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DIFFERENTIALS

DIFFERENTIALS. dy/dx represents the instantaneous change in a function as dx approaches zero dx is the amount x changes dy is the change in the function as a result of the change in x, dx The ratio of dy to dx, dy/dx, is equal to the derivative of the function if it exists. dy/dx = f’(x)

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DIFFERENTIALS

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  1. DIFFERENTIALS • dy/dx represents the instantaneous change in a function as dx approaches zero • dx is the amount x changes • dy is the change in the function as a result of the change in x, dx • The ratio of dy to dx, dy/dx, is equal to the derivative of the function if it exists. • dy/dx = f’(x) • The differential dx is an independent variable • The differential dy is dy = f’(x)dx

  2. DIFFERENTIALS • y = x5 + 37x dy = (5x4 + 37)dx • This gives you the change in y as x goes to x+dx • y = sin(3x) dy = (3cos3x)dx • We will be estimating the change in one variable as another one changes.

  3. ESTIMATING CHANGE WITH DIFFERENTIALS • Let’s look at an example: • The area of a circle will increase as the radius of the circle increases but how do those relate to each other. • How much will the area increase if the radius increases from 10 m to 10.1 m? • A = πr2 dr = .1 • dA = π(2r)dr = π(2)(10)(.1) = 2π m • So, the estimated increase in area was 6.28 m2 when the radius increased by .1 m. • Let’s check • This is the start to relating the rates of change for different variables

  4. DESCRIBING CHANGE • There are three ways to describe change. • Absolute – this is the actual amount of the change we calculated. Δf = f(a + dx) – f(a) Estimated: df=f’(a)dx • Relative – this is the ratio of the change to the value of the function at that point Δf/f(a) Estimated: df/f(a) • Percentage – this is the relative change stated as a percentage Δf/f(a) x 100 Estimated: df/f(a) x 100

  5. EXAMPLE • Calculate the percentage change in the circle’s area that we calculated earlier. • dA/A x 100 = 100 x (2π)/((2π)(10)2) = 2% • The true change 2.01π/100π = 2.01% • This is a pretty close estimate

  6. RELATED RATES • Any equation involving two or more variables that are differentiable functions of time can be used to find an equation that relates their corresponding rates. • These problems can be solved by first finding an equation that relates the variables. • Then this equation can be differentiated implicitly to relate the rates of each variable. • A = xy dA/dt = x(dy/dt) + y(dx/dt)

  7. STRATEGIES • Draw a picture of the action – label all variables • Write down what you are trying to find • Write down given information in terms of the variables on your picture • Find an equation that would relate the variables • Differentiate with respect to time • Plug in the known values of the variables. • Solve the equation

  8. VOLUME OF A CONE • Find how the volume of a cone changes as the radius or the height of the cone changes. • Write the equation that relates the volume of a cone in terms of radius and height. • How does dV/dt relate to dr/dt if the height is constant? • How does dV/dt relate to dh/dt if the radius is constant? • How does dV/dt relate to dr/dt and dh/dt if neither is constant

  9. V = πr2h • Height is constant: dV/dt = πh(2r)(dr/dt) • Radius is constant: dV/dt = πr2(dh/dt) • Neither constant: dV/dt = π((r2(dh/dt)+h(2r)(dr/dt)) • Now, if you know the rate of change of the radius plug it into the equation • dr/dt = 2 dV/dt = πh(2r)(2) = 4πhr

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