Q. 12 - Calculus

Question

12. If a function $f(x, y)$ is differentiable at $(a, b)$, explain how to use the gradient $\nabla f(a,b)$ to find the equation of the plane tangent to the graph of $f$ at $(a, b)$.

Step-by-Step Solution

Answer

The equation of the plan tangent to the graph of the function at $(a,b)$ is $z-f(a,b)=\nabla f(a,b)·⟨(x,y)-(a,b)⟩$

Introduction

 The given is the function is $f(x,y)$ is differentiable at the point $(a,b)$

The objective is to find the equation of the plane tangent 

Step 1

Let $f(x, y)$ be defined as a function of two variables on an open set including the point $(a, b)$ and let $\Delta z=f(a+\Delta x,b+\Delta y)-f(a,b)$. The function $f$ is said to be differentiable if $f_x(a, b) and f_y(a, b)$ both exists and

$\Delta z=f_x(a,b)\Delta x+f_y(a,b)\Delta y+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y$

where ${\epsilon }_1$ and ${\epsilon }_2$ are functions of $\Delta x$ and $\Delta y$, and they're both zero when $(\Delta x,\Delta y)\to (0,0)$.

Step 2

Substitute $\Delta x=x-a,\Delta y=y-b$ and $\Delta z=z-f(a,b)$ in $\left(1\right)$ and eliminate ${\epsilon }_1∆ x and {\epsilon }_2∆y$ terms.

$z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)$

$=\nabla f_{+}(a,b)·⟨(x,y)-(a,b)⟩$

As a result, the tangent plane to the surface equation $f(x, y)$ at $(a, b)$ is

$z-f(a,b)=\nabla f(a,b)·⟨(x,y)-(a,b)⟩$