EXAMPLE 4 Find the directional derivative of the function f(x, y) = x2y3 − 3y at the point (2, −1) in the direction of the vector v = 5i + 2j. SOLUTION First we compute the gradient vector at (2, −1).

-2

Step-by-step explanation:

The directional derivative of the function f(x, y) = x²y³ − 3y at the point (2, −1) in the direction of the vector v = 5i + 2j is expressed using the formula

∇f(x,y)•v where ∇f(x,y) is the gradient of the function f(x,y)

∇f(x,y) = δ/δx(x²y³ − 3y)i + δ/δy(x²y³ − 3y)j

∇f(x,y) = 2xy³i+(3x²y²-3)j

∇f(x,y) at (2,-1) = 2(2)(-1)³i + {3(2)²(-1)²-3}j

∇f(x,y) at (2,-1) = -4i+(12-3)j

∇f(x,y) at (2,-1) = -4i+9j

∇f(x,y)•v = (-4i+9j)(5i+2j)

Note that i•i = j•j = 1 and i•j = 0

∇f(x,y)•v = -4(5)+9(2)

∇f(x,y)•v = -20+18

∇f(x,y)•v = -2

you are saying it as if a fact, not a question, but yes, you would be correct? ; )

step-by-step explanation: