Question

Compute the directional derivative of the following function at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. f(x,y)= ln(2 +3x^2 +3y^2); P(1,-2); (3,1)

Answer 1

Answer:

The directional derivate is given by: $D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$

Step-by-step explanation:

The directional derivative at point (x,y) is given by:

$D_{u}(x,y) = f_{x}(x,y)*a + f_{y}(x,y)*b$

In which a is the x component of the unit vector and b is the y component of the unit vector.

Vector:

We are given the following vector: $v = (3,1)$

Its modulus is given by: $\sqrt{3^2 + 1^2} = \sqrt{10}$

The unit vector is given by each component divided by it's modulus. So

$v_u = (\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}})$

This means that $a = \frac{3}{\sqrt{10}}, b = \frac{1}{\sqrt{10}}$

Partial derivatives:

$f(x,y) = \ln{(2 + 3x^2 + 3y^2)}$

So

$f_x(x,y) = \frac{6x}{\ln{(2 + 3x^2 + 3y^2)}}$

$f_x(1,-2) = \frac{6(1)}{\ln{(2 + 3(1)^2 + 3(-2)^2)}} = \frac{6}{\ln{17}}$

$f_y(x,y) = \frac{6y}{\ln{(2 + 3x^2 + 3y^2)}}$

$f_y(1,-2) = \frac{6(-2)}{\ln{(2 + 3(1)^2 + 3(-2)^2)}} = -\frac{12}{\ln{17}}$

Directional derivative:

$D_{u}(x,y) = f_{x}(x,y)*a + f_{y}(x,y)*b$

$D_{u}(x,y) = \frac{6}{\ln{17}}\times\frac{3}{\sqrt{10}}-\frac{12}{\ln{17}}\times\frac{1}{\sqrt{10}}$

$D_{u}(x,y) = \frac{18}{\ln{17}\sqrt{10}} - \frac{12}{\ln{17}\sqrt{10}}[tex]

[tex]D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$

The directional derivate is given by: $D_{u}(x,y) = \frac{6}{\ln{17}\sqrt{10}}$