A Spot of Maths
I felt like doing math problems last night. They are a convenient alternative to puzzles or games like Sudoku and FreeCell, because there is so much variety to be found (calculus/geometry/linear algebra/etc.). Endless and useful fun! :o)
In hopes of maximizing the useful aspect, I decided I would brush up on some parts of math that I didn't learn properly the first time around. I ended up starting out with linear approximations, because my understanding of approximations and error calculations has always been a bit sketchy. Not sketchy enough, in truth, to warrant this review, but one must begin somewhere.
A sample problem from the fifth edition of Single-Variable Calculus, by Robert A. Adams:
"The acceleration a of gravity at an altitude of h miles above the surface of the earth is given by a = g[R/(R + h)]², where g ≈ 32 ft/s² is the acceleration at the surface of the earth, and R ≈ 3960 miles is the radius of the earth. By about what percentage will a decrease if h increases from 0 to 10 miles?"It's not a particularly interesting or difficult problem, I admit. I will try to find a better one and post it later on. In the meantime, here is my solution to the acceleration problem:
s = 0 miles = h at surface of the earthI had some trouble with the derivative of a, because I treated R as a variable. Tsk, tsk! In earlier problems, I found that I had forgotten the derivative of tan(x), and that my memory of the quotient rule and the 30-60-90 triangle was indistinct. Oh well. Those are all things that can easily be looked up and recommitted to memory.
a'(x) = derivative of a with respect to h at h = x
L(x) = linear approximation of a at h = x
Δa = decrease in a between h = 0 miles and h = 10 miles
P = percentage by which a decreases between h = 0 miles and h = 10 miles
a'(h) = [gR²/(R + h)²]'
= gR²[-2/(R + h)³]
= -2gR²/(R + h)³
a'(s) = a'(0) = -2g/R
a(s) = gR²/(R + s)² = g
L(h) = a(s) + a'(s)(h - s)
Δa ≈ L(h) - a(s)
= a(s) + a'(s)(h - s) - a(s)
= a'(s)(h - s)
= -2gh/R
P = (Δa/a(s)) × 100%
= ((-2gh/R)/g) × 100
= -200h/R
= -200(10)/3960
≈ 0.5%
If I actually stick to this, I may soon have a look at Taylor Polynomials, differential equations, partial derivatives, and proofs. Not particularly likely, I'm afraid. :o( I am a fickle creature.
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