EXAMPLE 5 If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k. SOLUTION (a) The gradient of f is ∇f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z)

Step-by-step explanation:

Given f(x, y, z) = x sin(yz), the formula for calculating the gradient of the function is expressed as ∇f(x, y, z) = fx(x, y, z)i+ fy(x, y, z)j+fz(x, y, z)k where;

fx, fy and fz are the differential of the functions with respect to x, y and z respectively.

a) ∇f(x, y, z) = sin(yz)i + xzcos(yz)j + xycos(yz)k

The gradient of f =  sin(yz)i + xzcos(yz)j + xycos(yz)k

b) Directional derivative of f at (1,2,0) in the direction of v = i + 4j − k is expressed as ∇f(1, 2, 0)*v

∇f(1, 2, 0) = sin(2(0))i +1*0cos(2*0)j + 1*2cos(2*0)k

∇f(1, 2, 0) = sin0i +0cos(0)j + 2cos(0)k

∇f(1, 2, 0) = 0i +0j + 2k

Given v = i + 4j − k

∇f(1, 2, 0)*v (note that this is the dot product of the two vectors)

∇f(1, 2, 0)*v =  (0i +0j + 2k)*(i + 4j − k )

Given i.i = j.j = k.k =1 and i.j=j.i=j.k=k.j=i.k = 0

∇f(1, 2, 0)*v = 0(i.i)+4*0(j.j)+2(-1)k.k

∇f(1, 2, 0)*v = 0(1)+0(1)-2(1)

∇f(1, 2, 0)*v =0+0-2

∇f(1, 2, 0)*v= -2

Hence, the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k is -2

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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