Find the directional derivative of f(x,y)=x^2y^3+2x^4y at the point (-1, 5) in the direction θ=5π/6.
The gradient of f is:
∇f=⟨ , ⟩
∇f(−1,5)=⟨ , ⟩
The directional derivative is:

Question

Find the directional derivative of f(x,y)=x^2y^3+2x^4y at the point (-1, 5) in the direction θ=5π/6. 
The gradient of f is:
∇f=⟨ , ⟩ 
∇f(−1,5)=⟨ , ⟩ 
The directional derivative is:

anonymous · Answer

i got the gradient as <2xy^3+8x^3y, 3x^2y^2+2x^4> making the gradient at (-1,5) = <-290,77>, but i got stuck on the directional gradient. formula my book gives is mag of the gradient times cos (theta) = direction, but when i plugged in sqrt(290^2+77^2)cos(5pi/6), webwork said it was wrong......please help?

phi · Answer

I have not worked it out, but my thought is to define  unit vector
u = < cos theta, sin theta>
and form the dot product with the gradient vector.

anonymous · Answer

i'll give that a try

anonymous · Answer

ok, that worked.  thanks @phi!!

gorv · Answer

@saiken2009  plzz provide the medal by clicking best response for helping you..if you r satisfird by @phi  's  help