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
Read moreTable of Contents
Textbook Solutions for Calculus: Early Transcendental Functions
Question
In Exercises 51 - 54, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f(x, y) = c at P, (c) find the tangent line to the level curve f(x, y) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane.
\(f(x, y)=4 x^{2}-y\)
c = 6, P(2, 10)
Text Transcription:
f(x, y) = 4x^2 - y
Solution
The first step in solving 13.6 problem number 51 trying to solve the problem we have to refer to the textbook question: In Exercises 51 - 54, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f(x, y) = c at P, (c) find the tangent line to the level curve f(x, y) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy-plane.\(f(x, y)=4 x^{2}-y\)c = 6, P(2, 10)Text Transcription:f(x, y) = 4x^2 - y
From the textbook chapter Directional Derivatives and Gradients you will find a few key concepts needed to solve this.
Visible to paid subscribers only
Step 3 of 7)Visible to paid subscribers only
full solution