6.5 Answer

6.5

1. The directional derivative is given by

$$f_{{\hat u}}'(x,y) = \nabla f(x,y) \cdot {\hat u}$$

Here, ${\hat u}$ is the directional unit vector in the direction of $u$.

(a) Find a unit directional vector ${\hat u}$. Then

$${\hat u} = \frac{(1,\sqrt{3})}{\Vert(1,\sqrt{3})\Vert} = \frac{(1,\sqrt{3})}{\sqrt{1 + 3}} = \frac{(1,\sqrt{3})}{2}$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (2x+1, 1)$$

Thus the directional derivative at $(0,0)$ is

$$f_{\hat u}'(0,0) = \nabla f(0,0) \cdot {\hat u} = (1,1)\cdot \frac{(1,\sqrt{3})}{2} = \frac{1}{2}(1 + \sqrt{3})$$

(b)

Find the unit vector ${\hat u}$. Then

$${\hat u} = \frac{(1,\sqrt{3})}{\Vert(1,\sqrt{3})\Vert} = \frac{(1,\sqrt{3})}{\sqrt{1 + 3}} = \frac{(1,\sqrt{3})}{2}$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (-\sin{x}, \cos{y})$$

Thus the directional derivative at $(0,0)$ is

$$f_{\hat u}'(0,0) = \nabla f(0,0) \cdot {\hat u} = (0,1)\cdot \frac{(1,\sqrt{3})}{2} = \frac{\sqrt{3}}{2}$$

2.

(a) We fin the unit directional vector ${\hat u}$. Then

$${\hat u} = (\cos{\frac{2\pi}{3}}, \sin{\frac{2\pi}{3}}) = (-\frac{1}{2}, \frac{\sqrt{3}}{2}).$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (\frac{x^2y - 2xy^2}{(x-y)^2}, \frac{x^3}{(x-y)^2})$$

Then the directional derivative at $(1,-1)$ is

$$f_{\hat u}'(1,-1) = \nabla f(1,-1) \cdot {\hat u} = (-\frac{3}{4},\frac{1}{4})\cdot (-\frac{1}{2},\frac{\sqrt{3}}{2}) = \frac{3 + \sqrt{3}}{8}$$

(b) We find the unit directional vector ${\hat u}$. Then

$${\hat u} = (\cos{\frac{2\pi}{3}}, \sin{\frac{2\pi}{3}}) = (-\frac{1}{2}, \frac{\sqrt{3}}{2}).$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (\frac{2x}{x^2 + y^2}, \frac{2y}{x^2 + y^2})$$

Thus the directional derivative at $(1,-1)$ is

$$f_{\hat u}'(1,-1) = \nabla f(1,-1) \cdot {\hat u} = (1,-1)\cdot (-\frac{1}{2},\frac{\sqrt{3}}{2}) = -\frac{1}{2}(1 + \sqrt{3})$$

3.

(a) We find the unit directional vector ${\hat u}$. Then

$${\hat u} = \frac{(-1,3)}{\Vert(-1,3)\Vert} = \frac{(-1,3)}{\sqrt{1 + 9}} = \frac{(-1,3)}{\sqrt{10}}$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (\log{y}, \frac{x+1}{y})$$

Then the directional derivative at $(0,1)$.

$$f_{\hat u}'(0,1) = \nabla f(0,1) \cdot {\hat u} = (0,1)\cdot \frac{(-1,3)}{\sqrt{10}} = \frac{3}{\sqrt{10}}$$

The directional derivative becomes max when the direction is the same as the direction of $\nabla f$. Thus,

$${\hat u} = (0,1)$$

(b) We find the unit directional vector ${\hat u}$. Then

$${\hat u} = \frac{(-1,3)}{\Vert(-1,3)\Vert} = \frac{(-1,3)}{\sqrt{1 + 9}} = \frac{(-1,3)}{\sqrt{10}}$$

$$\nabla f(x,y) = (f_{x},f_{y}) = (y^2 e^{xy} + (x-1)y^3 e^{xy}, 2(x-1)ye^{xy} + x(x-1)y^2 e^{xy})$$

Then the directional derivative at $(0,1)$. Thus

$$f_{\hat u}'(0,1) = \nabla f(0,1) \cdot {\hat u} = (0,-2)\cdot(\frac{(-1,3)}{\sqrt{10}} = \frac{-6}{\sqrt{10}}$$

The directional derivative becomes max when the direction is the same as the direction of $\nabla f$.

$${\hat u} = \frac{(0,-2)}{\Vert(0,-2)\Vert} = (0,-1)$$