What you will learn today

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1 What you will learn today Tangent Planes and Linear Approximation and the Gradient Vector Vector Functions 1/21

2 Recall in one-variable calculus, as we zoom in toward a point on a curve, the graph becomes indistinguishable from its tangent line. Similarly, if we zoom in toward a point on a surface, it looks more and more like a plane (called its tangent plane). Suppose a surface S is the graph of a continuous function f, which has first partial derivatives. Look at the two traces C 1 and C 2 passing through a point P on S, say by intersecting S with x = P x and y = P y. C 1 and C 2 has tangent line T 1 and T 2 at P. Then the tangent plane at P is the plane that contains both T 1 and T 2 at P Vector Functions 2/21

3 In the above, we could have replaced C 1 and C 2 by any two curves on S not sharing the same tangent. Suppose P is (x P, y P, z P ), the tangent plane at P has equation z z P = f x (x P, y P )(x x P ) + f y (x P, y P )(y y P ) Vector Functions 3/21

4 The tangent plane tells us that we can approximate the function by using z = z P + f x (x P, y P )(x x P ) + f y (x P, y P )(y y P ) near P. Here z P = f (x P, y P ). This is also called linearization of the function since we are using a linear function. Vector Functions 4/21

5 Find the linearization of the function f (x, y) = xe xy at (1,0). In general, we want the linear approximation to be good. Here is a precise definition of being good : Definition If the increment z = f (a + x, b + y) f (a, b) has the form z = f x (a, b) x + f y (a, b) y + ɛ 1 x + ɛ 2 y where ɛ 1, ɛ 2 0 as ( x, y) (0, 0), then we say f(x,y) is differentiable at (a, b). When the function is differentiable, the linear approximation is close to the surface. Otherwise it may not be a useful approximation. Vector Functions 5/21

6 Compare: In the one-variable case, a function f(x) is differentiable at x=a iff its derivative f x exists (see p126). However, for a function of two variables, even when both f x and f y exist at (a, b), the function may not be differentiable according to the above definition. Like the one-variable case, differentiability implies continuity. Vector Functions 6/21

7 f (x, y) = { xy x 2 +y 2 if (x, y) (0, 0) 0 if (x, y) = (0, 0) We can verify both partials exist at (0,0), however the function is not differentiable, in fact it is not even continuous. Vector Functions 7/21

8 Are there safe situations when we are guaranteed a function is differentiable? Yes. Theorem If both f x and f y are continuous near (a, b), then f is differentiable near (a,b). Vector Functions 8/21

9 1. Verify in the above bad example the partials are not continuous near (0,0). 2. Compare the condition with the theorem on mixed partials. (Clairaut s rule) 3. In the example f (x, y) = xe xy, use the linearization to estimate f(1.1, -0.1). Vector Functions 9/21

10 When a function is differentiable, we define the total differential dz = f x (x, y)dx + f y (x, y)dy The base radius and height of a right circular cone are measured as 10cm and 25cm respectively., with a possible error as much as 0.1cm in each. Find the maximum error in the volume calculation. Vector Functions 10/21

11 Recall the chain rule for one-variable calculus, when y(x(t)), dy dt = dy dx dx dt The two-variable case is similar, when z = f (x(t), y(t)), dz dt = f dx x dt + f dy y dt z = x 2 y + 3xy 4, x = sin 2t, y = cos t, find dz dt when t = 0. Vector Functions 11/21

12 When z = f (x(t, s), y(t, s)), You can draw a tree diagram. z t = f x x t + f y y t If g(s, t) = f (s 2 t 2, t 2 s 2 ), show that it satisfies the PDE t g s + s g t = 0 Vector Functions 12/21

13 Let W (s, t) = F (u(s, t), v(s, t)), and u(1, 0) = 2, u s (1, 0) = 2, u t (1, 0) = 6, F u (2, 3) = 1, v(1, 0) = 3, v s (1, 0) = 5, v t (1, 0) = 4, F v (2, 3) = 10. Find W s (1, 0) and W t (1, 0). If z = f (x, y), x = r 2 + s 2, y = 2rs, find z r and 2 z r 2. Vector Functions 13/21

14 Implicit differentiation: Suppose we have an equation of the form F (x, y, z) = 0 Taking partial derivatives w.r.t. x on both sides, w.r.t. y, F x + F z z x = 0 F y + F z z y = 0 As a theorem, when F z (a, b, c) 0, and F x, F y and F z are continuous near (a,b,c), z is given implicitly as a function of x and y. Vector Functions 14/21

15 Find z x and z y if x 3 + y 3 + z 3 + 6xyz = 1. Vector Functions 15/21

16 Recall in the definition of partial derivatives we only looked at the rate of change w.r.t to i and j direction. We will look at the rate of change w.r.t any direction given by a unit vector u = a, b towards a point (x 0, y 0 ). D u f (x 0, y 0 ) = lim h 0 f (x 0 + ah, y 0 + bh) f (x 0, y 0 ) h is the directional derivatives along u = a, b if it exists. When are we guaranteed that the directional derivatives exist? Theorem If f is a differentiable function of x and y, then f has directional derivatives along any direction u = a, b, and moreover, D u f (x, y) = f x a + f y b Vector Functions 16/21

17 Find the directional derivative D u of f (x, y) = x 3 3xy + 4y 2 and u has angle π 6 with i on the xy-plane. In the above theorem we can write more concisely as D u f = f x, f y u We define f := f x, f y called the gradient of f. Since u is a unit vector, the directional derivative is the scalar projection of the gradient onto u. Vector Functions 17/21

18 Just as in the situation of doing scalar projections, the directional derivative gets its maximum (not just max of the absolute value) when the direction u is the same as f. It is minimum when u is opposite to f, and 0 when they are perpendicular (or when u is along the tangent of the level curves of z.) Vector Functions 18/21

19 Similarly, for functions of three variables, if u = a, b, c is a unit vector, D u f = f x a + f y b + f z c f = f x, f y, f z D u f = f u If f (x, y, z) = xsin yz, find the directional derivative along the direction v = i + 2 j k. The temperature in space is given by T (x, y, z) = 80 1+x 2 +2y 2 +3z 2, in which direction does the temperature increase fastest at (1,1,-2)? And how much? Vector Functions 19/21

20 In the two variable case z = f (x, y), the gradient vector is perpendicular to the level curve (why?). In the three variable case u = f (x, y, z), the gradient vector f x, f y, f z is perpendicular to the level surface f (x, y, z) = k. In other words, it gives the normal direction of the surface f (x, y, z) = k at each point on the surface. In other words, at (x 0, y 0, z 0 ) on the surface f (x, y, z) k = 0, the equation of the tangent plane is F x (x 0, y 0, z 0 )(x x 0 )+F y (x 0, y 0, z 0 )(y y 0 )+F z (x 0, y 0, z 0 )(z z 0 ) = 0 Compare with implicit differentiation. What is equation of its normal line? Consider tangent plane of the ellipsoid x2 4 + y 2 + y 2 9 = 3 at ( 2, 1, 3). You can plot a gradient vector field. Vector Functions 20/21

21 What you have learned today Tangent Planes and Linear Approximation and the Gradient Vector Vector Functions 21/21