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6.1: Antiderivatives and Slope Fields First, a Little Review

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6.1: Antiderivatives and Slope Fields First, a Little Review 6.1: Antiderivatives and Slope Fields First, a little review: 2 2 Consider: yx= + 3 yx= − 5 or then: yx′ = 2 yx′ = 2 It doesn’t matter whether the constant was 3 or -5, since when we take the derivative the constant disappears. However, when we try to reverse the operation: Given: yx′ = 2 find y We don’t know what the constant is, so we put “C” in y =+xC2 the answer to remind us that there might have been a constant. → If we have some more information we can find C. Given: yx ′ = 2 and y = 4 when x = 1 , find the equation for y . y =+xC2 This is called an initial value 2 41=+C problem. We need the initial 3 = C values to find the constant. yx=+2 3 An equation containing a derivative is called a differential equation. It becomes an initial value problem when you are given the initial condition and asked to find the original equation. → Initial value problems and differential equations can be illustrated with a slope field. Slope fields are mostly used as a learning tool and are mostly done on a computer or graphing calculator, but a recent AP test asked students to draw a simple one by hand. → Generally, slope fields are used when: • the derivative of a function is known but the function is not. • when it is difficult (or impossible) to determine the function from its derivative analytically. A famous example is: dy 2 = e− X dx It is not possible to analytically find the antiderivative − X 2 of e It is possible for the derivative to be a function of both x and y: dy 2xy =− dx 1+ x2 It is therefore necessary to create a table with both x and y values and evaluate the derivative at each (x, y) coordinate. In the following example, the derivative is ONLY a function of x. Therefore, the slopes stay the same regardless of the value of y. Y’ = 2X X -2 -1 0 1 2 -4 -4 -2 0 2 4 -3 -4 -2 0 2 4 -2 -4 -2 0 2 4 -1 -4 -2 0 2 4 Y 0 -4 -2 0 2 4 1 -4 -2 0 2 4 2 -4 -2 0 2 4 3 -4 -2 0 2 4 4 -4 -2 0 2 4 → yx′ = 2 x y y′ Draw a segment 00 0 with slope of 2. 01 0 002 003 10 2 Draw a segment 11 2 with slope of 0. Draw a segment 20 4 with slope of 4. -1 0 -2 -2 0-4 → yx′ = 2 If you know an initial condition, such as (1,-2), you can sketch the curve. By following the slope field, you get a rough picture of what the curve looks like. In this case, it is a parabola. → For more challenging differential equations, we will use the calculator to draw the slope field. dy 2xy =− 2 On the TI-89: dx 1+ x Push MODEMODE and change the Graph type to DIFF EQUATIONS. Go to: Y= Press I and make sure FIELDS is set to SLPFLD. Go to: Y= and enter the equation as: yt12′ =− ∗ ∗y1/(1+t^2) (Notice that we have to replace x with t , and y with y1.) (Leave yi1 blank.) → yt12′ =− ∗ ∗y1/(1+t^2) Set the viewing window: WINDOW Then draw the graph: GRAPH → Be sure to change the Graph type back to FUNCTION when you are done graphing slope fields. → It turns out on your TI-89, you just plotted the slope field for the function: C y = 1+ x2 To sketch the slope field for the case in which C = 8, notice that this means y(0) = 8. Go to the y = screen, make sure t0 = 0 (note t represent x on your calculator). Then set yi1 = 8. Now plot. 4 Integrals such as xd 2 x are called definite integrals ∫1 because we can find a definite value for the answer. 4 xd2 x ∫1 1 4 x3 + C 3 1 The constant always cancels 1133 ⋅−41++CC⋅when finding a definite 33integral, so we leave it out! 64 1 63 +−CC− = = 21 33 3 2 Integrals such as ∫ xd x are called indefinite integrals because we can not find a definite value for the answer. ∫ xd2 x 1 x3 + C 3 When finding indefinite integrals, we always include the “plus C”. → Many of the integral formulas are listed on page 307. The first ones that we will be using are just the derivative formulas in reverse. On page 308, the book shows a technique to graph the integral of a function using the numerical integration function of the calculator (NINT). x yt= sin t dtor yx1N= INT( sinx,x,0,x) ∫0 This is extremely slow and usually not worth the trouble. A better way is to use the calculator to find the indefinite integral and plot the resulting expression. → To find the indefinite integral on the TI-89, use: ∫ ( xx∗sin , x) The calculator will return: sin ( xx) −⋅cos( x) Notice that it leaves out the “+C”. Use COPY and PASTE to put this expression in the Y= screen, and then plot the graph. → [-10,10] by [-10,10] π.
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