Question

5. If \( f(x, y)=x e^{y} \), find the rate of change of \( f \) (directional derivative) at the point \( P(2,0) \) in the direction from \( P \) to \( Q\left(\frac{1}{2}, 2\right) \).

Answer 1

The directional derivative of the function f(x, y) = x e^y at point P(2,0) in the direction towards Q(1/2, 2) is calculated first by determining the unit vector in the direction from P to Q, then finding the gradient of f at P, and finally taking the dot product of the gradient and the unit vector. The result is 5/2sqrt(6.25).

Step-by-step explanation:

To find the directional derivative of the function f(x, y) = x e^y at point P(2,0) in the direction towards Q(1/2, 2), we first need to find the unit vector that points in the direction from P to Q.

Let's start by finding the vector u, which is the difference between point Q and point P:

• u = Q - P = (1/2 - 2, 2 - 0) = (-3/2, 2)

Next, we normalize u to get the unit vector u_hat:

• u_hat = u / |u|

Where |u| is the magnitude of vector u. Calculate the magnitude:

• |u| = sqrt(((-3/2)^2) + (2^2)) = sqrt(6.25)

Now, normalize u:

• u_hat = (-3/2 / sqrt(6.25), 2 / sqrt(6.25))

The gradient of f, denoted as \\abla f, at point P(2,0) gives us the rates of change in the x and y directions:

• \\abla f = (∂f/∂x, ∂f/∂y)
• At P(2,0), \\abla f = (e^0, 2e^0) = (1, 2)

Finally, the directional derivative D_u f at P is the dot product of \\abla f and u_hat:

• D_u f = \\abla f . u_hat
• D_u f = (1, 2) . (-3/2 / sqrt(6.25), 2 / sqrt(6.25))
• D_u f = (1 * -3/2 / sqrt(6.25)) + (2 * 2 / sqrt(6.25))
• D_u f = -3/2sqrt(6.25) + 4/sqrt(6.25)
• D_u f = (-3/2 + 4)/sqrt(6.25)
• D_u f = (5/2) / sqrt(6.25)
• D_u f = 5/2sqrt(6.25)

The value of the directional derivative indicates how fast the function f(x,y) is changing at point P(2,0) in the direction towards point Q(1/2, 2).