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

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 \((-1.05,2.1)\):

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