SOMEONE ASKED 👇
Consider the following equation.
f(x, y) = y3/x, P(1, 3), u = 1/3(2i + 5j)…?
HERE THE ANSWERS 👇

Well,
f(x,y) = y^3 /x
(a)
∂f / ∂x = – y^3/x^2
∂f / ∂y = 3y^2/x
∇f(x, y) = ∂f / ∂x * i + ∂f / ∂y * j
therefore :
∇f(x, y) = – y^3/x^2 i 3y^2/x j
(b)
∇f(1,3) = – 27 i – 27 j
∇f(1,3) = 27 (i + j)
(c)
u = (1/3)(2i + 5j) is not an unit vector : u = (1/3) sqrt(2^2 + 5^2) = sqrt29 /3
therefore :
v = (1/u) u = (1/sqrt29)(2i + 5j) is a unit vector and we have :
Duf(1, 3) = ∇f(1,3) . v
= 27 (i + j) . (1/sqrt29)(2i + 5j)
= – (27/ sqrt29)(2 + 5)
= – 189 /sqt29
hope it’ ll help !!

Y Y3
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