سوال ۳۳ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

The directional derivative of $f$ at a point $P$ in the direction of a unit vector $\mathbf{u}$ is the dot product of the gradient with that direction: $D_{\mathbf{u}}f(P) = \nabla f(P)\cdot \mathbf{u}.$

Step 1 — Gradient. With $f(x,y) = x^{2} + e^{xy} - 3xy^{2}$, $f_x = 2x + y\,e^{xy} - 3y^{2},\qquad f_y = x\,e^{xy} - 6xy.$

Step 2 — Evaluate at the point. The ordered pair in the (right-to-left) Persian text is $(x,y) = (1,0)$. Then $f_x(1,0) = 2(1) + 0\cdot e^{0} - 3(0)^{2} = 2,$ $f_y(1,0) = 1\cdot e^{0} - 6(1)(0) = 1,$ so $\nabla f(1,0) = (2,\,1)$.

Step 3 — Direction. A unit vector at angle $\tfrac{\pi}{4}$ to the positive $x$-axis is $\mathbf{u} = \left(\cos\tfrac{\pi}{4},\ \sin\tfrac{\pi}{4}\right) = \left(\tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{2}}{2}\right).$

Step 4 — Dot product. $D_{\mathbf{u}}f = (2,1)\cdot\left(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2}\right) = 2\cdot\tfrac{\sqrt{2}}{2} + 1\cdot\tfrac{\sqrt{2}}{2} = \sqrt{2} + \tfrac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}.$

This is option 3.

Exam trick: Because $\mathbf{u}$ is a unit vector, never normalize again — just dot it with $\nabla f$. Reading the point at the wrong order, $(0,1)$, gives $\nabla f = (-2,0)$ and $D_{\mathbf{u}}f = -\sqrt{2}$ (option 2 in magnitude); the answer key fixes the intended point as $(x,y)=(1,0)$.