18 directional derivatives and gradient
- 1. Directional Derivatives and the Gradient
- 2. Directional Derivatives and the Gradient Given a vector v = <x, y>, we define the directional y slope of v to be |x|.
- 3. Directional Derivatives and the Gradient Given a vector v = <x, y>, we define the directional y slope of v to be |x|. Example A. Find the directional slope of <1, 2>, <–1, –2>, <1, –2> and <–1, 2>.
- 4. Directional Derivatives and the Gradient Given a vector v = <x, y>, we define the directional y slope of v to be |x|. Example A. Find the directional slope of <1, 2>, <–1, –2>, <1, –2> and <–1, 2>. The directional slope of y <–1, 2> <1, 2> a. <1, 2> is 2, b. <–1, 2> is 2, x
- 5. Directional Derivatives and the Gradient Given a vector v = <x, y>, we define the directional y slope of v to be |x|. Example A. Find the directional slope of <1, 2>, <–1, –2>, <1, –2> and <–1, 2>. The directional slope of y <–1, 2> <1, 2> a. <1, 2> is 2, b. <–1, 2> is 2, x c. <–1, –2> is –2, <–1, –2> <1, –2> d. <1, –2> is –2.
- 6. Directional Derivatives and the Gradient Given a vector v = <x, y>, we define the directional y slope of v to be |x|. Example A. Find the directional slope of <1, 2>, <–1, –2>, <1, –2> and <–1, 2>. The directional slope of y <–1, 2> <1, 2> a. <1, 2> is 2, b. <–1, 2> is 2, x c. <–1, –2> is –2, <–1, –2> <1, –2> d. <1, –2> is –2. The directional slope of v indicates the steepness of v and the vertical direction it’s pointing. Positive directional slope indicates v points upward and negative directional slope means v points downward.
- 7. Directional Derivatives and the Gradient y = f(x) Given a differentiable point p on the p curve y = f(x) and a tangent vector v at p, depending whether v points up or down, we obtain two possible directional slopes for v.
- 8. Directional Derivatives and the Gradient y = f(x) Given a differentiable point p on the p curve y = f(x) and a tangent vector v at p, depending whether v points up or down, we obtain two possible directional slopes for v. If v points to the right in the positive x–axis direction then its directional slope is f '(x), and if v points to the left then the directional slope of v is –f '(x).
- 9. Directional Derivatives and the Gradient y = f(x) Given a differentiable point p on the p curve y = f(x) and a tangent vector v at p, depending whether v points up or down, we obtain two possible directional slopes for v. If v points to the right in the positive x–axis direction then its directional slope is f '(x), and if v points to the left then the directional slope of v is –f '(x). For example, v = <1, –2> and u = <–2, 4> y=x 2 u = <–2, 4> are tangent to y = x 2 at (–1, 1), x v = <1, –2>
- 10. Directional Derivatives and the Gradient y = f(x) Given a differentiable point p on the p curve y = f(x) and a tangent vector v at p, depending whether v points up or down, we obtain two possible directional slopes for v. If v points to the right in the positive x–axis direction then its directional slope is f '(x), and if v points to the left then the directional slope of v is –f '(x). For example, v = <1, –2> and y=x 2 u = <–2, 4> are tangent to y = x 2 at (–1, 1), v points to the positive x–axis direction, so its directional slope is –2 = f '(x), x v = <1, –2>
- 11. Directional Derivatives and the Gradient y = f(x) Given a differentiable point p on the p curve y = f(x) and a tangent vector v at p, depending whether v points up or down, we obtain two possible directional slopes for v. If v points to the right in the positive x–axis direction then its directional slope is f '(x), and if v points to the left then the directional slope of v is –f '(x). For example, v = <1, –2> and u = <–2, 4> y=x 2 u = <–2, 4> are tangent to y = x 2 at (–1, 1), v points to the positive x–axis direction, so its directional slope is –2 = f '(x), and u points in x the negative x–axis direction with directional slope 2 = –f '(x). v = <1, –2>
- 12. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle.
- 13. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. In the x–axis direction z = f(x, y) P=(a, b, c) y x <–1, 0> (a, b) <1, 0>
- 14. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, In the x–axis direction z = f(x, y) P=(a, b, c) y x <–1, 0> (a, b) <1, 0>
- 15. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, In the x–axis direction z = f(x, y) P=(a, b, c) y x <–1, 0> (a, b) <1, 0>
- 16. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M y x x <–1, 0> (a, b) <1, 0>
- 17. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, so M = df/dx|p is also called the directional derivative in the direction of <1, 0> In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M y x x <–1, 0> (a, b) <1, 0>
- 18. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, so M = df/dx|p is also called the directional derivative in the direction of <1, 0> and an opposite tangent has directional derivative –M. In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M y x x <–1, 0> (a, b) <1, 0>
- 19. Directional Derivatives and the Gradient Let P = (a, b, c) be a differentiable point on the surface defined by z = f(x, y). The plane y = b is aligned in the two directions of <1, 0> or <–1, 0> on the unit circle. The partial derivative df/dx|p= M is the directional slope in the direction <1, 0> at P, so M = df/dx|p is also called the directional derivative in the direction of <1, 0> and an opposite tangent has directional derivative –M. In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M directional slope y = –df/dx|p = – M x x <–1, 0> (a, b) <1, 0>
- 20. Directional Derivatives and the Gradient In other words, if we’re standing at P with a compass, the directional derivative in direction of <1, 0>, i.e. to the east, which is M = df/dx|p, In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M y x x <–1, 0> <1, 0> (a, b)
- 21. Directional Derivatives and the Gradient In other words, if we’re standing at P with a compass, the directional derivative in direction of <1, 0>, i.e. to the east, which is M = df/dx|p, indicates the up or down and the steepness of the trail that is heading eastward. In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M y x x <–1, 0> <1, 0> (a, b)
- 22. Directional Derivatives and the Gradient In other words, if we’re standing at P with a compass, the directional derivative in direction of <1, 0>, i.e. to the east, which is M = df/dx|p, indicates the up or down and the steepness of the trail that is heading eastward. Likewise, –M = –df/dx|p indicates the up or down and the steepness of the trail that is heading in the <–1, 0> direction or to the west. In the x–axis direction z y=b z = f(x, y) P P=(a, b, c) directional slope = df/dx|p = M directional slope y = –df/dx|p = – M x x <–1, 0> <1, 0> (a, b)
- 23. Directional Derivatives and the Gradient In general, given any 2D (unit) directional vector u = <a, b>, y x (a, b) u = <a, b>
- 24. Directional Derivatives and the Gradient In general, given any 2D (unit) directional vector u = <a, b>, the plane that contains the point P aligned in the direction of u, makes a trace on the surface z = f(x, y). y x (a, b) u = <a, b>
- 25. Directional Derivatives and the Gradient In general, given any 2D (unit) directional vector u = <a, b>, the plane that contains the point P aligned in the direction of u, makes a trace on the surface z = f(x, y). In the u–direction P=(a, b, c) y x (a, b) u = <a, b>
- 26. Directional Derivatives and the Gradient In general, given any 2D (unit) directional vector u = <a, b>, the plane that contains the point P aligned in the direction of u, makes a trace on the surface z = f(x, y). We define the directional derivative in the direction of u at P, denoted as D u(P), to be the directional slope of any vector tangent at P in the direction u. In the u–direction P=(a, b, c) y x (a, b) u = <a, b>
- 27. Directional Derivatives and the Gradient In general, given any 2D (unit) directional vector u = <a, b>, the plane that contains the point P aligned in the direction of u, makes a trace on the surface z = f(x, y). We define the directional derivative in the direction of u at P, denoted as D u(P), to be the directional slope of any vector tangent at P in the direction u. z In the u–direction P=(a, b, c) P=(a, b, c) y Du(P) = the directional derivative x (a, b) = the directional slope of any tangent, u = <a, b> in the direction of u.
- 28. Directional Derivatives and the Gradient Again using the trail–analogy that if we’re standing at P, with the compass needle pointing in the direction of u = <a, b>, Du(P) indicates the up or down and the steepness of the trail that is heading in the u–direction. z In the u–direction P=(a, b, c) P=(a, b, c) y Du(P) = the directional derivative x (a, b) = the directional slope of any tangent, u = <a, b> in the direction of u.
- 29. Directional Derivatives and the Gradient Again using the trail–analogy that if we’re standing at P, with the compass needle pointing in the direction of u = <a, b>, Du(P) indicates the up or down and the steepness of the trail that is heading in the u–direction. Note that D–u(P), the opposite directional derivative, satisfies D–u(P) = –Du(P). z In the u–direction P=(a, b, c) P=(a, b, c) y –u Du(P) = the directional derivative x (a, b) = the directional slope of any tangent, u = <a, b> in the direction of u.
- 30. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane itself consists of all the tangent vectors to all the traces at P
- 31. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P
- 32. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P and Du(P) are precisely the directional x slopes of these vectors.
- 33. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P and Du(P) are precisely the directional x slopes of these vectors. Hence let’s study the directional derivatives of a plane.
- 34. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P and Du(P) are precisely the directional x slopes of these vectors. Hence let’s study the directional derivatives of a plane. We observe that the directional derivatives in a plane depend only on the direction u but not on the location,
- 35. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P and Du(P) are precisely the directional x slopes of these vectors. Hence let’s study the directional derivatives of a plane. We observe that the directional derivatives in a plane depend only P u on the direction u but not on the In a plane D (P) = D (Q) location, u u Q u
- 36. Directional Derivatives and the Gradient The problem of calculating Du(P) may be reduced to the surface’s tangent plane at P because the tangent plane P itself consists of all the tangent vectors to all the traces at P and Du(P) are precisely the directional x slopes of these vectors. Hence let’s study the directional derivatives of a plane. We observe that the directional derivatives in a plane depend only P u on the direction u but not on the In a plane location, i.e. for a fixed u, P and Q D (P) = D (Q) Q u u in a plane that Du(P) = Du(Q). u
- 37. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's.
- 38. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular we may reconstruct the plane using M and L as shown.
- 39. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular we may reconstruct the plane using M and L as shown. Mdx x dx
- 40. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular we may reconstruct the plane using M and L as shown. Ldy Mdx x y dy dx
- 41. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular we may reconstruct the plane using M and L as shown. Ldy Mdx x y dy dx
- 42. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular we may reconstruct the plane using M and L as shown. Ldy Mdx The Gradient y x The vector <M, L> is called the dy dx gradient (vector) of T.
- 43. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular The gradient ∇ f(P) = <M,L> we may reconstruct the plane using M and L as shown. Ldy Mdx The Gradient y <M,L> x The vector <M, L> is called the M dy dx gradient (vector) of T. L
- 44. Directional Derivatives and the Gradient We also note that the directional derivatives in the plane are determined by the way the plane sits in space. For example, a flat plane defined by z = c has the Du(P) = 0 for all u's and P's. The direction that a plane T faces is determined by its partials derivative df/dx = M and df/dy = L. In particular The gradient ∇ f(P) = <M,L> we may reconstruct the plane using M and L as shown. Ldy Mdx The Gradient y <M,L> x The vector <M, L> is called the M dy dx gradient (vector) of T. L If T is the tangent plane at P to the surface z = f(x, y), then we call <M, L> the gradient at P, and is denoted as∇ f(P), i.e. ∇ f(P) = (fx(P), fy(P)).
- 45. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb.
- 46. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>.
- 47. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>. We need ∇f(3, 2) = (fx(3, 2), fy(3, 2)) where fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 .
- 48. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>. We need ∇f(3, 2) = (fx(3, 2), fy(3, 2)) where fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . So M = –1/2 and L = –1/3 and the gradient at P is ∇f(3, 2) = <–1/2, –1/3>.
- 49. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>. We need ∇f(3, 2) = (fx(3, 2), fy(3, 2)) where fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . So M = –1/2 and L = –1/3 and the gradient at P is ∇f(3, 2) = <–1/2, –1/3>. The directional derivative Duf(2, 3) with u = <4/5, –3/5> is ∇f(3, 2) • u
- 50. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>. We need ∇f(3, 2) = (fx(3, 2), fy(3, 2)) where fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . So M = –1/2 and L = –1/3 and the gradient at P is ∇f(3, 2) = <–1/2, –1/3>. The directional derivative Duf(2, 3) with u = <4/5, –3/5> is ∇f(3, 2) • u = <–1/2, –1/3> • <4/5, –3/5> = –1/5.
- 51. Directional Derivatives and the Gradient Gradient–Directional Derivative Theorem: Given that z = f(x, y) is differentiable at P, the directional derivative at P in the direction of u = <a, b> with |u| = 1 is Duf(x, y) = ∇f(x, y) • u = <M, L> • <a, b> = Ma + Lb. Example B. Let f(x, y) = √49 – x2 – y2 and P = (3, 2, 6). Find the gradient at P, and the directional derivative with u = < 4/5, 3/5>. We need ∇f(3, 2) = (fx(3, 2), fy(3, 2)) where fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . So M = –1/2 and L = –1/3 and the D f(2, 3)=–1/5 u (3, 2, 6) gradient at P is ∇f(3, 2) = <–1/2, –1/3>. The directional derivative Duf(2, 3) y with u = <4/5, –3/5> is ∇f(3, 2) • u x (3, 2, 0) = <–1/2, –1/3> • <4/5, –3/5> = –1/5. u=<4/5, –3/5>
- 52. Directional Derivatives and the Gradient Zero Gradient Theorem: Let f(x, y) be differentiable at P = (a, b, c) a. Regular Saddle Point Maximum Minimum Monkey Saddle Point, etc..
- 53. Directional Derivatives and the Gradient Zero Gradient Theorem: Let f(x, y) be differentiable at P = (a, b, c) a. If ∇f(a, b) = 0, then all directional derivatives are 0
- 54. Directional Derivatives and the Gradient Zero Gradient Theorem: Let f(x, y) be differentiable at P = (a, b, c) a. If ∇f(a, b) = 0, then all directional derivatives are 0 and P could be a maximum, Maximum
- 55. Directional Derivatives and the Gradient Zero Gradient Theorem: Let f(x, y) be differentiable at P = (a, b, c) a. If ∇f(a, b) = 0, then all directional derivatives are 0 and P could be a maximum, or a minimum, Maximum Minimum
- 56. Directional Derivatives and the Gradient Zero Gradient Theorem: Let f(x, y) be differentiable at P = (a, b, c) a. If ∇f(a, b) = 0, then all directional derivatives are 0 and P could be a maximum, or a minimum, or a saddle point. Regular Saddle Point Maximum Minimum Monkey Saddle Point, etc..
- 57. Directional Derivatives and the Gradient Gradient–Steepness Theorem: Let f(x, y) be differentiable at P = (a, b, c) and ∇f(a, b) = <M, L> (≠ 0) then <M, L> gives the direction of maximum directional derivative (steepest climb) with value equal to |<M, L>|. The opposite direction <–M, –L> gives the direction of minimum directional derivative (steepest descend) with value –|<M, L>|.
- 58. Directional Derivatives and the Gradient Gradient–Steepness Theorem: Let f(x, y) be differentiable at P = (a, b, c) and ∇f(a, b) = <M, L> (≠ 0) then <M, L> gives the direction of maximum directional derivative (steepest climb) with value equal to |<M, L>|. The opposite direction <–M, –L> gives the direction of minimum directional derivative (steepest descend) with value –|<M, L>|. The gradient ∇ f(P) = <M,L> Mdx Ldy <M,L> x y M dy dx L
- 59. Directional Derivatives and the Gradient Gradient–Steepness Theorem: Let f(x, y) be differentiable at P = (a, b, c) and ∇f(a, b) = <M, L> (≠ 0) then <M, L> gives the direction of maximum directional derivative (steepest climb) with value equal to |<M, L>|. The opposite direction <–M, –L> gives the direction of minimum directional derivative (steepest descend) with value –|<M, L>|. Here is an example of the steepest direction in a plane. The gradient The Steepest Climb (0, 1, 1/2) ∇ f(P) = <M,L> Mdx Ldy 1/2 <M,L> x (1, 0, 1/4) y M 1 1/4 y dy dx 1 L <1/4, 1/2>=∇f x The Steepest Direction
- 60. Directional Derivatives and the Gradient Example: Let f(x, y) = √k – x2 – y2, with k > 0. Draw the gradient in the R2 plane at (a, b) in the domain.
- 61. Directional Derivatives and the Gradient Example: Let f(x, y) = √k – x2 – y2, with k > 0. Draw the gradient in the R2 plane at (a, b) in the domain. We have fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 .
- 62. Directional Derivatives and the Gradient Example: Let f(x, y) = √k – x2 – y2, with k > 0. Draw the gradient in the R2 plane at (a, b) in the domain. We have fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . At (a, b) in the domain, after clearing the denominator, we've the direction the gradient or the direction of the steepest climb to be <–a, –b>.
- 63. Directional Derivatives and the Gradient Example: Let f(x, y) = √k – x2 – y2, with k > 0. Draw the gradient in the R2 plane at (a, b) in the domain. We have fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . At (a, b) in the domain, after clearing the denominator, we've the direction the gradient or the direction of the steepest climb to be <–a, –b>. The surface is a hemisphere centered at the origin hence <–a, –b> is the direction to the top which is also the direction of the steepest climb.
- 64. Directional Derivatives and the Gradient Example: Let f(x, y) = √k – x2 – y2, with k > 0. Draw the gradient in the R2 plane at (a, b) in the domain. We have fx = –x/√49 – x2 – y2, fy = –y/√49 – x2 – y2 . At (a, b) in the domain, after clearing the denominator, we've the direction the gradient or the direction of the steepest climb to be <–a, –b>. (a,b) The surface is a hemisphere centered at the origin hence <–a, –b> is the direction to the <–a,–b> top which is also the direction of the steepest climb.
- 65. Directional Derivatives and the Gradient Gradient–Level Theorem: Let f(x, y) be differentiable at P with a nonzero gradient then the gradient is perpendicular to the level curve.
- 66. Directional Derivatives and the Gradient Gradient–Level Theorem: Let f(x, y) be differentiable at P with a nonzero gradient then the gradient is perpendicular to the level curve. As in last example, f(x, y) = √49 – x2 – y2. The level curves are concentric circles with (0,0) as the center.
- 67. Directional Derivatives and the Gradient Gradient–Level Theorem: Let f(x, y) be differentiable at P with a nonzero gradient then the gradient is perpendicular to the level curve. As in last example, <x, y> f(x, y) = √49 – x2 – y2. The level curves are <–x, –y> concentric circles with (0,0) as the center. The gradient <–x, –y> is perpendicular to the levels curves at all points.