Q. 12. - Calculus

Question

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

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

Step 1: Given information

Let, $f(x, y)$is a function of two variables defined on an open set containing the point $(a, b)$

Let $\Delta z=f(a+\Delta x,b+\Delta y)-f(a,b)$

 If both $f_x(a,b)$ and $f_y(a,b)$ exist, the function $f$ is said to be differentiable and

$\Delta z=f_x(a,b)\Delta x+f_y(a,b)\Delta y+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y\dots \dots \text{ (1) }$.where ${\epsilon }_1$ and${\epsilon }_2$  are functions of $\Delta x$and $\Delta y$and

both are zero when $(\Delta x,\Delta y)\to (0,0)$

Step 2: The objective is to find the equation of the plane tangent to the graph of f ( x ,   y )   at ( a ,   b )  

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

$$\begin{gathered} z-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b) \\ \Rightarrow ∆z=\left(f_x\right(a,b),f_y(a,b\left)\right)·\left(\right(x-a),(y-b\left)\right) \\ =\nabla f(a,b)·\left(\right(x,y)-(a,b\left)\right) \end{gathered}$$

Hence, the equation of tangent plane to the surface $f(x, y)$at $(a, b)$is:

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