Contents • 9.5 Directional Derivatives • 9.6 Tangent Planes and Normal Lines • 9.7 Divergence and Curl • 9.8 Lines Integrals • 9.9 Independence of Path
9.5 Directional Derivative • IntroductionSee Fig 9.26.
The Gradient of a Function • Define the vector differential operator asthen (1) (2)are the gradients of the functions.
Example 1 Compute Solution
Example 2 If F(x, y, z) = xy2 + 3x2 – z3, find the gradient at (2, –1, 4). Solution
DEFINITION 9.5 Directional Derivatives The directional derivative of z = f(x, y) in the direction of a unit vector u = cos i + sin jis (4)provided the limit exists.
THEOREM 9.6 ProofLet x, y and be fixed, then g(t) = f(x + t cos , y + t sin ) is a function of one variable. If z = f(x, y) is a differentiable function of xand y, and u = cos i + sin j, then (5) Computing a Directional Derivative
First (6) Second by chain rule
Here the subscripts 1 and 2 refer to partial derivatives of f(x + t cos , y + t sin ) w.r.t. (x + t cos )and (y + t sin ), respectively. When t = 0, x + t cos and y + t sin are simply x and y, respectively, then (7) becomes (8) Comparing (4), (6), (8), we have
Example 3 Find the directional derivative of f(x, y) = 2x2y3 + 6xyat (1, 1) in the direction of a unit vector whose angle with the positive x-axis is /6. Solution
Example 3 (2) Now, = /6, u = cos i + sin j becomesThen
Functions of Three Variables • w=F(x,y,z)where , , are the direction angles of the vector u measured relative to the positive x, y, z axis. But as before, we can show that (9)
Since u is a unit vector, from (10) in Sec 7.3 thatIn addition, (9) shows
Example 5 Find the directional derivative of F(x,y, z) = xy2 – 4x2y + z2at (1, –1, 2) in the direction 6i + 2j + 3k. SolutionSincewe have
Example 5 (2) Since ‖6i + 2j + 3k‖= 7, then u = (6/7)i + (2/7)j + (3/7)k is a unit vector. It follows from (9) that
Maximum Value of the Direction Derivative • From the fact thatwhere is the angle between and u. Becausethen
In other words, The maximum value of the direction derivative is and it occurs when uhas the same direction as (when cos = 1), (10)andThe minimum value of the direction derivative is and it occurs when uhas opposite direction as (when cos = −1)(11)
Example 6 • In Example 5, the maximum value of the directional derivative at (1, −1, 2) is and the minimum value is .
Gradient points in Direction of Most Rapid Increase of f • Put it in another way, (10) and (11) state that:The gradient vector points in the direction in which f increase most rapidly, whereas points in the direction of the most rapid decrease of f.
Example 8 The temperature in a rectangular box is approximated by If a mosquito is located at ( ½, 1, 1), in which the direction should it fly up to cool off as rapidly as possible?
Example 8 (2) SolutionThe gradient of T is Therefore, To cool off most rapidly, it should fly in the direction −¼k, that is, it should dive to the floor of the box, where the temperature is T(x, y, 0) = 0
9.6 Tangent Plane and Normal Lines • Geometric Interpretation of the Gradient : Functions of Two VariablesSuppose f(x, y) = c is the level curve of z = f(x, y) passes through P(x0,y0),that is, f(x0, y0) = c.If x = g(t), y = h(t) such that x0 = g(t0), y0= h(t0),then the derivative of f w.r.t. t is (1)When we introduce
then (1) becomes When at t = t0, we have (2)Thus, if , is orthogonal to at P(x0, y0).See Fig 9.30.
Example 1 Find the level curves of f(x, y) = −x2+ y2 passing through (2, 3). Graph the gradient at the point. Solution Since f(2, 3) = 5, we have −x2 + y2= 5.NowSee Fig 9.31.
Geometric Interpretation of the Gradient : Functions of Three Variables • Similar to the concepts of two variables, the derivative of F(f(t), g(t), h(t)) = c implies (3)In particular, at t = t0, (3) is (4)See Fig 9.32.
Example 2 Find the level surfaces of F(x, y, z) = x2 + y2 + z2 passing through (1, 1, 1). Graph the gradient at the point. SolutionSince F(1, 1, 1) = 3，then x2 + y2 + z2 = 3See Fig 9.33.
DEFINITION 9.6 Tangent Plane Let P(x0, y0, z0) be a point on the graph of F(x, y, z) = c, where F is not 0. The tangent plane at P is a plane through P and is perpendicular to F evaluated at P.
That is, . See Fig 9.34. THEOREM 2.1 Let P(x0, y0, z0) be a point on the graph of F(x, y, z) = c, where F is not 0. Then an equation of the tangent plane at P isFx(x0, y0, z0)(x – x0) + Fy(x0, y0, z0)(y – y0) + Fz(x0, y0, z0)(z – z0) = 0 (5) Criterion for an Extra Differential
Example 3 Find the equation of the tangent plane to x2 – 4y2 + z2 = 16 at (2, 1, 4). SolutionF(2, 1, 4) = 16, the graph passes (2, 1, 4). Now Fx(x, y, z) = 2x, Fy(x, y, z) = – 8y, Fz(x, y, z)= 2z, thenFrom (5) we have the equation: 4(x – 2) – 8(y – 1) + 8(z – 4) = 0 or x – 2y + 2z = 8.
Surfaces Given by z = f(x, y) • When the equation is given by z = f(x, y), then we can set F = z – f(x, y) or F = f(x, y) – z, and F=0 represents the equation.
Example 4 Find the equation of the tangent plane to z = ½x2 + ½ y2 + 4 at (1, –1, 5). SolutionLet F(x, y, z) = ½x2 + ½ y2 – z + 4. This graph did pass (1, –1, 5), since F(1, –1, 5)= 0. Now Fx = x, Fy = y, Fz = –1, then From (5), the desired equation is (x - 1) – (y + 1) – (z – 5) = 0or x - y - z = -3
Normal Line • Let P(x0, y0, z0) is on the graph of F(x, y, z) = c, where F 0. The line containing P that is parallel to F(x0, y0, z0) is called the normal line to the surface at P.
Example 5 Find parametric equations for the normal line to the surface in Example 4 at (1, –1, 5). SolutionA direction vector for the normal line at (1, –1, 5) is F(1, –1, 5) = i – j – kthen the desired equations arex = 1 + t, y = –1– t, z = 5 – t
9.7 Divergence and Curl • Vector Functions of two or three variables • F(x, y) = P(x, y)i+ Q(x, y)j F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)kare also called vector fields. • The concept of velocity field and force field plays an important role in mechanics, electricity and magnetism.
Example 1: Two-dimensional VF Graph the vector field F(x, y) = – yi + xj SolutionSinceletFor and k = 2, we have(i) x2 + y2 = 1：at (1, 0), (0, 1), (–1, 0), (0, –1), the corresponding vectors j, –i,– j , ihave the same length 1.
Example 1 (2) (ii) x2 + y2 = 2：at (1, 1), (–1, 1), (–1, –1), (1, –1), the corresponding vectors – i + j, – i – j, i – j, i + jhave the same length . (iii) x2 + y2 = 4：at (2, 0), (0, 2), (–2, 0), (0, –2), the corresponding vectors 2j, –2i, –2j, 2ihave the same length 2. See Fig 9.38.
DEFINITION 9.7 • In practice, we usually use this form: (1) Curl The curl of a vector field F = Pi + Qj + Rk is the vector field
DEFINITION 9.8 • Observe that we can also use this form: (4) Divergence The divergence of a vector field F = Pi + Qj + Rk is the scalar function
Example 2 IfF = (x2y3 – z4)i + 4x5y2zj – y4z6 k, find curl Fand div F。 Solution