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Chapter 14

Chapter 14. Section 14.5 Curvilinear Motion, Curvature. Acceleration Acceleration measures the change in velocity (which is a vector) over time. This will be another vector. The derivative of velocity is acceleration or the second derivative of position. Position Velocity Acceleration .

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Chapter 14

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  1. Chapter 14 Section 14.5 Curvilinear Motion, Curvature

  2. Acceleration Acceleration measures the change in velocity (which is a vector) over time. This will be another vector. The derivative of velocity is acceleration or the second derivative of position. Position Velocity Acceleration Curvature Curvature measures the amount of “bend” a curve has. This is done by measuring the total change in the direction (and only the direction) for each unit of length you move around the curve. Computing it this way is difficult though. Curvature Using the chain rule we can take the derivative of the unit tangent vector with respect to time. Taking the length of this vector and keeping in mind speed is a scalar we get the identity to the right. Keeping in mind the unit tangent vector T is the velocity divided by the speed multiplying the unit tangent by speed given velocity.

  3. Acceleration, Speed, Curvature, Unit Tangent and Unit Normal Vectors The values of acceleration, speed, curvature, unit tangent and unit normal vectors can be related by a single equation. This can be done by taking the derivative of velocity and apply the product rule. In the second term on the right side divide numerator and denominator by and substitute in the value from the previous slide. Remember the derivative of the tangent divided by its length is the unit normal. From this we get the fundamental motion equation that gives the amount of forced directed in both the tangential and normal direction. These values are very often referred to as tangential component of acceleration and the normal component of acceleration.

  4. The motion equation can be used to derive a formula for curvature by crossing each side with velocity. Distribute the velocity. Velocity is parallel to the unit tangent vector so the cross product is zero. Take the length of both sides keeping in mind curvature and speed are positive scalars and can be factored out. Apply the identity for length of cross product where  is the angle between the vectors which is . The length of the velocity vector is the speed and the unit normal has length 1. Solve this equation to get the curvature is the length of velocity crossed with acceleration divided by speed cubed.

  5. Example Find the curvature for the curve given to the right at the point on the curve where . Compute velocity, acceleration and find the values at time along with the value of the speed. Cross the velocity and acceleration and compute the length of this vector. Divide this number by the speed cubed to get the curvature.

  6. z Example Compute the curvature for each of the circles pictured to the right parameterized by the vector functions and . y x In each case notice that the curvature is constant (a circle is “bent” the same way no matter where you look at it) and the value is the reciprocal of the radius of the circle.

  7. Geometric Interpretation of Curvature The curvature at a point on a curve is the reciprocal of the circle of “best fit” to the curve at that point. If the curve is nearly straight the circle will need to be big so the reciprocal of the radius is small and if the curve has a lot of bend the circle is small so the reciprocal of the radius is big. large curvature small curvature Example Find the curvature of any line. Any line with direction vector and point has parameterization: Curvature of a plane curve Curvature is a purely geometric (mathematical) concept. It can be computed for any curve in the ordinary xy-plane. We can describe the curve so that this will have the parameterization given to the right. Substitute this in the formula and replace t by x.

  8. Example A car is negotiating a curve in a road that has the shape . The car is weighted so it begins to side sideways if the normal component of acceleration exceeds 3600 where distances are measured in miles and time in hours. What is a safe speed to negotiate the turn? Recall the normal component of acceleration is given to the right. To find the place where the curvature is greatest (i.e. the turn is the tightest) in this situation is where the derivative is zero. If the car does not slide here it will not slide anywhere on this curve. Compute the curvature at this position. Substitute this value in for and solve for the speed. We see keeping the car under 30 mph will safely negotiate the turn.

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