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- https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/14%3A_Partial_Derivatives/14.06%3A_Directional_Derivatives_and_the_Gradient_Vector\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat...\vecs∇f(x,y,z)=fx(x,y,z)ˆi+fy(x,y,z)ˆj+fz(x,y,z)ˆk=(2e−2zcos2xcos2y)ˆi+(−2e−2zsin2xsin2y)ˆj+(−2e−2zsin2xcos2y)ˆk=2e−2z(cos2xcos2yˆi−sin2xsin2yˆj−sin2xcos2yˆk).
- https://math.libretexts.org/Bookshelves/Calculus/Vector_Calculus_(Corral)/02%3A_Functions_of_Several_Variables/2.04%3A_Directional_Derivatives_and_the_GradientFor a function z=f(x,y), we learned that the partial derivatives ∂f/∂x and∂f/∂y represent the (instantaneous) rate of change of f in the positive x and y directions, respectively. What about other dir...For a function z=f(x,y), we learned that the partial derivatives ∂f/∂x and∂f/∂y represent the (instantaneous) rate of change of f in the positive x and y directions, respectively. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative.
- https://math.libretexts.org/Courses/De_Anza_College/Calculus_IV%3A_Multivariable_Calculus/01%3A_Differentiation_of_Functions_of_Several_Variables/1.06%3A_Directional_Derivatives_and_the_GradientA function z=f(x,y) has two partial derivatives: ∂z/∂x and ∂z/∂y. These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change ...A function z=f(x,y) has two partial derivatives: ∂z/∂x and ∂z/∂y. 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). Similarly, ∂z/∂y represents the slope of the tangent line parallel to the y-axis. Now we consider the possibility of a tangent line parallel to neither axis.
- https://math.libretexts.org/Courses/Misericordia_University/MTH_171-172%3A_Calculus_-_Early_Transcendentals_(Stewart)/14%3A_Partial_Derivatives/14.06%3A_Directional_Derivatives_and_the_Gradient_Vector\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat...\vecs∇f(x,y,z)=fx(x,y,z)ˆi+fy(x,y,z)ˆj+fz(x,y,z)ˆk=(2e−2zcos2xcos2y)ˆi+(−2e−2zsin2xsin2y)ˆj+(−2e−2zsin2xcos2y)ˆk=2e−2z(cos2xcos2yˆi−sin2xsin2yˆj−sin2xcos2yˆk).
- https://math.libretexts.org/Bookshelves/Scientific_Computing_Simulations_and_Modeling/Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/13%3A_Continuous_Field_Models_I__Modeling/13.02%3A_Fundamentals_of_Vector_CalculusIn order to develop continuous field models, you need to know some basic mathematical concepts developed and used in vector calculus.
