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 | Functions of one independent variable
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 | In Calculus I and II you learn to differentiate and integrate functions of one independent variable, generally written y = f(x) = x4 + x3 - 5x2 + 2x - 1. Graphs of such functions are drawn in two dimensions (a plane) since we only have two variables, the independent variable x (horizontal axis) and the dependent variable y (vertical axis) which depends on the choice of x.
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| Functions of two independent variables
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| In Calculus III you learn how to work with functions with more than one independent variable. For example, with two independent variables, x and y, we get equations of the form: z = f(x,y) = 2 - x2 - 2y2. We now have three variables, two of which (x and y) are independent, and one (z) which depends on the choice of x and y. Graphs of such functions require three dimensions.
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| Differential changes for functions of one independent variable:
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| If we have a function of the form y = f(x), we can find the change in y when we make a differential change (a tiny amount) in x. We write dy = f'(x) dx. f'(x) is the derivative of the function f(x) (i.e., the slope of the tangent to the function at x). Multiplying the slope by a small step in the x-direction, dx, gives us the corresponding differential change in y, dy.
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| Differential changes for functions of two independent variables:
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| If we have a function of the form z = f(x,y), to find the change in z when we make differential (tiny) changes in x and y, we have to find out the rate of change of the function in the x direction times the differential dx and add the rate of change of the function in the y direction times the differential dy. To find the rate of change of the function in the x direction, we need to hold y constant and just vary x. This is called the partial derivative of f with respect to x and is written either as ∂f/∂x or as fx. To find the rate of change of the function in the y direction, we hold x constant and just vary y. This gives us the partial derivative of f with respect to y and is written either as ∂f/∂y or fy. Hence, to find the differential change in z, dz, we write:
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| The graph below shows the plane function z = f(x,y) = x/2 + y/2 + 1. To find fx, we hold y constant and take the derivative with respect to x, so fx = 1/2. To find fy, we hold x constant and take the derivative with respect to y, so fy = 1/2. If we start at the point (1,-1,1) and take a step of size dx = 1, we arrive at the point (2,-1,1.5). Notice the little right triangle under the plane as we go from (1,-1,1) to (2,-1,1.5). The base of the triangle is 1, the slope of the hypotenuse is 1/2, and the rise in the triangle is 1/2. That's the change in z due to the change in x. Now we move in the y direction. Again, notice the right triangle below the plane as we move from (2,-1,1.5) to (2,2,3). The change in y, dy = 3, and the corresponding change in z is 1.5. Using the expression above: dz = fxdx + fydy = (1/2)(1) + (1/2)(3) = 1/2 + 3/2 = 2.
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| If we had a function of the form z = f(x,y) = C (a constant), then dz = fxdx + fydy = 0.
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| This is the form of a differential equation called an "exact" differential equation, usually written P(x,y) dx + Q(x,y) dy = 0, where P(x,y) and Q(x,y) are functions of independent variables x and y. We identify P(x,y) = fx and Q(x,y) = fy, and the solution is a function of the form f(x,y) = C, where C depends on initial conditions.
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| Shortcut for determining if an ODE is exact
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| Note: we can find out immediately if an ODE is exact. If an ODE is exact, then P(x,y) = fx and Q(x,y) = fy. If you take the second partial derivative but with respect to the other variable (called the "mixed" partial second derivative) you get fxy or fyx, depending on the order of the derivatives. However, it turns out that it doesn't matter which derivative you take first. That is, fxy = fyx. So, if P(x,y) = fx and Q(x,y) = fy, then for the ODE to be exact, fxy = ∂P/∂y and fyx = ∂Q/∂x, and since fxy = fyx, ∂P/∂y = ∂Q/∂x. If this condition is not met, then the ODE is not exact.
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| If we think this is an exact differential equation then the coefficient of dx should be fx and the coefficient of dy should be fy. Hence, fx = x tells us f(x,y) = x2/2 + g(y). We took the antiderivative of x to get x2/2, but instead of a constant of integration, we got a function of y. This is because fx is a partial derivative and was found by holding y constant. Now taking the partial derivative of f(x,y) with respect to y, we get fy = g'(y) and we equate that to y (the coefficient of dy in the DE). Thus, g'(y) = y and g(y) = y2/2. Combining this with our earlier result, we get: f(x,y) = x2/2 + y2/2 = C. This is an implicit equation (not solved explicitly for y) for a family of circles with radius equal to sqrt(2C). If we solve it explicitly for y, we get y = ±sqrt(2C - x2), two solutions.
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| Example: (2x + y) dx + (x - 6y) dy = 0
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| We assume fx = 2x + y and get f(x,y) = x2 + xy + g(y). We take the partial derivative of this with respect to y and set it equal to x - 6y: x + g'(y) = x - 6y. This tells us that g'(y) = -6y, thus g(y) = -3y2. Hence the final solution is: f(x,y) = x2 + xy - 3y2 = C.
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| Read pp. 63-68 of the text and do exercises p. 75 (1,3,5,7,9,11,13,15,17,19,21).
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