14.4 Tangent planes and Linear Approximation

Tangent vector

• Much like calc 1’s tangent line of a function, but now we are doing it in 3 dimensions

• Definniton

  • $z=f(a,b)+{\partial }_{x}f(a,b)(x-a)+{\partial }_{y}f(a,b)(y-b)$

    • This is simple, this is like the other form of the tangent line, but make sure to only use (a,b) NOT (x,y)

Linear Approximation:

• much like newtons method linera apporximation in calc 1, we use the tangent plane to apporxmiate the curve close to the point of the plane

Differential Geometry

• Its using the tangent plane to under stand properties of other surfaces

  • $df=({\partial }_{x}f(a,b))\Delta x+({\partial }_{y}f(a,b)\Delta y)$
    • to get the differential of f at $(a,b)$ is defined by
      • $df{|}_{(a,b)}=({\partial }_{x}f(a,b))dx+({\partial }_{y}f(a,b))dy$

• If there are more than 2 varibles

  • linear approxmitation
    • $w(x,y,z)\approx {\partial }_{x}w(a,b,c)(x-a)+{\partial }_{y}w(a,b,c)(y-b)+{\partial }_{z}w(a,b,c)(z-c)$

Examples

• Find the equation of the tangent plane to the surface $z=\sqrt{4-x^2-2y^2}$ at the point $(1,-1,1)$

• Linear Approxmitaion

  • Let $f(x,y)=x{e}^{xy}$ use the linear approximation to find an approximate value of $f(1.01,-0.02)$

$$\begin{array}{r}\text{using point}f(1,0)\\ f(1.01,-0.02)-f(1,0)\approx {\partial }_{x}f(1,0)(1.01-1)+{\partial }_{y}f(1,0)(-0.02-0)\\ {\partial }_{x}f=e^{xy}+xy{e}^{xy}⟹{\partial }_{x}f(1,0)=1\text{ using product rule}\\ {\partial }_{y}f=x^2{e}^{xy}⟹{\partial }_{y}f(1,0)=1\\ f(1.01,-0.02)-f(1,0)\approx (0.01)-(0.02)\end{array}$$

• Differentials

  • For $f(x,y)=2\sin x+3\cos y$ find $f\left(\frac{\pi }{2},0\right)$ and use the differential df to find the approximate value of $f\left(\frac{29\pi }{60},\frac{\pi }{50}\right)$