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Gradients and Directional Derivatives Calculator

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Example
Created on 2024-06-20Asked by Olivia Nguyen (Solvelet student)
Find the gradient and the directional derivative of the function f(x,y)=x2+y2 f(x, y) = x^2 + y^2 at the point (1,1) (1, 1) in the direction of the vector v=⟨1,2⟩ \mathbf{v} = \langle 1, 2 \rangle .

Solution

To find the gradient of the function f(x,y)=x2+y2 f(x, y) = x^2 + y^2 at the point (1,1) (1, 1) : ∇f(x,y)=⟨∂f∂x,∂f∂y⟩ \nabla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle =⟨2x,2y⟩ = \left\langle 2x, 2y \right\rangle ∇f(1,1)=⟨2⋅1,2⋅1⟩ \nabla f(1, 1) = \left\langle 2 \cdot 1, 2 \cdot 1 \right\rangle =⟨2,2⟩ = \langle 2, 2 \rangle To find the directional derivative in the direction of v=⟨1,2⟩ \mathbf{v} = \langle 1, 2 \rangle : First, normalize v \mathbf{v} : u=v∥v∥=⟨1,2⟩12+22=⟨1,2⟩5=⟨15,25⟩ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{\langle 1, 2 \rangle}{\sqrt{1^2 + 2^2}} = \frac{\langle 1, 2 \rangle}{\sqrt{5}} = \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle The directional derivative Duf D_{\mathbf{u}} f is given by: Duf=∇f⋅u D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} =⟨2,2⟩⋅⟨15,25⟩ = \langle 2, 2 \rangle \cdot \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle =2⋅15+2⋅25 = 2 \cdot \frac{1}{\sqrt{5}} + 2 \cdot \frac{2}{\sqrt{5}} =25+45 = \frac{2}{\sqrt{5}} + \frac{4}{\sqrt{5}} =65 = \frac{6}{\sqrt{5}} Solved on Solvelet with Basic AI Model
Some of the related questions asked by Camila Perez on Solvelet
1. Compute the gradient of the function f(x,y,z)=x2+y2+z2f(x, y, z) = x^2 + y^2 + z^2 at the point (1,2,−3)(1, 2, -3),2. Find the directional derivative of the function g(x,y)=xy2+2xg(x, y) = xy^2 + 2x at the point (1,−1)(1, -1) in the direction of the vector v=2i+3j\mathbf{v} = 2\mathbf{i} + 3\mathbf{j}.,
DefinitionThey are concepts of multi-variable calculus and you can consider them as a measure of how fast a function changes in such or such direction. The gradient is a vector that points to where the function increases the most, with a norm to measure of how fast the function is increasing. On the other hand, the directional derivative is the rate of change of a function as one moves in a specified direction. The gradient is equal to taking the unit vector that points in that direction. The directional derivative is then just the dot product of the gradient with that unit vector who’s taking is along that path. The tools of a gradient and a directional derivative are more natural than other tools that deal better with optimization theory, physical theory, and engineering to study the functions and how they change. Description: if you take a function f(x,y), where its gradient is ∇f=(∂xf,∂yf) and the function is moving in the direction of a unit vector u, so the directional derivative is ∇f⋅u.
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