Directional Derivatives and Gradient The gradient vector is calculated using partial derivatives of the function f(x,y). For a function f(x,y), the gradient vector, denoted as f (pronounced grad f ) is created by using the first partial derivatives: f f The Gradient Vector: f = i + j. x y Ex) Find f for f( x, y) 4 = x x + y. Ex) Evaluate f(3,4) for f( x, y) 4 = x x + y. So, calculating the gradient vector only involves using partial derivatives of the function f(x,y) but what is the gradient vector?
Imagine you re standing on the surface z = f( x, y) above the point ( x0, y 0) just below you in the xy plane. The gradient vector acts like a compass which points in the direction of the steepest ascent from where you re standing. On a 3-D surface, the gradient is a 2-D vector which points in the direction of the steepest ascent from any point on the surface. So from where you re standing, the blue route in front of you tells you the steepest climb because the gradient vector (in the xy-plane below) is pointing in that direction. The gradient vector tells you two things: 1. The direction (as a vector) of the maximum rate of change (steepest ascent) from any point on the surface of z = f (x,y) 2. The magnitude of the gradient vector represents the literal slope of the maximum rate of change (slope of the steepest ascent).
Overlapping the gradient vectors with the level curves for a z = f( x, y) surface gives you a 2D representation of the surface s shape., This is a contour map (set of level curves) for some function z = f( x, y). The red vectors represent gradient vectors for this particular surface. NOTEWORTHY STUFF: 1. The gradient vectors are perpendicular to each level curves. 2. The gradient vectors point uphill so this indicates that the concentric contours on the left represent a the concentric contours on the right represent a 3. The gradient vectors lengths represent the amount of steepness at those particular points on the surface. Ex) Make a rough sketch of the surface from the contour map shown above.
Ex) For the functionf( x, y) = ln( x + y ), calculate the direction of the maximum 1 rate of change at the point (1, 2) and also, calculate the value of the maximum rate of change at that point. Graph of the surface The level curves for f( x, y) = ln( x + y ) f( x, y) = ln( x + y ) with the gradient vector 1 drawn from the point (1, 2).
Directional Derivatives The gradient vector is essentially a derivative measure on the surface but it only tells you the maximum slope from a given point. What if you wanted the slope in a direction other than the gradient? If you know the 1) gradient vector 2) the point where you re measuring this new slope ( x0, y 0) from AND 3) a unit vector indicating the new direction you re measuring this slope... then you can calculate the directional derivative for the function. The Directional Derivative for the function f( x, y ) at the point ( x0, y 0) heading in the direction of the unit vector u is denoted as Du f( x, y) calculated by using a dot product: Duf ( x, y) = f( x0, y0) i u Ex) Calculate the directional derivative on the surface 1 point (1, 2) in the direction of the angle θ = π /3. f x y (, ) ln( x y ) = + at the
Contour map of f x y (, ) = ln( x + y ) 1 The blue vector is the gradient vector evaluated at the point (1, 2). The red vector is the directional derivative we just evaluated. It has a smaller length than the gradient vector because climbing along the surface in the red vector s direction is a less steep climb then going in the gradient s direction. Ex) Evaluate the directional derivative for the function the point (1,2) in the direction of the vector v = i 2j. f( x, y) 2xy 3x y 2 4 3 = + at
Tangent Planes In order to find the equation of a plane you ll need a point and a vector NORMAL (perpendicular) to the plane s surface. Therefore, in order to find the equation of a TANGNET plane to a surface in 3 space, you ll need a point on the surface and a vector NORMAL to the surface itself. BUT... how do we get the normal vector to a surface? Ex) Find the equation of the tangent plane to the surface the point (1, 1,7). x + y + 2z = 16 at Ex) Now find the parametric equations for the normal line to the surface x + y + 2z = 16 at the point (1, 1,7).
SUPPLEMENTAL HOMEWORK: GRADIENT AND DIRECTIONAL DERIVATIVES Problems 1 5: a) Find the gradient of f b) Evaluate the gradient at the point P c) Find the directional derivative of f at the point P in the direction of the unit vector u. 1. f( x, y) x y x y 3 4 4 3 = +, ( 1, 1 ) 3 1 P, u = i + j x 3 4 2. f( x, y) = e cosy, P ( 0, 0 ), u =, 5 5 3. 2 f( x, y) y / x 2 =, P ( 1, 2 ), u is the unit vector for the direction angle θ = π 3 4. f( x, y, z) = ln(3x + 6y + 9 z), P ( 1, 1, 1 ), u is parallel to the vector v = 4i + 12j + 6k 5. f( x, y, z) = xy + yz + xz, P ( 1, 1, 3 ), u is parallel to the vector PQ with Q ( 2, 4, 5 ). Problems 6 9: The level curves of a set of functions are graphed below. Sketch the direction and relative magnitude of the gradient vectors on the level curves at the points labeled A, B and C. 6. 7. 8. 9.
Problems 10 12: For the surfaces given in each problem, determine the (a) equations of the tangent plane at the point P and (b) the parametric equations of the normal line at the point P. 10. 11. 12. x + y + z =, P (3,3,5) 2 2( 2) ( 1) ( 3) 10 y = x z, P (4,7,3) 2 xyz = 6, P (3,2,1)