سوال ۲۸ کنکور ارشد مهندسی کامپیوتر ۱۴۰۴

We have a space curve $\vec{r}(y) = (x(y),\, y,\, z(y))$ parametrized by $y$, whose coordinates satisfy $x^2 + y^3 \sin z + y = 2, \qquad 2x^3 y z - y^2 e^{z} + 1 = 0.$ We want $\dfrac{dz}{dy}$ at the point of the curve.

Step 1 — locate the point. The relevant point is the triple $(x,y,z)$ satisfying BOTH equations. Test $(x,y,z) = (1,1,0)$:

• Equation 1: $1^2 + 1^3 \sin 0 + 1 = 1 + 0 + 1 = 2$ ✓
• Equation 2: $2(1)^3(1)(0) - 1^2 e^{0} + 1 = 0 - 1 + 1 = 0$ ✓

Both hold, so we evaluate at $(x,y,z)=(1,1,0)$, i.e. $y=1$ with $x=1$ and $z=0$.

Step 2 — differentiate along the curve. Treat $x=x(y)$ and $z=z(y)$, and write $x' = \dfrac{dx}{dy}$, $z' = \dfrac{dz}{dy}$.

Differentiate the second equation $2x^3 y z - y^2 e^{z} + 1 = 0$ with respect to $y$: $2\big(3x^2 x'\,yz + x^3 z + x^3 y\,z'\big) - \big(2y\,e^{z} + y^2 e^{z}\,z'\big) = 0.$ At $(1,1,0)$ we have $z=0$ and $e^{z}=1$. The first two terms inside the parentheses both carry a factor $z=0$ and vanish, so $x'$ drops out entirely: $2\big(0 + 0 + (1)(1)z'\big) - \big(2(1)(1) + (1)^2(1)z'\big) = 0$ $2z' - (2 + z') = 0 \;\Longrightarrow\; 2z' - 2 - z' = 0 \;\Longrightarrow\; z' = 2.$

Thus $\frac{dz}{dy}\Big|_{(1,1,0)} = 2.$

Step 3 — (optional) recover $x'$. Differentiating the first equation, $2x x' + 3y^2\sin z + y^3\cos z\,z' + 1 = 0$; at $(1,1,0)$ with $z'=2$ this gives $2x' + 0 + (1)(1)(2) + 1 = 0 \Rightarrow x' = -\tfrac{3}{2}$. This is not needed for the answer.

The value is $2$, which matches the official key (option 1).

Exam trick: Although the point appears written as $(0,1,1)$, the curve is parametrized by $y$, so the actual point on the curve is the triple $(x,y,z)=(1,1,0)$ that satisfies BOTH equations — verify it by substitution before differentiating. At $z=0$ all the $x^3$-type terms in the second equation carry a factor of $z$ and vanish, so $z'$ is found directly from $2z' - 2 - z' = 0$ without solving for $x'$.