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

The directional derivate is given by:
`D_(u)(x,y) = (6)/(ln(17)√(10))`

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:
`√(3^2 + 1^2) = √(10)`

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

`v_u = ((3)/(√(10)), (1)/(√(10)))`

This means that
`a = (3)/(√(10)), b = (1)/(√(10))`

Partial derivatives:

`f(x,y) = ln((2 + 3x^2 + 3y^2))`

So

`f_x(x,y) = (6x)/(ln((2 + 3x^2 + 3y^2)))`

`f_x(1,-2) = (6(1))/(ln((2 + 3(1)^2 + 3(-2)^2))) = (6)/(ln(17))`

`f_y(x,y) = (6y)/(ln((2 + 3x^2 + 3y^2)))`

`f_y(1,-2) = (6(-2))/(ln((2 + 3(1)^2 + 3(-2)^2))) = -(12)/(ln(17))`

Directional derivative:

`D_(u)(x,y) = f_(x)(x,y)*a + f_(y)(x,y)*b`

`D_(u)(x,y) = (6)/(ln(17))*(3)/(√(10))-(12)/(ln(17))*(1)/(√(10))`

`D_(u)(x,y) = (18)/(ln(17)√(10)) - (12)/(ln(17)√(10))[tex]</p><p>[tex]D_(u)(x,y) = (6)/(ln(17)√(10))`

The directional derivate is given by:
`D_(u)(x,y) = (6)/(ln(17)√(10))`