Example 3.4 Find the directional derivative of g(x, y) = ex2cos y at (1, π)in the direction −3i + 4j.

Sure, here are the steps to find the directional derivative:

Step 1: Compute the gradient of g(x, y) = e^(x^2)cos(y)

The gradient of a function is a vector that contains the first derivative of the function with respect to each variable. In this case, we have two variables, x and y.

The partial derivative of g with respect to x is: dg/dx = 2x * e^(x^2) * cos(y)

The partial derivative of g with respect to y is: dg/dy = -e^(x^2) * sin(y)

So, the gradient of g is ∇g = (2x * e^(x^2) * cos(y))i - (e^(x^2) * sin(y))j

Step 2: Evaluate the gradient at the point (1, π)

∇g(1, π) = (2 * e * cos(π))i - (e * sin(π))j = -2ei - 0j = -2ei

Step 3: Normalize the direction vector -3i + 4j

The direction vector is v = -3i + 4j. The magnitude of v is ||v|| = sqrt((-3)^2 + 4^2) = 5

The normalized direction vector is v_hat = v / ||v|| = (-3/5)i + (4/5)j

Step 4: Compute the directional derivative

The directional derivative of g at (1, π) in the direction v is given by the dot product of the gradient of g at (1, π) and the normalized direction vector v_hat.

D_v g(1, π) = ∇g(1, π) . v_hat = -2e * (-3/5) + 0 * (4/5) = 6e/5

So, the directional derivative of g(x, y) = e^(x^2)cos(y) at (1, π) in the direction -3i + 4j is 6e/5.