Question

Find the directional derivative of f =xyz along the tangent to the curve x=t,y=t^2,z=t^3at the point( 1,1,1)

Answer 1

Answer:

Let f(x,y)=x2y.f(x,y)=x2y. (a) Find ∇f(3,2)∇f(3,2). (b) Find the derivative of ff in the direction of (1,2) at the point (3,2).

Solution: (a) The gradient is just the vector of partial derivatives. The partial derivatives of ff at the point (x,y)=(3,2)(x,y)=(3,2) are:

∂f∂x(x,y)∂f∂x(3,2)=2xy=12∂f∂y(x,y)∂f∂y(3,2)=x2=9

∂f∂x(x,y)=2xy∂f∂y(x,y)=x2∂f∂x(3,2)=12∂f∂y(3,2)=9

Therefore, the gradient is

∇f(3,2)=12i+9j=(12,

I think this is the answer please mark as a brain list