alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# We say that the curve yx envelopes this family z i x

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Unformatted text preview: → 0 ≠ 0 . It only TENDS TO zero as P → Q . Now we take δy ---- = -----δx L imit { 3x2 + 3x δ x + δ x2 } = 3x2 δx → 0 99 we saying It is impor tant to note tha t in no w a y w e ar e sa ying the FIRST DERIVA alw straight DERIVATIVE is al w a ys a str aight line function of the ffor m m x + c . or , Example: Let y(x) = x 3 . Then y (x) = 3x 2 i s not a straight line function. Draw the curve of the function y(x) = x 3. Let P be a point on the curve of the function x 3. Let x 1 be the x co-ordinate of the point P. The value of the FIRST DERIVATIVE at x = x 1 of the function x 3 is: , y ( x 1) = 3 x 1 2 . This numerical value 3x 12 = tan θ1, where θ1 is the angle that the tangent tangent at the point P on the curve of the function x 3 makes with the x-axis. Draw the tangent z 1(x) to the cur ve at the point y(x 1 = 1) = 1 3 = 1. tangent tang Measure the angle θ 1 t hat the tangent z 1(x) makes with the x-axis. tangent It should be just about 72 . And tan θ1 = t an 72 = 3 .078. At the point on the cur ve x 3 = 1 t he x co-ordinate is x1 = 1 . Evaluate the FIRST DERIVATIVE of x 3 at x 1 = 1 . , SLOPE curv at 1 tan 72 = y (x1) = 3 = SLOPE of TANGENT to the cur ve at x 1 = 1.. Try this again at the point on the curve y(x2 = 2) = 23 = 8. tangent Draw the tangent z 2(x) to the curve at this point. tang Measure the angle θ 2 t hat the tangent z 2 (x) makes with the x-axis. tangent It should be a little more than 85 . And tan θ2 = t an 85 = 1 2. At the point on the cur ve x 3 = 8 t he x co-ordinate is x2 = 2 . Evaluate the FIRST DERIVATIVE of x 3 at x2 = 2 . , SLOPE curv at 2 tan 85 = y (x2) = 12 = SLOPE of TANGENT to the cur ve at x2 = 2.. 100 zi ( x i ) = y (x i ) At x = xi slope of tangent zi (x) = y ’(x i ) slope zi (x) = y ’(xi ) . x + c i is the tangent to y (x) at x = xi tangent The constant c i = zi ( x i ) − y ’(xi ) . xi In point -slope form the linear equation of the tangent is : pointtangent p oint t ang zi (x) − zi ( x i ) = y ’(x i ) . (x − x i ) At x = 1 the equation of tangent to y = x 3 is : z1(x) = 3x − 2 t angent At x = 2 the equation of tangent to y = x3 is : z2(x) = 12x...
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## This note was uploaded on 11/29/2012 for the course PHYSICS 105 taught by Professor Tamerdoğan during the Fall '09 term at Middle East Technical University.

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