Find $\frac{\partial y}{\partial z}$ of the surface $g(s,t)=(s^2+2t,s+t,e^{st})$ near $g(1, 1) = (3, 2, e)$.

Question

Consider the surface given by $g(s, t) = (s^2 + 2t, s + t, e^{st})$.

Think of $y$ as a function of $x$ and $z$. Find $\dfrac{\partial y}{\partial z}(3,e)$ near $g(1, 1) = (3, 2, e)$.

really tough question because i cant understand how it is related to implicit functions like what should be my $F(x,y,z)$.

thanks a lot for your help.

Answer 1

Note that $(x,y,z)=\big(s^2+2t,s+t,\exp(st)\big)$ implies $$\text{d}x=2s\,\text{d}s+2\,\text{d}t\,,\tag{1}$$ $$\text{d}y=\text{d}s+\text{d}t\,,\tag{2}$$ and $$\text{d}z=t\exp(st)\,\text{d}s+s\exp(st)\,\text{d}t\,.\tag{3}$$ Thus, the differentials at $(s,t)=(1,1)$ are $$\text{d}x=2\,\text{d}s+2\,\text{d}t\,,$$ $$\text{d}y=\text{d}s+\text{d}t\,,$$ and $$\text{d}z=\text{e}\,\text{d}s+\text{e}\,\text{d}t\,.$$ To compute $\dfrac{\text{d}y}{\text{d}z}(s=1,t=1)$, we require that $\text{d}x$ and $\text{d}z$ be linearly independent, which is not the case here. Therefore, this suggests that $\dfrac{\text{d}y}{\text{d}z}(s=1,t=1)$ may not exist (see a comment below).

Here is a way to compute $\dfrac{\text{d}y}{\text{d}z}(s,t)$ for regular points $(s,t)$ (which you shall learn later that it means $s^2\neq t$). Using (1) and (3), we obtain $$\text{d}s=\frac{s\exp(st)\,\text{d}x-2\,\text{d}z}{2\,(s^2-t)\,\exp(st)}$$ and $$\text{d}t=\frac{2s\,\text{d}z-t\exp(st)\,\text{d}x}{2\,(s^2-t)\,\exp(st)}\,.$$ By (2), we get $$\text{d}y=\text{d}s+\text{d}t=\frac{s-t}{2\,(s^2-t)}\,\text{d}x+\frac{s-1}{(s^2-t)\,\exp(st)}\,\text{d}z\,.$$ Thus, $$\frac{\partial y}{\partial x}(s,t)=\frac{s-t}{2\,(s^2-t)}\text{ and }\frac{\partial y}{\partial z}(s,t)=\frac{s-1}{(s^2-t)\,\exp(st)}\,.$$ When $\sigma^2=\tau$, we see that $$\lim_{(s,t)\to(\sigma,\tau)}\,\frac{\partial y}{\partial x}(s,t)\text{ and }\lim_{(s,t)\to(\sigma,\tau)}\,\frac{\partial y}{\partial z}(s,t)$$ do not exist. Hence, $\dfrac{\partial y}{\partial x}(s,t)$ and $\dfrac{\partial y}{\partial z}(s,t)$ do not exist along the curve $s^2=t$.

As you can see from the surface plot below, the "ridge" of this surface is formed along the curve given by $s^2=t$. The folding of the surface there means that $y$ cannot be expressed as a function of $x$ and $z$ near this curve.

Answer 2

Hint: the cross product $\frac{\partial g}{\partial s} \times \frac{\partial g}{\partial t}$ must be a normal vector to the surface. On the other hand, if you view the surface as defining a function $y = f(x, z)$, then the curve $(x_0, f(x_0, t), t)$ lying along the surface must have tangent vector $(0, \frac{\partial y}{\partial z}, 1)$ which must therefore be perpendicular to this normal vector.