Beam Abs
calculus madness -_____-?
1) The strength of a rectangular beam is proportional to its width times the square of its depth. Find the dimensions of the strongest beam that can be cut from a 12 in. diameter log.
( diagram : a circle with a 12 in. diameter, with a rectangle fit in the middle, the diameter is the diagonal of the rectangle)
2) The curve f(x) – x^2 – 2x
a) Determine whether the derivative of absolute value of f(x) exists at x = 0. Justify your answer.
b) determine whether f( abs(x)) is continuous at x=0. Justify your answer.
1. The strength is S = k x y^2, and y = sqrt(R^2 – x^2). Substitute this into the first equation, and equate the derivative with respect to x to zero to find the minimax point.
2. There is a typo here; I assume that we have f(x) = x^2 -2x. The derivative of this would be 2x-2. But absolute value can be treated as the square root of the square, i.e. sqrt(x^4 – 4x^3 + 4x^2). Since the thing inside the parens is continuous everywhere, so is the square root. But a bit of graphing shows that it is not differentiable at x = 0.
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