Answer to Question #175410 in Calculus for kenneth

Question

Find the directional derivative of the function at P in the direction of v. f(x, y) =

x/y P(1,1) v = −j

Answer 1

Let's find the partial derivatives:

$\frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {\frac{x}{y}} \right) = \frac{1}{y} \Rightarrow \frac{{\partial f}}{{\partial x}}\left( P \right) = 1$

$\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {\frac{x}{y}} \right) = - \frac{x}{{{y^2}}} \Rightarrow \frac{{\partial f}}{{\partial y}}\left( P \right) = - \frac{1}{{{1^2}}} = - 1$

find the direction cosines:

$\left| {\overline v } \right| = \sqrt {{0^2} + {{( - 1)}^2}} = 1$

$\cos \alpha = \frac{0}{1} = 0,\,\,\cos \beta = - \frac{1}{1} = - 1$

Then

$\frac{{df}}{{dv}} = \frac{{\partial f}}{{\partial x}}\left( P \right)\cos \alpha + \frac{{\partial f}}{{\partial y}}\left( P \right)\cos \beta = 1 \cdot 0 - 1 \cdot \left( { - 1} \right) = 1$

Answer: 1