Question

Find the directional derivative, $f_v$, of the function $f(x, y) = 4 + 2x\sqrt{y}$ at the point $P(2, 1)$ in the direction of the vector $\vec{v} = \langle 3, -4 \rangle$. 1. $f_v = -\frac{2}{5}$ 2. $f_v = -\frac{3}{5}$ 3. $f_v = -\frac{1}{5}$ 4. $f_v = \frac{1}{5}$ 5. $f_v = 0$

          Find the directional derivative, $f_v$, of the function
$f(x, y) = 4 + 2x\sqrt{y}$
at the point $P(2, 1)$ in the direction of the vector
$\vec{v} = \langle 3, -4 \rangle$.
1. $f_v = -\frac{2}{5}$
2. $f_v = -\frac{3}{5}$
3. $f_v = -\frac{1}{5}$
4. $f_v = \frac{1}{5}$
5. $f_v = 0$

Find the directional derivative, fv, of the function
f(x, y) = 4 + 2x√(y)
at the point P(2, 1) in the direction of the vector
v⃗ = ⟨ 3, -4 ⟩.
1. fv = -(2)/(5)
2. fv = -(3)/(5)
3. fv = -(1)/(5)
4. fv = (1)/(5)
5. fv = 0

Added by Jesse G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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The gradient of a function f(x,y) is given by the vector (∂f/∂x, ∂f/∂y). In this case, f(x,y) = 4 + 2xy, so ∂f/∂x = 2y and ∂f/∂y = 2x. Therefore, the gradient of f(x,y) is (∂f/∂x, ∂f/∂y) = (2y, 2x). Show more…

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Find the directional derivative, fv, of the function f(x,y) = 4 + 2xy at the point P(2,1) in the direction of the vector v(3,-4). 2. fv CIH 4. fv 5. fv = 0

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