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Wize University Calculus 2 Textbook > Multivariable Functions

Tangent plane & Linear Approximation

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Tangent Plane

Given a surface SS with equation z=f(x,y)z=f\left(x,y\right), the tangent plane at the point P(a,b,c)P\left(a,b,c\right) on the surface is the plane that contains all tangent lines at the point PP, it is defined as
z=c+fx(a,b)(x−a)+fy(a,b)(y−b)z=c+f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left(y-b\right)
or
z=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)z=f\left(a,b\right)+f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left(y-b\right)

Example
Find the tangent plane to the curve z=x2y−3x+y2z=x^2y-3x+y^2 at the point (1, 1, −1)\left(1,\ 1,\ -1\right).
Let f(x,y)=z=x2y−3x+y2f\left(x,y\right)=z=x^2y-3x+y^2
  • fx=2xy−3 → fx(1,1)=−1f_x=2xy-3\ \to\ f_x\left(1,1\right)=-1
  • fy=x2+2y → fy(1,1)=3f_y=x^2+2y\ \to\ f_y\left(1,1\right)=3
So, using our equation for tangent planes, we get:
z=−1+(−1)(x−1)+(3)(y−1)z=-1+\left(-1\right)\left(x-1\right)+\left(3\right)\left(y-1\right)
z=−x+3y−3z=-x+3y-3

Practice: Tangent Plane

Find an equation of the plane tangent to the curve f(x,y,z)=x2+y2−4z2f(x,y,z)=x^2 +y^2 - 4z^2 at the point P(1,1,0)P\left(1,1,0\right) .
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Linear Approximations

Given a surface defined by the function z=f(x, y)z=f\left(x,\ y\right), the linearization of ff at the point (a,b)\left(a,b\right) is L(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)L\left(x,y\right)=f\left(a,b\right)+f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left(y-b\right)

The linear approximation or tangent plane approximation at that point is f(x,y)≈f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)f\left(x,y\right)\approx f\left(a,b\right)+f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left(y-b\right)

Example
Given that f(x,y)=eycos⁡(xy)f\left(x,y\right)=e^y\cos\left(xy\right), use linear approximation to estimate f(0.1, −0.1)f\left(0.1,\ -0.1\right)
Find the first partial derivatives:
  • fx=−yeysin⁡(xy)f_x=-ye^y\sin\left(xy\right)
  • fy=eycos⁡(xy)−xeysin⁡(xy)f_y=e^y\cos\left(xy\right)-xe^y\sin\left(xy\right)
The linearization at the point (0,0)\left(0,0\right) is
L(x,y)L\left(x,y\right)
=f(0,0)+fx(0,0)(x−0)+fy(0,0)(y−0)=f\left(0,0\right)+f_x\left(0,0\right)\left(x-0\right)+f_y\left(0,0\right)\left(y-0\right)
=1+y=1+y

Therefore, the linear approximation of f(0,1,−0,1)≈1+(−0.1)=0.9f\left(0,1,-0,1\right)\approx1+\left(-0.1\right)=0.9

Practice: Linear approximation

Use linear approximation to estimate the value of function f(x,y)=x2exyf\left(x,y\right)=x^2e^{xy} at the point (1.01, 1.01).