The directional derivative of a scalar function $f$ at a point $P$ in the direction of a vector $\mathbf{a}$ is given by:
$$D_{\mathbf{a}}f(P) = \nabla f(P) \cdot \mathbf{a}$$
where $\nabla f$ is the gradient of $f$.
In this case, we have $f(x, y, z) = x^2 + 2y^2 + z$ and $\mathbf{a} = 3\hat{\imath} – 4\hat{\jmath}$. Therefore, the directional derivative is:
$$D_{\mathbf{a}}f(P) = \nabla f(P) \cdot \mathbf{a} = (2x + 0y + 1z)(3\hat{\imath} – 4\hat{\jmath}) = 3 – 4 = -1$$
Therefore, the correct answer is $\boxed{-1}$.
To explain each option in brief:
- Option A is incorrect because $D_{\mathbf{a}}f(P) = -1$, not $-4$.
- Option B is incorrect because $D_{\mathbf{a}}f(P) = -1$, not $-2$.
- Option C is incorrect because $D_{\mathbf{a}}f(P) = -1$, not $-1$.
- Option D is correct because $D_{\mathbf{a}}f(P) = -1$.