$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Tangent Planes and Linear Approximation

## Tangent Planes

An equation of the plane tangent to the surface $$z=f(x,y)$$ at the point $$(a,b,f(a,b))$$ is $\ans{z=f_x (a,b)\ (x-a) + f_y (a,b)\ (y-b) + f(a,b).}$

Find the tangent plane to the surface $$z=3x^2+3y^2$$ at $$(1,1,6)$$.

#### Solution

Since the partial derivatives are $$f_x = 6x$$ and $$f_y = 6y$$, the tangent plane is \ans{\begin{align} z &= 6(x-1) + 6(y-1) + 6\\ &= 6(x+y-1) \end{align}}

We can graph this surface and tangent plane with Matlab:

        	>> hold on
>> fsurf(@(x,y) 3*x.^2+3*y.^2)
>> fsurf(@(x,y) 6.*x+6.*y-6)


#### Try it yourself:

(click on a problem to show/hide its answer)

1. Find the tangent plane to $$z=4-2x^2-y^2$$ at the point $$(2,2,-8)$$.
2. $$z=-8(x-2) - 4(y-2) - 8$$

3. Find the tangent plane to $$z=e^{xy}$$ at the point $$(1,0,1)$$.
4. $$z=y+1$$

5. Find the tangent plane to $$z=\ln (1+xy)$$ at the point $$(1,2,\ln 3)$$.
6. $$z=\dfrac{2}{3}(x-1) + \dfrac{1}{3}(y-2) + \ln 3$$

## Linear Approximation

Just as we can use tangent lines to approximate functions of one variable, we can use the tangent plane as a linear approximation for a function of two variables.

If $$f(x,y) = \dfrac{5}{x^2+y^2}$$, use the linear approximation at the point $$(-1,2,1)$$ to estimate the value of $$f(-1.05,2.1)$$.

#### Solution

Find the tangent plane to $$f$$ at $$(-1,2,1)$$:

\begin{align} f_x &= \dfrac{-80x}{(x^2+y^2)^2} \textrm{ and } f_y = \dfrac{-10y}{(x^2+y^2)^2}\\ f_x (-1,2) &= \dfrac{10}{25} = \dfrac{2}{5} \textrm{ and } f_y (-1,2) = \dfrac{-20}{25} = -\dfrac{4}{5}\\ \\ L(x,y) &= f_x (a,b)\ (x-a) + f_y (a,b)\ (y-b) + f(a,b)\\ &= \dfrac{2}{5}(x+1) - \dfrac{4}{5}(y-2) + 1 \end{align}

Then use this tangent plane to estimate the value of the function at $$(-1.05,2.1)$$:

$\ans{L(-1.05,2.1) = 0.9 \approx f(-1.05,2.1)}$