## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 801: 21

#### Answer

$$\frac{44}{5}$$

#### Work Step by Step

Given $$f(x, y)=x^{2}+y^{3}, \quad \mathbf{v}=\langle 4,3\rangle, \quad P=(1,2)$$ Find the direction vector: \begin{align*} \mathbf{u}&=\frac{\mathbf{v}}{\|\mathbf{v}\|}\\ &=\frac{\langle 4,3\rangle}{\sqrt{16+9}}\\ &=\left\langle\frac{4}{5}, \frac{3}{5}\right\rangle \end{align*} and \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle\\ &=\left\langle 2 x, 3 y^{2}\right\rangle \\ \nabla f_{(1,2)}&=\langle 2,12\rangle \end{align*} Then the directional derivative is given by \begin{align*} D_{\mathbf{u}} f(1,2)&=\nabla f_{(1,2)} \cdot \mathbf{u}\\ &=\langle 2,12\rangle \cdot\left\langle\frac{4}{5}, \frac{3}{5}\right\rangle\\ &=\frac{8}{5}+\frac{36}{5}\\ &=\frac{44}{5} \end{align*}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.