Partial fraction decomposition is a way of "re-writing" a function that appears to be non-integrable as one that is easily recognized to be integrable. There's really nothing special about partial fraction decomposition, and you definitely don't need to understand it to understand 1-variable Calculus. The special part is the integration that comes afterwards - that's the actual "Calculus" part. The most obvious applications of integration are afforded by The Fundamental Theorem of Calculus, which tells us (amongst other things) that finding areas under or between curves is really just integrating. So one extremely useful application of integration is finding areas under curves, or areas between curves. In other words, if you have a weird 2-dimensional shape that can be bounded by curves, then you can use integration to calculate the area of that shape.
Integration is also used for a variety of applications in chemistry, physics, economics, computer science, and engineering. Differentiation and Integration are two of the most important and useful concepts in all of mathematics.
Unfortunately most algebra and calc classes don't teach about conic sections anymore. For those interested, here's a video showing what Marathon is talking about:
[quote=Weatherby270]Partial fraction decomposition is a way of "re-writing" a function that appears to be non-integrable as one that is easily recognized to be integrable. There's really nothing special about partial fraction decomposition, and you definitely don't need to understand it to understand 1-variable Calculus. The special part is the integration that comes afterwards - that's the actual "Calculus" part. The most obvious applications of integration are afforded by The Fundamental Theorem of Calculus, which tells us (amongst other things) that finding areas under or between curves is really just integrating. So one extremely useful application of integration is finding areas under curves, or areas between curves. In other words, if you have a weird 2-dimensional shape that can be bounded by curves, then you can use integration to calculate the area of that shape.
Integration is also used for a variety of applications in chemistry, physics, economics, computer science, and engineering. Differentiation and Integration are two of the most important and useful concepts in all of mathematics.
Calculus lets you do math with things that are changing. How long does it take to get from A to B if your speed is constant VS. how long does it take to get from A to B if your acceleration is constant. When you accelerate, your speed is changing and the math is harder.
Calculus lets you switch back and forth between first order, second order, or third order functions. position= first order. When your position changes over time that's velocity - second order. When your velocity changes over time that's acceleration - third order.
Calculus lets you do math with one dimensional shapes in two dimensions or two dimensional shapes in three dimensions. Want to know the volume of a donut? A circle (2D) moving through space (3D) in a straight line makes a cylinder - easy. What if it rotates around a point instead of moving in a straight line? That's a donut. Calculus lets you move that circle through space and calculate the volume the circle leaves behind as it moves.
I think it's good to learn stuff - exercises your brain and helps you think in new ways. There's value in learning complicated stuff you won't necessarily use.
An engineer friend of mine used calculus to help me make a dip stick to measure fuel in a 550 gallon cylindrical barrel that was also tilted about 10°. Had been 30 years since my calculus days.
I have two sons,,and both are engineers,and had to take all them fancy math courses.I told them I quit taking math when the teacher said Pie R Squared,,,and I knew very well pies are round.They just say,,,go weed the garden again Dad.
Most folks think of a cord as being 128 ft3. In Minnesota, and probably other places; a cord of:
cut
split
ranked
wood is legally 120 ft3 which is a slightly easier number to work with. I consider the volume to be exactly 1 cord.
My math assumes
the shape is a parabola
the area of a parabola is Area=2/3(a)(b)
"a" = the length of the horizontal line at the base of the parabola
"b" = the length of the vertical bisector of said line "a"
Therefore the area = 2/3(7.5 ft)(6 ft) =30 ft2
The stack is 48"deep, or 4 ft
Vol = Area x Depth, or (30 ft2 X 4 ft)=120 ft3
The appropriate volume for stacked cut and split cordwood is 120 ft3 per cord. Therefore; 120 ft3/120 ft3 per cord = 1cord
Assuming the more commonly considered 128 ft3 per cord: 120 ft3 /128 ft3 per cord = .9375 cord.
My point is that you don't need to have a working knowledge of calculus for a lot of everyday problems. If you can recognize the shape, it should be a simple matter to find a formula in which you just need to plug in a few values.
That is how it works out for a parabola with a length (surface area) of 16 ft and a base of 7 1/2 ft. For that, I did not show my work, and should probably loose a couple points. actually, I think it is called cheating. I should have given that along with the base and length to save that important step.
The surface area of a parabola is a bunch more complicated. I simply measured the heights as I changed the base dimension. I then settled on a decent ratio.
Again, sorry, I should have supplied that 6 ft height.
Ok. I just treated it as a half of a circle. But it looked like what I could haul with my old pickup when they didn't have club cans and had a decent sized bed.
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