In Exercises 1 - 4, use Theorem 13. 9 to find the directional derivative of the function at P in the direction of the unit vector \(u=\cos \theta i+\sin \theta j\). \(f(x, y)=x^{2}+y^{2}, \quad P(1,-2), \quad \theta=\frac{\pi}{4}\) Text Transcription: u = cos theta i +sin theta j f(x, y) = x^2 + y^2, P(1, -2), theta = pi / 4
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Textbook Solutions for Calculus: Early Transcendental Functions
Question
In Exercises 19 - 22, use the gradient to find the directional derivative of the function at P in the direction of v.
\(f(x, y, z)=x^{2}+y^{2}+z^{2}, \quad P(1,1,1), \quad \mathbf{v}=\frac{\sqrt{3}}{3}(\mathbf{i}-\mathbf{j}+\mathbf{k})\)
Text Transcription:
f(x, y, z) = x^2 + y^2 + z^2, P(1, 1, 1), v = sqrt{3} / 3(i - j + k)
Solution
The first step in solving 13.6 problem number 21 trying to solve the problem we have to refer to the textbook question: In Exercises 19 - 22, use the gradient to find the directional derivative of the function at P in the direction of v.\(f(x, y, z)=x^{2}+y^{2}+z^{2}, \quad P(1,1,1), \quad \mathbf{v}=\frac{\sqrt{3}}{3}(\mathbf{i}-\mathbf{j}+\mathbf{k})\)Text Transcription:f(x, y, z) = x^2 + y^2 + z^2, P(1, 1, 1), v = sqrt{3} / 3(i - j + k)
From the textbook chapter Directional Derivatives and Gradients you will find a few key concepts needed to solve this.
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