How to Use Calculus to Find the Total Distance Traveled

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You can find the total distance traveled by an object in two- or multi-dimensional space by using integral calculus. Integration is a mathematical tool for finding distances, volumes and areas of curves and shapes. For example, if you are conducting a science experiment in which the velocity of an object is defined by a mathematical function, you can apply integral calculus to the function to find the distance traveled by the object.

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• Identify the velocity function and the time interval over which the distance traveled by an object must be calculated. If this function is not available, you might have to derive it from a graph or use software tools to determine the distance traveled. For illustration purposes, assume that the velocity function, v(t), is 2t^2 - t - 6 and the time interval is from t = 0 to t = 5.

• Note whether the velocity function changes direction over the time interval. If the object changes direction one or more times in the time interval, then the distance traveled is the sum of the distances traveled in each sub-interval. In other words, if an object moves 5 meters to the left and then 10 meters to the right, the total distance traveled is 15 meters (5 meters + 10 meters). In the example, it is clear that v(t) is less than zero for t = 0 and greater than zero for t = 5; it therefore changes direction at least once. Although the distance from the starting point is 5 meters, you add each sub-interval distance together to find total distance traveled.

• Determine where the object changes direction by solving the function. Use trial and error to find and isolate common terms. If this does not work, you might need to use more complex algorithms; this is also known as factoring or finding the zeroes or roots of a function. In the example, rewrite v(t) as 2t^2 - 4t + 3t - 6. Regroup the terms to get 2t(t - 2) + 3(t - 2) and then (2t + 3)(t - 2). Set each polynomial to zero to solve the function. Thus, the zeroes of the function are at t = 2 and t = -3/2. Since the time interval cannot be negative, there is only one directional change at t = 2. Consequently, the time interval t = 0 to 5 has two sub-intervals: t = 0 to 2 and t = 2 to 5. The function is negative for t between 0 and 2, and positive for t = 2 and above.

• Compute the integral of the velocity function using the basic rules of integration. In the example, the integral of 2t^2 - t - 6 is (2/3)t^3 - t^2/2 - 6t + k. The constant term, "k," is not used in the distance calculation.

• Calculate the distance traveled over each sub-interval. In the example, the distance from t = 0 to 2 is (2/3)(2^3 - 0) - (1/2)(2^2 - 0) - 6(2 - 0), or -26/3. The distance from t = 2 to 5 is (2/3)(5^3 - 2^3) - (1/2)(5^2 - 2^2) - 6(5 - 2), or 99/2. Remember that the velocity function is negative from t = 0 to 2 and positive from t = 2 to 5. Thus, the total distance traveled is -(-26/3) + 99/2, or 349/6.

References

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