to learn and understand calculus. Once you learn the material , answering problems becomes easy ( or at least doable) and the excitement of being able to find the answer emerges.
Learning Calculus can be fun to though , once you finally understand a new definition , or theorem it enables you to do something new in math that you did not know how to do before. It opens new doors in your ability to solve problems , and things that seemed impossible now become second nature.
For example , say you knew basic algebra and how to find the slope of a graph, but then you think of a curved graph ( say ln(x) ) and wonder how to find out its slope at a defined point. You can increasingly approximate the slope at the point but not quite reach it using algebra, but with differential calculus you can literally find the slop of a point of a curved graph. For example at x=2 dy/dx of y=ln(x) or the infinemestial slope at the point x=2 of the natural log of x is equal to the 1/2.
While I was learning how to take the volume of irregular solids, I wondered if you could somehow use related rates to find the rate of change of the volume of an irregular solid where the upper integrand is changing. So what this means is that say you have a graph of a function ( for example y=x2) and then you rotate the graph around an axis ( say the x ) so what you get is a solid with a volume that can be determined using integration and the property that each slice of the solid is a circle. The area of each of the circles is 𝛑r² and the one thing that changes between the each of the circles is the radius. The radius at each point or for each circle is equal to the value of the function at that point ( since the diameter is twice the function value ) or x2
If you sum the areas of every circle in the range that you want ( say from 0 to 2 , which numbers are called the lower and upper bounds of integration, or the integrands ) then you get the volume. To sum the areas of every circle you have to use an integral which is and the limit as the number of circles approaches infinity or the limit as the width of the cylinders approaches zero of the sum of each of the cylinders. It would be 𝛑∫(x²)²dx since 𝛑(x²)² is the area of one circle and the sum of all of them is the integral. Using integration rules ( ∫xn=
xn+1
from 0 to 2. The rate of change of this volume was a bit hard for me to determine how to find when given that the upper integrand was changing at some rate, but I eventually got it , with Heavenly Father's help. So the fundamental theorem of Calculus state that the derivative of an integral of a function with respect to the changing upper integrand is equal to the function of the integrand.
By the chain rule the derivative of a composite function f(g(x)) with respect to x is equal to the derivative of f(g(x)) with respect to g(x) multiplied by the derivative of g(x) with respect to x. So say that you were given the rate of change of the upper integrand and asked for the rate of change of an irregular volume produced by rotating f(x) around the x axis. The way to do this is multiply f(x) by dx/dt or the function of x multiplied by the rate of change in x. This is a simpler way than what I did. I used a differential equation to solve for the upper integrand in terms of t and then plugged that into the integral as the integrand and than applied the fundamental theorem and the chain rule.
Once you have mastered Calculus answering problems gives you a "zing" to the brain and it can get mildly addictive, so do not do it to much.
No comments:
Post a Comment