### 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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