į x ( x, y, z ) = −2 e −2 z cos 2 x cos 2 y, f y ( x, y, z ) = −2 e −2 z sin 2 x sin 2 y and f z ( x, y, z ) = −2 e −2 z sin 2 x cos 2 y, so ∇ f ( x, y, z ) = f x ( x, y, z ) i + f y ( x, y, z ) j + f z ( x, y, z ) k = ( 2 e −2 z cos 2 x cos 2 y ) i + ( - 2 e - 2 z sin 2 x sin 2 y ) j + ( 2 e - 2 z sin 2 x cos 2 y ) k = 2 e −2 z ( cos 2 x cos 2 y i − sin 2 x sin 2 y j − sin 2 x cos 2 y k ). and b., we first calculate the partial derivatives f x, f y, f x, f y, and f z, f z, then use Equation 4.40.į x ( x, y, z ) = 10 x − 2 y + 3 z, f y ( x, y, z ) = −2 x + 2 y − 4 z and f z ( x, y, z ) = 3 x − 4 y + 2 z, so ∇ f ( x, y, z ) = f x ( x, y, z ) i + f y ( x, y, z ) j + f z ( x, y, z ) k = ( 10 x − 2 y + 3 z ) i + ( −2 x + 2 y − 4 z ) j + ( 3 x - 4 y + 2 z ) k. These three cases are outlined in the following theorem.įor both parts a. In the first case, the value of D u f ( x 0, y 0 ) D u f ( x 0, y 0 ) is maximized in the second case, the value of D u f ( x 0, y 0 ) D u f ( x 0, y 0 ) is minimized. If φ = π, φ = π, then cos φ = −1 cos φ = −1 and ∇ f ( x 0, y 0 ) ∇ f ( x 0, y 0 ) and u u point in opposite directions. If φ = 0, φ = 0, then cos φ = 1 cos φ = 1 and ∇ f ( x 0, y 0 ) ∇ f ( x 0, y 0 ) and u u both point in the same direction. Recall that cos φ cos φ ranges from −1 −1 to 1. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at ( x 0, y 0 ) ( x 0, y 0 ) multiplied by cos φ. The ‖ u ‖ ‖ u ‖ disappears because u u is a unit vector. Therefore, the z-coordinate of the second point on the graph is given by z = f ( a + h cos θ, b + h sin θ ). The distance we travel is h h and the direction we travel is given by the unit vector u = ( cos θ ) i + ( sin θ ) j. We measure the direction using an angle θ, θ, which is measured counterclockwise in the x, y-plane, starting at zero from the positive x-axis ( Figure 4.39). Given a point ( a, b ) ( a, b ) in the domain of f, f, we choose a direction to travel from that point. We start with the graph of a surface defined by the equation z = f ( x, y ). Now we consider the possibility of a tangent line parallel to neither axis. Similarly, ∂ z / ∂ y ∂ z / ∂ y represents the slope of the tangent line parallel to the y -axis. For example, ∂ z / ∂ x ∂ z / ∂ x represents the slope of a tangent line passing through a given point on the surface defined by z = f ( x, y ), z = f ( x, y ), assuming the tangent line is parallel to the x-axis. These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). A function z = f ( x, y ) z = f ( x, y ) has two partial derivatives: ∂ z / ∂ x ∂ z / ∂ x and ∂ z / ∂ y. In Partial Derivatives we introduced the partial derivative. 4.6.5 Calculate directional derivatives and gradients in three dimensions.4.6.4 Use the gradient to find the tangent to a level curve of a given function.4.6.3 Explain the significance of the gradient vector with regard to direction of change along a surface.4.6.2 Determine the gradient vector of a given real-valued function.4.6.1 Determine the directional derivative in a given direction for a function of two variables.Where we cannot use Einstein notation, since it is impossible to avoid the repetition of more than two indices. In vector calculus, the gradient of a scalar-valued differentiable function f ), The values of the function are represented in greyscale and increase in value from white (low) to dark (high). Multivariate derivative (mathematics) The gradient, represented by the blue arrows, denotes the direction of greatest change of a scalar function.
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