Question

Use the gradient to find the directional derivative of the
function at *P* in the direction of

*PQ*.

*f*(*x*, *y*) = 3*x*^{2} -
*y*^{2} + 4, *P*(7,
7), *Q*(2, 6)

Answer #1

Use the gradient to find the directional derivative of the
function at P in the direction of
PQ.
g(x, y,
z) =
xye2z, P(5,
10, 0), Q(0, 0, 0)

Find the gradient ∇f and the directional derivative at the point
P (1,−1,2) in the direction a = (2,−1,1) for the function f (x,y,z)
= x^3z − y(x^2) + z^2. In which direction is the directional
derivative at P decreasing most rapidly and what is its value?

Find the gradient of the function and the max value of the
directional derivative at the given point.
f(x,y) = y2ex^(-2)y
+ x3y at the point (1,1)

Find the directional derivative of the function at P in the
direction of v. f(x, y) = x3 − y3, P(8, 5), v = 2 2 (i + j)

Find the directional derivative of the function at the given
point in the direction of the vector v.
f(x, y, z) = x2y + y2z, (2, 7, 9), v = (2,
−1, 2)
Dvf(2, 7, 9) =

find the directional derivative of f(x,y) = x^2y^3 +2x^4y at the
point (3,-1) in the direction theta= 5pi/6
the gradient of f is f(x,y)=
the gradient of f (3,-1)=
the directional derivative is:

f(x,y,z) = xey+z.
(a) Find the gradient of f, ∇f.
(b) Find the directional derivative of f at the point (2, 1, 2)
in the direction of ? = 3? + 4?.

Find the directional derivative of the function at the given
point in the direction of the vector v.
f(x,y,z)= x2y3+2xz+yz3
(-2,1,-1) v= <1,-2,2>
Use the chain rule to find dz/dt. z=sin(x,y) x= scos(t)
y=2t+s3

Find the value of the directional
derivative of the function w = f ( x , y , z ) = 2 x y + 3 y z
- 4 x z
in the direction of the vector v =
< 1 , -1 , 1 > at the point P ( 1 , 1 , 1 ) .

Find the directional derivative of the function
f(x,y,z)=ln(x2+y2−1)+y+6z at the point (1,1,0) in the direction of
the vector v→=i→−2j→+2k→

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