Question

6. Find the directions in which the directional derivative of $f(x, y) = x^2 + xy^3$ at the point $(2, 1)$ has value 2. 7. Find the maximum rate of change of $f$ at the given point and the direction in which it occurs: $f(x, y) = 4y\sqrt{x}, (4, 1)$

          6. Find the directions in which the directional derivative of $f(x, y) = x^2 + xy^3$ at the point $(2, 1)$ 
has value 2.
7. Find the maximum rate of change of $f$ at the given point and the direction in which it occurs:
$f(x, y) = 4y\sqrt{x}, (4, 1)$

6. Find the directions in which the directional derivative of f(x, y) = x^2 + xy^3 at the point (2, 1)
has value 2.
7. Find the maximum rate of change of f at the given point and the direction in which it occurs:
f(x, y) = 4y√(x), (4, 1)

Added by William A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

03/09/2023

Step 1

In this case, f(x, y) = 2 + y^3. So, ∂f/∂x = 0 and ∂f/∂y = 3y^2. Step 2: Evaluate the gradient at the point (2, 1) Substituting x = 2 and y = 1 into the partial derivatives, we get: ∂f/∂x = 0 and ∂f/∂y = 3(1)^2 = 3. Step 3: Find the directional derivative The Show more…

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6. Find the directions in which the directional derivative of f(, ) = 2 + y3 at the point (2, 1) has value 2. 7. Find the maximum rate of change of f at the given point and the direction in which it occurs: f (, ) = 4g/, (4, 1)