Properties of the gradient • k∇f(p)k is maximum rate of increase at p. • ∇f(p) Recall that for a function f(x, y) to understand lo- points in the direction which gives the p cally how the function is behaving we need to un- maximum rate of increase at . derstand the partial derivatives. For convenience we • −k∇f(p)k is maximum rate of decrease at p. have lumped these partial derivatives together into a vector which we call the gradient vector and denote • −∇f(p) points in the direction which gives the as ∇f. In particular, the gradient is a vector valued maximum rate of decrease at p. function defined as In particular the gradient encodes information about  ∂f ∂f  how to move to achieve maximum rates of increase ∇f(x, y) = (x, y), (x, y) . and/or decrease. ∂x ∂y On the other hand Du~ f(p) = 0 along level curves If we have a function of three variables, for example or level surfaces. This is because on a level curve or g(x, y, z) then the gradient becomes level surface the function is constant (i.e., unchang- ing). So if ~u is tangent to a level curve we can con- ∂g ∂g ∂g  clude that ∇f(p) is perpendicular to ~u. Or put more ∇g(x, y, z) = (x, y, z), (x, y, z), (x, y, z) . ∂x ∂y ∂z succinctly:

∂f ∇f(p) is perpendicular to level curves/surfaces. If we look at the individual terms we see that ∂x is indicating how the function is behaving in the x- direction. That is to say that if we start from an initial This gives an easy way to find the normal for tangent point and move in the (positive) x-direction then the planes to a surface, namely given a surface described F( ) = k ∇F( ) initial rate of change of the function is given by this by p we use p as the normal vector. ∂f partial derivative. Similarly we have that ∂y indi- Chain rule cates how the function is behaving in the y-direction. But what if we want to know how the function is The chain rule works for when we have a function behaving in some other direction? inside of a function. For example z = f(x, y), x = x(t) To answer this kind of question we will use direc- and y = y(t). Then z is ultimately a function of t and tional derivatives, so it is natural to ask how does z vary as we vary t, or in other words what is dz . From differentiability f(p + h~u) − f(p) dt Du~ f(p) = lim . we can rearrange the terms to get h 0 h ∂z ∂z ∆z = ∆x + ∆y + ERROR Where Du~ f(p) is “the→ directional derivative of f in ∂x ∂y the direction ~u from the point p = (x0, y0)”. To in- dicate direction we will always use unit vectors. If where ∆z = f(x, y) − f(x0, y0), ∆x = x − x0 and we think of f as elevation and p as indicating our ∆y = y − y0. Also we note that the error term is latitude/longitude position and ~u indicating which very small, much smaller than our other terms and (x y ) direction we want to go, then Du~ f(p) indicates how becomes insignificant near 0, 0 . If we now divide steep it is from our current location if we move in the both sides by ∆t and take a limit as ∆t 0 we have ~u direction indicated by . dz ∂z dx ∂z dy We have already mastered doing directional = + . → dt ∂x dt ∂y dt derivatives in the x and y directions (these are our old friends, the partial derivatives). The nice thing In the above expression we have used both “dz” and is that when our function is differentiable finding di- “∂z”. We use the “d” when we are treating the func- rectional derivatives in any direction is just as easy. tion as depending on a single variable and the “∂” In particular we have the following: when we are treating the function as depending on two or more variables; however they both are asking Du~ f(p) = ∇f(p) · ~u. to do the same thing, namely take a derivative. The same idea works in more complicated situa- (Usually the hardest thing about these problems is to tions. For instance we could have that z = f(x, y) and remember to make sure that ~u is a unit vector.) that x = x(s, t) and y = y(s, t) so that z depends on Recall that ~u · ~v = k~ukk~vk cos θ, and since ~u is a both s and t. In this case we have unit vector we have D f(p) = k∇f(p)k cos θ. Since u~ ∂z ∂z ∂x ∂z ∂y −1 cos θ 1 this gives bounds on how large the = + 6 6 ∂s ∂x ∂s ∂y ∂s directional derivative can be, and in which direction ∂z ∂z ∂x ∂z ∂y we achieve maximum and minimum values. In par- = + ticular we have the following: ∂t ∂x ∂t ∂y ∂t In general whenever we deal with functions within or in differential form functions (possibly within functions themselves, an dz = f dx + f dy inception of functions if you will) then we can find x y . how the variables change with respect to one another. These types of formula are useful when we want One easy way to keep track is to form a tree showing to approximate the change in output given that we the dependencies among the variables. Then to find know the approximate changes of our input. In par- the derivative of the top variable with respect to one ticular this can be used for error tolerance but we of the leaves we simply add up the product of the can also use this to give approximate values for the partial derivatives of the paths (as discussed in class). function near a point that we can easily evaluate the One special case that we have is when we have im- function. plicit relationships among the variables. For exam- Tangent planes are trying to mimic the function ple, F(x, y) = 0 defines y as a function of x. So F so that it agrees locally with the function in both depends on x and y but the relationship also shows the value of the function and the first order partial that y is a function of x. Taking the derivative of both derivatives of the function. We can also try to find a sides with respect to x using the chain rule we get function that matches the value, the first order partial derivatives and the second order partial derivatives. dy dy −Fx Fx + Fy = 0 or = . Taylor polynomials dx dx F These are done by using the . The y second order Taylor polynomials are shown below Similarly we get the following: (where the function and derivatives are all evaluated at the point (x0, y0)):

∂z −Fx ∂z −Fy 1 2 F(x, y, z) = k = = and = . z = f + fx(x − x0) + fy(y − y0) + 2 fxx(x − x0) + ∂x Fz ∂y Fz 1 2 fxy(x − x0)(y − y0) + 2 fyy(y − y0) ⇒ Tangent planes and other miscellany Review problems (One to appear on Quiz 6) We have seen tangent planes done in two different 1. On the surface z = x3 − 3xy + y2 a marble is ways. When we did differentiability for a function placed over the point (1, 2). When released the z = f(x, y) we said that a function locally looks like marble initially move in the direction of steepest a plane along with some possible error, the plane we decrease. Find a vector pointing ha, bi pointing got was as follows: in the direction the marble will initially move. ∂f ∂f 2. Find the directional derivative for the function z = f(x , y )+ (x , y )(x−x )+ (x , y )(y−y ). 2 0 0 ∂x 0 0 0 ∂y 0 0 0 f(x, y, z) = xz − 3xy + 2xyz − 3x + 5y − 17 from the point (2, −6, 3) in the direction of the origin. The second way we have seen these tangent planes is when dealing with F(x, y, z) = k, which we can 3. On the amazing can machine, which has the abil- think of as a level surface. In this case for a point ity to alter the dimensions of cylindrical cans, h = 5 p = (x , y , z ) we can use the properties of gradient the gauges currently read as follows: , 0 0 0 dV = −6π dr = − 1 dh = 1 V h to note that ∇F(p) will be our normal vector so that dt , dt 2 , and dt , where , , r our tangent plane is are volume, height and radius. But the gauge for r is broken. Find the possible value(s) for r. ∇F(p) · hx − x0, y − y0, z − z0i = 0. 4. Given the implicitly relationship z + sin z = xy find ∂2z/∂x∂y only in terms of z. or if we expand out the above we get the following: 5. Suppose that f(x, y) is differentiable and satisfies 3 2 4 3 ∂F ∂F ∂F f(t −t+1, 2−t ) = t −4t +4t+6. Find fx(1, 1) (p)(x − x0) + (p)(y − y0) + (p)(z − z0) = 0 ∂x ∂y ∂z and fy(1, 1). 6. Let f(x, y) = ex−y1 + sin(2x + 3y)
. Find an It is important to note that these two definitions are approximate value for ∆z = f(0.2, 0.1) − f(0, 0). compatible, i.e., if we wanted the tangent plane for z = f(x, y) we would get the same plane as if we 7. Find an equation of the tangent plane to the sur- worked with the function F(x, y, z) = f(x, y) − z = 0. face x2 + xy + y2 + z2 = 16 at the point (1, 2, 3). We can rearrange our terms above to get the fol- 8. Find the second order Taylor polynomial for lowing: 2 f(x, y) = ex+y + x sin y at the point (0, 0). ∆z ≈ fx∆x + fy∆y.