What is the directional derivative of xy2 yz3 at point (2,-1,1) in the direction of the notmal to the surface xlogz-y2=-4 at (-1,2,1)?

Question

Answer

Directional Derivative

The directional derivative is the rate at which a function changes as one moves in a specified direction. It is used in vector calculus to study functions of several variables.

Given Function

The given function is f(x,y,z) = xy²yz³. We need to find the directional derivative of this function at point (2,-1,1) in the direction of the normal to the surface xlogz-y²=-4 at (-1,2,1).

Gradient Vector

The gradient vector of a function is a vector that points in the direction of the greatest rate of increase of the function at a given point. The magnitude of the gradient vector is the rate of increase in that direction.

The gradient vector of the given function is:

grad(f) = (y²z³, 2xyz³, 3xy²z²)

At point (2,-1,1), the gradient vector is:

grad(f)(2,-1,1) = (1, -4, 12)

Normal Vector

The normal vector to a surface at a given point is a vector that is perpendicular to the tangent plane at that point.

The normal vector to the surface xlogz-y²=-4 at (-1,2,1) is:

n = (-d/dx(xlogz-y²), -d/dy(xlogz-y²), d/dz(xlogz-y²))

At point (-1,2,1), the normal vector is:

n(-1,2,1) = (-2, -1, 1)

Directional Derivative Formula

The directional derivative of a function f in the direction of a unit vector u at a point (x,y,z) is:

D_u(f)(x,y,z) = grad(f)(x,y,z) · u

where · denotes the dot product.

Unit Vector in the Direction of the Normal

The unit vector in the direction of the normal to the surface at (-1,2,1) is:

u = n(-1,2,1) / ||n(-1,2,1)||

where ||n(-1,2,1)|| is the magnitude of the normal vector.

||n(-1,2,1)|| = sqrt(6)

So, the unit vector is:

u = (-2/sqrt(6), -1/sqrt(6), 1/sqrt(6))

Directional Derivative

Substituting the values in the directional derivative formula, we get:

D_u(f)(2,-1,1) = grad(f)(2,-1,1) · u

D_u(f)(2,-1,1) = (1, -4, 12) · (-2/sqrt(6), -1/sqrt(6), 1/sqrt(6

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