10Eapplications

= Applications of Antidifferentiation =


 * Finding c **


 * Recall that when we antidifferentiate, we always have to add a constant, c
 * This is because any constant term is lost when we differentiate,
 * So when we undo the differentiation, we cannot know what the constant term was


 * Because of the + c term, when we graph the antiderivative of a function
 * we can draw the shape of the graph, but we don't know where vertically to place it on the axes


 * In order to find the value of c, we need to be given a point on the axes that the graph passes through
 * Sometimes this extra information is called
 * Initial Conditions .. or
 * Boundary Conditions


 * Given the coordinates of a point, we simply substitute (x, y) into the antiderivative and solve for c


 * Example 1 **

... ... Find the antiderivative of 3x 2 – 6x + 2, ... ... Given that the point passes through (1, 5)


 * Solution:**

math . \qquad y = \displaystyle{ \int \; 3x^2 - 6x + 2 \; dx } \\.\\ . \qquad y = \dfrac{3x^3}{3} - \dfrac{6x^2}{2} +2x + c \\.\\ . \qquad y = x^3 - 3x^2 + 2x + c math


 * Substitute (1, 5)

math . \qquad 5 = (1)^3 -3(1)^2 + 2(1) + c \\.\\ . \qquad 5 = 1 - 3 + 2 + c \\.\\ . \qquad 5 = c math


 * Hence c = 5, so we can state the final solution for y

math . \qquad y = x^3 - 3x^2 + 2x + 5 math


 * Rate of Change Problems **


 * Example 2 **

math . \qquad \dfrac{dx}{dt} = 30t - t^2 \quad m/s math
 * The rate of change of position (which is called velocity)
 * of a racing car travelling down a straight stretch of road is given by:


 * Find the velocity after 15 seconds
 * Find the position down the track after 15 seconds


 * Solution:**


 * To find the velocity after 15 seconds, substitute t = 15 into the velocity rule

math . \qquad \dfrac{dx}{dt} = 30 \big( 15 \big) - \big( 15 \big)^2 \\.\\ . \qquad \dfrac{dx}{dt} = 450 - 225 \\.\\ . \qquad \dfrac{dx}{dt} = 225 \quad m/s math


 * Position (x) will be given by the antiderivative of velocity with respect to t

math . \qquad x(t) = \displaystyle{ \int \; 30t - t^2 \; dt} \\.\\ . \qquad x(t) = \dfrac{30t^2}{2} - \dfrac{t^3}{3} + c \\.\\ . \qquad x(t) = 15t^2 - \dfrac{t^3}{3} + c math


 * after 0 seconds, position is 0 ( at t = 0, x = 0)

math . \qquad x(0) = 15\big( 0 \big)^2 - \dfrac{ \big( 0 \big)^3}{3} + c = 0 \\.\\ . \qquad c = 0 \\.\\ . \qquad \text{so} \\.\\ . \qquad x(t) = 15t^2 - \dfrac{t^3}{3} math


 * Find the position after 15 seconds

math . \qquad x(15) = 15 \big( 15 \big)^2 - \dfrac{ \big( 15 \big)^3}{3} \\.\\ . \qquad x(15) = 3375 - 1125 \\.\\ . \qquad x(15) = 2250 \quad m math


 * Alternatively, we could have calculated the definite integral of the velocity between t = 0 and t = 15

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