10/31/2002, DIRECTIONAL DERIVATIVE Math 21aHomework:11.5 10, 32, 4311.6 8, 16, 20, 26, 30, 42, 44HAPPY HALLOWEEN! Here are some Halloween surfaces:DIRECTIONAL DERIVATIVE.If f is a function of several variables and v is a vector, then ∇f · v is called the directional derivative of fin the direction v. We write ∇vf.∇vf(x, y, z) = ∇f(x, y, z) · vIt is usually assumed that v is a unit vector. Using the chain rule, one can writeddtDvf = f(x + tv).EXAMPLE. PARTIAL DERIVATIVES ARE SPECIAL DIRECTIONAL DERIVATIES.If v = (1, 0, 0), then ∇f · v = fx.If v = (0, 1, 0), then ∇f · v = fy.If v = (0, 0, 1), then ∇f · v = fz.The directional derivative is a generalization of thepartial derivatives. Like the partial derivatives, it isa scalar.EXAMPLE. DIRECTIONAL DERIVATIVE ALONG A CURVE.If f is the temperature in a room and r(t) is a curve with velocity r0(t), then ∇f(r(t)) · r0(t) is the temperaturechange, the body moving on a curve r(t) experiences: the chain rule told us that this is d/dtf (r(t)).GRADIENTS AND LEVEL SURFACES.Gradients areorthogonal to thelevel surfaces.The level surface of thelinear approximation istangent to the surface.Every vector ~x − ~x0in thissurface ∇f · (~x − ~x0) = 0is orthogonal to ∇f.STEEPEST DECENT. The directional derivative satisfies|Dvf| ≤ |∇f||v|because ∇f · v = |∇f||v|| cos(φ)| ≤ |∇f||v|. The direction v = ∇f is the direction, where f increases most, thedirection −∇f is the direction of steepest decent.IN WHICH DIRECTION DOES f INCREASE? If v = ∇f, then the directional derivative is ∇f · ∇f = |∇f|2.This means that f increases, if we move into the direction of the gradient!EXAMPLE. You are on a trip in a zeppelin at (1, 2) and want toavoid a thunderstorm, a region of low pressure. The pressure isgiven by a function p(x, y) = x2+ 2y2. In which direction do youhave to fly so that the pressure change is largest?Parameterize the direction by v = (cos(φ), sin(φ)). The pressure gradient is ∇p(x, y) = (2x, 4y). The directionalderivative in the φ-direction is ∇p(x, y) · v = 2 cos(φ) + 4 sin(φ). This is maximal for −2 sin(φ) + 4 cos(φ) = 0which means tan(φ) = 1/2.ZERO DIRECTIONAL DERIVATIVE. The rate of change in every direction is zero if and only if ∇f(x, y) = 0because if ∇f is not 0, we can choose v = ∇f and get D∇ff = |∇f|2.We will see later that points with ∇f = 0 are candidates for local maxima or minima of f. Points (x, y) where∇f(x, y) = (0, 0) are called stationary points or critical points. Knowing the critical points is important tounderstand the function f.PROPERTIES OF THE DIRECTIONAL DERIVATIVE. The directional derivative Dvhas all the propertiesof a derivative:Dv(f + g) = Dv(f) + Dv(g)Dv(fg) = Dv(f)g + fDv(g)THE MATTERHORN is a popular climbing mountain in the Swiss alps. Itsheight is 4478 meters (14,869 feet). It is quite easy to climb with a guide. Thereare ropes and ladders at difficult places. Even so, about 3 people die each yearfrom climbing accidents at the Matterhorn, this does not stop you from tryingan ascent. In suitable units on the ground, the height f(x, y) of the Matterhornis approximated by f(x, y) = 4000 − x2− y2. At height f (−10, 10) = 3800,at the point (−10, 10, 3800), you rest. The climbing route continues into thenorth-east direction v = (1, −1). Calculate the rate of change in that direction.We have ∇f(x, y) = (−2x, −2y), so that (20, −20)·(1, −1) = 40. This is a place,with a ladder, where you climb 40 meters up when advancing 1m forward.THE VAN DER WAALS (1837-1923) equation for real gases is(p + a/V2)(V − b) = RT (p, V ) ,where R = 8.314J/Kmol is a constant called the Avogadro number. Thislaw relates the pressure p, the volume V and the temperature T of a gas. Theconstant a is related to the molecular interactions, the constant b to the finiterest volume of the molecules.The ideal gas law pV = nRT is obtained when a, b are set to 0. The levelcurves or isotherms T (p, V ) = const tell much about the properties of the gas.The so called reduced van der Walls law T (p, V ) = (p + 3/V2)(3V − 1)/8is obtained by scaling p, T, V depending on the gas. Calculate the directionalderivative of T (p, V ) at the point (p, V ) = (1, 1) into the direction v = (1, 2).We have Tp(p, V ) = (3V − 1)/8 and TV(p, V ) = 3p/8 − (9/8)1/V2− 3/(4V3).Therefore, ∇T (1, 1) = (1/4, 0) and DvT (1, 1) =
View Full Document