alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

But the distance travelled 2r distance c 2 dimensions

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Unformatted text preview: x Limit Limit dθ = n → ∞ δ θ = n → ∞ 2π/ n In the diagram above an infinitesimal element o f area δ A is a triangle i nfinitesimal (wedge shape) of magnitude / / δ A = 1 2(R . δθ ).R = 1 2R 2 δ θ Limit / Limit / / dA = n → ∞ { 1 2(R . δθ ).R } = n → ∞ { 1 2R 2 δ θ } = 1 2R 2 d θ is an instantaneuos element of area. instantaneuos 179 1: E x a m p l e 1 : Find the ar ea under the cur v e f ( x ) = 2 x o v e r t h e ar inter v al [ 0, 1 ]. f(x) f(x) 3 (0 , 0 ) 654321 654321 654321 654321 654321 654321 654321 654321 654321 654321 654321 654321 1 f (x) = 2x 2a 1a 1 2 7654321 7654321 7654321 7654321 7654321 7654321 7654321 7654321 7654321 7654321 7654321 7654321 2 - a (0 , 0 ) 3x x+a f (x) = x +a a x Using Integral Calculus: F ’(x) = f(x) = 2x The INTEGRAL F(x) = ANTIDERIVATIVE of f(x) What is the ANTIDERIVATIVE of f(x) = 2x ? x2 + C F(x) = ∫ f(x) dx = ∫ 2xdx = x 2 + C In this case the constant C = 0. Can you think why ? T he ar ea under the cur v e f (x) = 2x fr om x = 0 to x = 1 is: ar 1 1 1 -∫ 0 f(x)dx = ∫ 0 2xdx = [ x2 ] 0 = 1 2 -- 0 2 = 1 . Evaluation of the INTEGRAL F(x) = x 2 i n the positive direction o ver the p ositive inter val [0, 1] is the CHANGE [ F(x) ]1 = [ x 2 ] 1 = 1 2 -- 0 2 = 1 . -0 0 1 1 0 0 Note: [ F C(x) ] = [ x 2 +C] = [1 2 +C] -- [0 2 +C] = again 1. Exer ercise: Exercise: What is area under f(x) = x + a over the interval [0,a]. Ref. page72. 180 energ ork In Physics the most important entity we are interested in is ener g y . Any wor k ener energ energ done r equires ener gy and causes CHANGE and any CHANGE involves ener gy . ener ener The sense of the word CHANGE in Calculus is not synonymous with the word CHANGE ork energ ( = wor k done = energy ) in Physics. ener Example 2: A car weighing one ton and starting from rest (0 kmph) accelerates steadily due EAST (along the x-axis) for 10 secs and levels off at 90 kmph (25 m/sec). What is the distance travelled from t = 10 secs to t = 20 secs distance and also from t = 0 secs to t = 10 secs ?...
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