alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

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Unformatted text preview: a cur v e where y is a function of x. 4: Example 4 What is the length of the curve y(x) over the interval [x1, x2] ? dx = infinitesimal CHANGE along the x-axis. dy = infinitesimal CHANGE along the y-axis. ds = infinitesimal piece of the cur ve. y(x) ds dx x1 (0, 0) dy x2 x dy ---- = y’(x) . So dy = y’(x).dx . dx 2½ ½ By Pythagoras theorem: ds = [(dx) 2 + ( dy) 2 ] = [(dx)2 + ( y’(x).dx) ] . = s= 2 ½ . dx = dy 2 [ 1 + (y’(x)) ] x2 ∫ ds = ∫x [ 1 + ( ----) ] dx ½ dy 2 [ 1 + ( ----) ] dx ½ . dx . dx 1 Integ We should now look up the Table of Integr als to find the appropriate integral where the integrand from the given WELL-BEHAVED y(x) is in the above form. Apply this method to find the length of the curve where f(x) = 2x over the interval [0, 1] as in example 1 (see page 180). Here f ’(x) = 2. The result should tally with the result √5 that we get using the Pythagoras Theorem. 186 Example 5: What is the CHANGE in height y(t) of the ball over [ 0, T ] ? y y 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 10987654321 1098765432110987654321 2 1098765432110987654321 2 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 210987654321 y’ (0,0) T t Height y(t) = u . s i n θ . t -- 1 g t 2 . -2 Vertical speed y ’(t) = u . s i n θ -- g t . What is the shaded area under y ’ (t) over [ 0, T ] ? The area of the two shaded triangles is zero. y( T ) -- y ( 0 ) = 0. The CHANGE in height is zero. This does not necessarily mean that the present height above the ground level ( = zero ) at t = T seconds is zero. If initially at t = 0 seconds the ball was at height 5 meters above the ground level ( = zero ), then the difference or CHANGE in height from t = 0 secs to t = T secs is zero. So the ball is back at its initial height of 5 meters above the ground level ( = zero). ∫ y’(t) . dt = y(t) + C In this case the constant of integration C = + 5 meters. 187 188 Geometrically speaking, the...
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