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math20c - Math 20C Supplement Curvature and acceleration in...

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(8/18/08) Math 20C Supplement Curvature and acceleration in the plane If a plane curve has the vector equation R ( t ) = ( x ( t ) ,y ( t ) ) , then at any point P where the velocity vector v ( t ) = R prime ( t ) = ( x prime ( t ) ,y prime ( t ) ) is not zero, the vector T ( t ) = v ( t ) | v ( t ) | is he vector parallel to the tangent line at P that points in the direction of the curve’s orientation. It is the unit tangent vector to the curve at that point. The unit normal vector N = N ( t ) is the unit vector that is perpendicular to points to the left of T ( t ). N can be found from T with the following rule. Rule 1 The unit normal vector to a curve at a point can be obtained from the unit tangent vector by interchanging its components and then multiplying the first component by - 1 . Thus, if T = ( a,b ) , then N = (- b,a ) (Figure 1). x y T = ( a,b ) N = (- b,a ) P C a b b a x 2 4 y 2 4 T = ( 1 , - 1 ) 2 N = ( 1 , 1 ) 2 FIGURE 1 FIGURE 2 Example 1 Find the x - and y -components of T and N at x = 2 on the parabola y = 5 - 1 4 x 2 , oriented from left to right. Then draw the vectors with the curve, using the scales on the axes to measure the components. Solution Orient the curve from left to right by using x as parameter. The position vector at x is R ( x ) = ( x, 5 - 1 4 x 2 ) , for which v ( x ) = R prime ( t ) = d dx ( x, 5 - 1 4 x 2 ) = (bigg d dx ( x ) , d dx (5 - 1 4 x 2 ) )bigg = ( 1 , - 1 2 x ) . Setting x = 2 gives v (2) = ( 1 , - 1 ) . The length of this velocity vector is |( 1 , - 1 )| = 1 2 + 1 2 = 2, so the unit tangent vector is T = ( 1 , - 1 ) / 2. Interchanging the components and then multiplying the first component by - 1 gives the unit normal vector N = ( 1 , 1 ) / 2. The curve and the two vectors are shown in Figure 2. square 1
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p. 2 (8/18/08) Section 0.1, Curvature and acceleration in the plane Curvature of plane curves Suppose that a curve has parametric equations, C : x = x ( s ) ,y = y ( s ) with arclength s along the curve as parameter and that x = x ( s ) and y = y ( s ) have continuous derivatives in the s -interval being considered. Let T ( s ) denote the unit tangent vector at (( x ( s ) ,y ( s )) on the curve and let φ ( s ) (phi) be the angle of inclination of T ( s ) (Figure 3). FIGURE 3 FIGURE 4 FIGURE 5 Definition 1 The curvature κ (kappa) of C : x = x ( s ) ,y = y ( s ) at (( x ( s ) ,y ( s )) is the derivative with respect to arclength s of the angle of inclination φ = φ ( s ) of the unit tangent vector: κ ( s ) = ds vextendsingle vextendsingle vextendsingle s . (1) If the curvature (1) is positive in an open s -interval, then φ = φ ( s ) is increasing on the interval and that portion of the curve is bending to the left, as in Figure 3. If the curvature is zero in an interval, then φ = φ ( s ) is constant in the interval and that portion of the curve is a straight line, as in Figure 4. If the curvature is negative, then φ = φ ( s ) is decreasing and the curve is bending to the right, as in Figure 5. Moreover, the curve bends sharply where | κ ( s ) | is large and bends gradually where | κ ( s ) | is small.
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