2Tangent PlanesSuppose a surface S has equation 𝑧=𝑓(𝑥,𝑦), where 𝑓 has continuous first partial derivatives, and let 𝑃( 𝑥 0 , 𝑦 0 , 𝑧 0 ) be a point on S.Let 𝐶 1 and 𝐶 2 be the curves of intersection of S with the planes 𝑦= 𝑦 0 and 𝑥= 𝑥 0 respectively.The direction of the tangent line to the curve 𝐶 1 at P is given by 𝐓 1 = 1,0, 𝑓 𝑥 ( 𝑥 0 , 𝑦 0 )The direction of the tangent line to the curve 𝐶 2 at P is given by 𝐓 2 = 0,1, 𝑓 𝑦 ( 𝑥 0 , 𝑦 0 )The tangent plane to S at P is the plane containing the tangent vectors 𝐓 1 and 𝐓 2 .Normal to the tangent plane: 𝐧=𝐓 1 × 𝐓 2 = − 𝑓 𝑥 𝑥 0 , 𝑦 0 ,− 𝑓 𝑦 𝑥 0 , 𝑦 0 ,1Plane through P with normal n:− 𝑓 𝑥 𝑥 0 , 𝑦 0 (𝑥− 𝑥 0 ) − 𝑓 𝑦 𝑥 0 , 𝑦 0 𝑦− 𝑦 0 +𝑧− 𝑧 0 =0Rearranging terms:Equation of the tangent plane to the surface 𝑧=𝑓(𝑥,𝑦) at 𝑃( 𝑥 0 , 𝑦 0 , 𝑧 0 )z= 𝑧 0 + 𝑓 𝑥 𝑥 0 , 𝑦 0 (𝑥− 𝑥 0 ) + 𝑓 𝑦 𝑥 0 , 𝑦 0 𝑦− 𝑦 0
3Tangent Plane ExampleFind an equation of the tangent plane to the paraboloid 𝑧= 3𝑦 2 + 𝑥 2 at (2,1,7)Simplifying:The paraboloid and its tangent plane at P:Zoom inZoom in
4Linear ApproximationThe tangent plane to the graph of 𝑧=𝑓(𝑥,𝑦) at 𝑃( 𝑥 0 , 𝑦 0 , 𝑧 0 ) is𝑧=𝑓( 𝑥 0 , 𝑦 0 )+ 𝑓 𝑥 𝑥 0 , 𝑦 0 (𝑥− 𝑥 0 ) + 𝑓 𝑦 𝑥 0 , 𝑦 0 𝑦− 𝑦 0if 𝑓 𝑥 and 𝑓 𝑦 are continuous at P.The linear function𝐿(𝑥,𝑦)=𝑓( 𝑥 0 , 𝑦 0 )+ 𝑓 𝑥 𝑥 0 , 𝑦 0 (𝑥− 𝑥 0 ) + 𝑓 𝑦 𝑥 0 , 𝑦 0 𝑦− 𝑦 0is called the linearization of 𝑓 at (𝑥 0 , 𝑦 0 )The approximation 𝑓 𝑥,𝑦 ≈𝐿 𝑥,𝑦 is called the linear approximation or tangent plane approximation of 𝑓 at (𝑥 0 , 𝑦 0 )The linear approximation is a good approximation when (𝑥,𝑦) is near (𝑥 0 , 𝑦 0 ) provided that the partial derivatives 𝑓 𝑥 and 𝑓 𝑦 exist and are continuous at (𝑥 0 , 𝑦 0 ), that is, provided that the function 𝑓 is differentiable.
5Linear Approximation - Example Consider the function 𝑓 𝑥,𝑦 = 𝑥+ 𝑒 2𝑦 .Explain why the function is differentiable at 8,0 and find thelinearization 𝐿(𝑥,𝑦)Both partial derivatives are continuous at the point, so 𝑓 is differentiable.𝐿(𝑥,𝑦)=𝑓(8,0)+ 𝑓 𝑥 8,0 (𝑥−8) + 𝑓 𝑦 8,0 𝑦(b) Use the linearization to approximate the function at 7.5, 0.2 .Compare with the actual value 𝑓 7.5,0.2 = 𝑒 0.4 ≈2.9986
6DifferentialsThe tangent plane at 𝑃 𝑎,𝑏, 𝑓 𝑎,𝑏 is an approximation to the function 𝑧=𝑓(𝑥,𝑦) for 𝑥,𝑦 near (𝑎,𝑏)𝑓(𝑥,𝑦)≈𝑓(𝑎,𝑏)+ 𝑓 𝑥 𝑎,𝑏 (𝑥−𝑎) + 𝑓 𝑦 𝑎,𝑏 𝑦−𝑏𝑓(𝑥,𝑦)−𝑓 𝑎,𝑏 ≈ 𝑓 𝑥 𝑎,𝑏 (𝑥−𝑎) + 𝑓 𝑦 𝑎,𝑏 𝑦−𝑏∆𝑓∆𝑥∆𝑦The quantity 𝑓 𝑥 𝑎,𝑏 ∆𝑥+ 𝑓 𝑦 (𝑎,𝑏)∆𝑦 is an approximation to ∆𝑓 and it represents the change in height of the tangent plane when (𝑎,𝑏) changes to (𝑎+∆𝑥, 𝑏+∆𝑦)Letting ∆𝑥 and ∆𝑦 approach zero, yields the following definition:Let 𝑧=𝑓(𝑥,𝑦) be a differentiable function, then the differential of the function at 𝑎,𝑏 is𝑑𝑓= 𝑓 𝑥 𝑎,𝑏 𝑑𝑥+ 𝑓 𝑦 𝑎,𝑏 𝑑𝑦
7Differentials Example Let 𝑧=5 𝑥 2 + 𝑦 2 .Find the differential 𝑑𝑧Use the differential to estimate the change in the function, ∆𝑧, when (𝑥,𝑦) changes from (1,2) to 1.05,2.1 .(a)(b)Actual value:
8Differentials 3D -Example The dimensions of a rectangular box are measured to be 70 cm, 55 cm and 30 cm and each measurement is correct to within 0.1 cm.Use differentials to estimate the largest possible error when the surface area of the box is calculated from these measurements.Let x, y and z be the dimensions of the box.Surface Area: 𝑆=2𝑥𝑦+2𝑦𝑧+2𝑥𝑧Differential:𝑑𝑆= 𝑆 𝑥 𝑑𝑥+ 𝑆 𝑦 𝑑𝑦+ 𝑆 𝑧 𝑑𝑧= 2𝑦+2𝑧 𝑑𝑥+ 2𝑥+2𝑧 𝑑𝑦+ 2𝑦+2𝑥 𝑑𝑧We are given ∆𝑥 ≤0.1, ∆𝑦 ≤0.1, and ∆𝑧 ≤0.1To find the largest error in the surface area we use 𝑑𝑥=𝑑𝑦=𝑑𝑧=0.1together with 𝑥=70, 𝑦=55 and 𝑧=30.∆𝑆≈ 2(55)+2(30 )0.1+ 2(70)+2(30) (55)+2(70) 0.1=62 cm 2