Explaining why the tangent plane at a point $(a, b, f(a, b))$ consist of points $(a, b, f(a, b)) + (h, k, T(h, k))$

Question

This is the exercise 4.5.1 from Calculus and Analysis in Euclidean Space by Shurman et al. The exercise asks us to explain why a tangent plane at point $(a, b, f(a, b))$ of a differentiable function $f$ consists of the points $(a, b, f(a, b)) + (h, k, T(h, k))$, where $T(h, k) = \varphi'(a)h + \psi'(b)k$.

My reasoning is that as the tangent plane at $(a, b, f(a, b))$ is spanned by the tangent lines parallel to the coordinate axes, who themselves are are given by the vectors $(h, b, \varphi'(a)h)$ and $(a, k, \psi'(b)k)$, it follows that all points in the plane can be represented with the linear combination $(a + h, b + k, \varphi'(a)h + \psi'(b)k + f(a, b)) = (a, b, f(a, b)) + (h, k, \varphi'(a)h + \psi'(b)k)$, where $(0, 0, f(a, b))$ is added in order for the plane to touch the graph of $f$ at $(a, b)$.

Edit: $\varphi$ and $\psi$ are, to my understanding from the problem's description, the partial derivatives of $f$ w.r.t. the first and second variable, respectively.

Answer 1

Let $\alpha\colon(-\varepsilon,\varepsilon)\longrightarrow\left\{\bigl(x,y,f(x,y)\bigr)\,\middle|\,x,y\in\Bbb R\right\}$ be a curve such that $\alpha(0)=\bigl(a,b,f(a,b)\bigr)$. Then, for each $t\in(-\varepsilon,\varepsilon)$, $\alpha(t)=\bigl(\alpha_1(t),\alpha_2(t),f\bigl(\alpha_1(t),\alpha_2(t)\bigr)\bigr)$, and therefore\begin{align}\alpha'(0)&=\left(\alpha_1'(0),\alpha_2'(0),\frac{\partial f}{\partial x}\bigl(\alpha_1(0),\alpha_2(0)\bigr)\alpha_1'(0)+\frac{\partial f}{\partial y}\bigl(\alpha_1(0),\alpha_2(0)\bigr)\alpha_2'(0)\right)\\&=\left(\alpha_1'(0),\alpha_2'(0),\frac{\partial f}{\partial x}\bigl(a,b\bigr)\alpha_1'(0)+\frac{\partial f}{\partial y}\bigl(a,b\bigr)\alpha_2'(0)\right),\end{align}which is of the form $\left(h,k,\frac{\partial f}{\partial x}h+\frac{\partial f}{\partial y}k\right)$. So, every tangent vector to the surface at $\bigl(a,b,f(a,b)\bigr)$ belongs to the plane$$\left\{\left(h,k,\frac{\partial f}{\partial x}(a,b)h+\frac{\partial f}{\partial y}(a,b)k\right)\,\middle|\,h,k\in\Bbb R\right\},$$and therefore this plane is the tangent plane that you are looking for.