Q. 13 - Calculus

Question

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

Step-by-Step Solution

Answer

The equation of the hyperplane tangent to the graph of the function is determined as $w-f(a,b,c)=\nabla f(a,b,c)·⟨(x,y,z)-(a,b,c)⟩$

Introduction

The given is the function $(x,y,z)$differentiable at $(a,b,c)$

The objective is to find the equation for the hyperplane tangent 

Step 1

Let $w=f(x, y, z)$ be defined as a function of two variables on an open set including the point $(a, b, c)$ and consider $\Delta w=f(a+\Delta x,b+\Delta y,c+\Delta z)-f(a,b,c)$.

The function $f$ is said to be differentiable if $f_x(a,b,c),f_y(a,b,c)$ and $f_z(a,b,c)$ exists and

$\Delta w=f_x(a,b,c)\Delta x+f_y(a,b,c)\Delta y+f_z(a,b,c)\Delta z+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y+{\epsilon }_3\Delta z$

Here, ${\epsilon }_1, {\epsilon }_2 and {\epsilon }_3$ are functions of $∆x, ∆y and ∆z$, and are zero when $(\Delta x,\Delta y)\to (0,0)$.

Step 2

Substitute $\Delta x=x-a,\Delta y=y-b,\Delta z=z-c$and $\Delta w=f(x,y,z)-f(a,b,c)$in $\left(1\right)$ and delete ${\epsilon }_1\Delta x,{\epsilon }_2\Delta y$ and ${\epsilon }_3\Delta z$ terms.

$∆ w=f_x(a, b, c) ∆ x+f_y(a, b, c)∆ y+f_z(a, b, c) ∆ z +{\epsilon }_1 ∆x+{\epsilon }_2 ∆y+{\epsilon }_3 ∆z$

$=\nabla f(a,b,c)·⟨(x,y,z)-(a,b,c)⟩$

[As  ]

As a result, the tangent plane to the surface equation

$w=f(x, y, z) at (a, b, c)$

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