Tangent Planes and Linear Approximation

Tangent Planes

An equation of the plane tangent to the surface z=f(x,y)z=f(x,y) at the point (a,b,f(a,b))(a,b,f(a,b)) is z=fx(a,b) (x−a)+fy(a,b) (y−b)+f(a,b).\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=3x2+3y2z=3x^2+3y^2 at (1,1,6)(1,1,6).

Solution

Since the partial derivatives are fx=6xf_x=6x and fy=6yf_y=6y, the tangent plane is z=6(x−1)+6(y−1)+6=6(x+y−1)\ans{\begin{aligned} z &= 6(x-1) + 6(y-1) + 6\ &= 6(x+y-1) \end{aligned}}

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)

Surface with tangent plane attached

  1. Find the tangent plane to z=4−2x2−y2z=4-2x^2-y^2 at the point (2,2,−8)(2,2,−8).

    2. z=−8(x−2)−4(y−2)−8z=-8(x-2) - 4(y-2) - 8

  2. Find the tangent plane to z=exyz=e^{xy} at the point (1,0,1)(1,0,1).

    3. z=y+1z=y+1

  3. Find the tangent plane to z=ln⁡(1+xy)z=\ln (1+xy) at the point (1,2,ln⁡3)(1,2,\ln 3).

    4. z=23(x−1)+13(y−2)+ln⁡3z=\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)=5x2+y2f(x,y) = \dfrac{5}{x^2+y^2}, use the linear approximation at the point (−1,2,1)(−1,2,1) to estimate the value of f(−1.05,2.1)f(−1.05,2.1).

Solution

Find the tangent plane to ff at (−1,2,1)(−1,2,1): fx=−80x(x2+y2)2 and fy=−10y(x2+y2)2fx(−1,2)=1025=25 and fy(−1,2)=−2025=−45L(x,y)=fx(a,b) (x−a)+fy(a,b) (y−b)+f(a,b)=25(x+1)−45(y−2)+1\begin{aligned} 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{aligned}

Then use this tangent plane to estimate the value of the function at (−1.05,2.1)(−1.05,2.1): L(−1.05,2.1)=0.9≈f(−1.05,2.1)\ans{L(-1.05,2.1) = 0.9 \approx f(-1.05,2.1)}