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April 5, 2017 | Mathematics | By admin | 0 Comments

By Niels Walet

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1: The definition of the differential. 31 32 CHAPTER 4. 2: The definition of the differential. 26-27 There exists a simple rule to calculate the differential of a product, d(uv) dv du =u +v . 5 . 28-30 In the same way we can find a relation for the differential of a quotient, d( uv ) v du − u dv = dx 2 dx . 6 . 5-18 Often we take a function of a function. In such a case, where y = f (g(x)) we put z = g(x), and find dy dy dz = . dx dz dx This rule is sometimes expressed in words as “the differative of the function, times the derivative of its argument”, and you may know it as dg(f (x)) = f (g(x))g (x).

0= and we have the desired results. 14: Find the velocity of a partocle that moves from r1 = (1, 2, 3) to r2 = (3, 6, 7) in 2 s along a straight line with constant velocity. Also find the position 5 s after passing r1 , Solution: Clearly r = r1 + vt if the particle is at point 1 at t = 0, We get, substituting t = 2; (3, 6, 7) = (1, 2, 3) + v2, from which we conclude (solving for each component separately) that v = (1, 2, 2). At time t = 5 we have r = (1, 2, 3) + (1, 2, 2)5 = (6, 12, 13). , specified by r(t) and θ(t).

At the start, where x = 0, z = x2 = 0. 8268 . , the object being integrated) changes from x sin(x2 ) to (1/2) sin z. Part of this change is due to the change from dx to dz. Note: The integration limits change (for definite integrals only). 46 CHAPTER 5. 5: dx . a2 + x 2 Calculate the indefinite integral I = Solution: Use substitution, and take z = (x/a), dz = (1/a)dx, x = az. = I = = = = dx 1 a dx 2 +x a dx adz a2 + a2 z 2 a dz a2 (1 + z 2 ) dz 1 a (1 + z 2 ) 1 tan−1 z + c. a a2 Finally we must substitute back using z = x/a, I= 1 x tan−1 + c.

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