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MATH 2B Lecture Notes - Lecture 5: Antiderivative

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27 Oct 2017

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$MATH 2B Full Course Notes$

MATH 2B Full Course Notes

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Math2b lecture 5 5. 4 net change & indefinite integrals. Reminder from last lecture notes: ftc: (cid:1858) (cid:4666)(cid:4667) (cid:1856)=(cid:1858)(cid:4666)(cid:4667)|(cid:3028)(cid:3029) Fi(cid:374)di(cid:374)g the area u(cid:374)der the cur(cid:448)e (cid:449)ould ulti(cid:373)atel(cid:455) give us the anti-derivative. Once we add all the areas a+b+c together, we would get 0 as the answer. Anti-derivative of v(t) gives us the displacement over [a,b] (cid:4666)(cid:884)(cid:1872) (cid:883)(cid:884)(cid:1872)(cid:2870)(cid:4667)|(cid:2868)(cid:2872)=((cid:884)(cid:4666)(cid:886)(cid:4667) (cid:883)(cid:884)(cid:4666)(cid:886)(cid:4667)(cid:2870)) ((cid:884)(cid:4666)(cid:882)(cid:4667) (cid:883)(cid:884)(cid:4666)(cid:882)(cid:4667)(cid:2870)) The total distance traveled must be > 0. Example: a) if (cid:4666)(cid:1872)(cid:4667)=(cid:884) (cid:1872), what is the displacement over [0,4]? (cid:2872) (cid:2872) =(cid:4666)(cid:884)(cid:1872) (cid:883)(cid:884)(cid:1872)(cid:2870)(cid:4667)|(cid:2868)(cid:2870) (cid:4666)(cid:884)(cid:1872) (cid:883)(cid:884)(cid:1872)(cid:2870)(cid:4667)|(cid:2870)(cid:2872)=(cid:4666)(cid:886) (cid:884)(cid:4667) (cid:4666)(cid:882)(cid:4667) [(cid:4666)(cid:890) (cid:890)(cid:4667) (cid:4666)(cid:886) (cid:884)(cid:4667)]=(cid:886: we cannot take the anti-derivative of an absolute value. Therefore, we split the interval up at where the function is equal to 0: afterwards, we determine whether the function will be positive or negative. Well, we plugged in a number from the interval and checked to see if the solution we got was positive or negative. For instance, from [0,2], we can plug in 1 to |2-t| and get +1.

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Related Questions

Sketch using these guidelines

pul aill al thes information together to sketch graphs that reveal the important You might ask: Why don't we just use a graphing calculator or computer to graph a It's true that modern technology is capable of producing very accurate graphs. But K 21+1+2 features of functions curve? Why do we need to use calculus? even the best graphing devices have to be used intelligently. It is easy to arrive at a misleading graph, or to miss important details of a curve, when relying solely on tech- nology. (See "Graphing Calculators and Computers" at www.stewartcalcelus.com, espe 4 discover the most interesting aspects of graphs and in many cases to and minimum points and inflection points exocily instead of approximately cially Examples 1, 3.4, and 5. See also Section 4.6.) The use of calculus enables us to calculate maximum FIGURE 1 For instance, Figure I shows the gr aph of /(x)-è¨-21x, + 18x + 2.At first glance it seems reasonable: It has the same shape as cubic curves like y x and it appears to have no maximum or minimum point. But if you compute the derivative, you will see that there is a maximum when x-0.75 and a minimum when x 1. Indeed if we zoom in to this portion of the graph, we see that behavior exhibited in Pigure 2 Without calculus, we could easily have overlooked it. 6a, FIGURE 2 In the next section we will graph functions by using the interaction between calculus and graphing devices. In this section we draw graphs by first considering the foilowing information. We don't assume that you have a graphing device, but if you do have one you should use it as a check on your work. â Guidelines for Sketching a Curve every item is relevant to every function. (For instance, a to make a sketch that displays the most important aspects of the function. The following checklist is intended as a guide to sketching a curve y- f(x) by hand. Not given curve might not have an asymptote or possess symmetry.) But the guidelines provide all the information you need A. Domain It's often useful to start by determining the domain D of f, that is, the set B. Intercepts The y-intercept is f(o) and this tells us where the curve intersects the of values of x for which f(x) is defined y-axis. To find the x-intercepts, we set y-0and solve for xã(You can omit this step if the equation is difficult to solve.) ry (a) Even fusction: reflectional symmetry all x in D, that is, the equation of the curve is unchanged when x is replacea by -x, then f is an even function and the curve is symmetric about the y-axis. This means that our work is cut in half. If we know what the curve looks like for 0, then we need only reflect about the y-axis to obtain the com- plete curve [see Figure 3(a). Here are some examples: y-ey-ry x). and for all x in D, then f is an odd function and the curve is symmetric about the origin. Again we can obtain the complete curve if we know what it looks like for x0. [Rotate 180Â° about the origin; see Figure 3(b).] Some b) Oud function rotational symmetry FIGURE 3 simple examples of odd functions are y-xy-x'y - and y-sin

Page 3 of 5 which gives us y as a function of Theoretically. the integration on the right of t= 1/ - g + alpha f(V)dv can be performed and the result could provide velocity. v,. as a function of t; Then we could perform the integration in y=y0 + V/- g + alpha f(V)dv and substitute v(t) to obtain y as a function of t. Theoretically, that is. Let's look at in example An Example Suppose the air resistance model for a particular object in free fall has where units for v area feet/second Note how f is defined in terms of the sign of v. As usual. downward has been selected as the negative direction so velocities v of an object in free fall will have negative values Thus -v will be positive and the square root makes. sense Similarly if the object is moving upward, v> 0. the value of f(v)) is negative. The object is dropped from a height: of 500 feet with no initial velocity Using g = 32, the integral models from above becomes and where we have used the definition of f(v) for negative values of v since the object is in free fall With a little hard work or the use of a CAS, the first equation can be written as Page 4 of 5 and the second as Now if we could solve the first expression for v in terms of r and substitute the result for v in the expression for y, the result would be very unwieldy to work with. Actually it is impossible to solve the first expression to get a nice closed form for v. Thanks to technology. however, these expressions can still be used to study the object's motion Choose any value of v. say v=-1. Substituting -1 for v in the equations for t and y tells us that t = 0.032 and y = 499.984 so the point (0.032, 499.984) would lie on a graph of distance versus time Similarly, setting v = -50 provides the point (1 839, 452 462) By using a wide range of values for v. we can generate the complete graph of the distance function as a function of time Fortunately, most CAS provides the capability of plotting the points (t(v), y(v)) foe a range of values of v without having to compute each point separately Such a graph appears below. One final note. It is also possible that the integrals representing t and y cannot be integrated into a closed form No problem In this case, simply use the expressions t= 1/ - g + alpha f(V)dV y=y0+ V/ - g + alpha f(V)dV Page 5 of 5 and the numerical integrators provided by a CAS to determine values of t und y Exercises In all of the exercises you will need a CAS or related tool In the example above the graph indicates the object strikes the ground some time between 6 and 7 seconds from release Determine this time to three decimal places In the case where alpha = 1/2 ,f(v) = -v, v0 = 0, y0=100 determine closed form. expressions for v and y as function. Of t. An object is thrown upward at 40 ft/sec from a height of 100 feet The air resistance can be modeled by alpha =0.001 and f(v) = - |v| v Using the trajectory model determine the speed of the object when it strike the ground, determine the time the object strikes the ground after takeoff, construct the graph of the object's height versus time.