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**Tangent Planes and Linear Approximations**

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Tangent Planes Suppose a surface S has equation π§=π(π₯,π¦), where π has continuous first partial derivatives, and let π( π₯ 0 , π¦ 0 , π§ 0 ) be a point on S. Let πΆ 1 and πΆ 2 be the curves of intersection of S with the planes π¦= π¦ 0 and π₯= π₯ 0 respectively. The direction of the tangent line to the curve πΆ 1 at P is given by π 1 = 1,0, π π₯ ( π₯ 0 , π¦ 0 ) The direction of the tangent line to the curve πΆ 2 at P is given by π 2 = 0,1, π π¦ ( π₯ 0 , π¦ 0 ) The tangent plane to S at P is the plane containing the tangent vectors π 1 and π 2 . Normal to the tangent plane: π§=π 1 Γ π 2 = β π π₯ π₯ 0 , π¦ 0 ,β π π¦ π₯ 0 , π¦ 0 ,1 Plane through P with normal n: β π π₯ π₯ 0 , π¦ 0 (π₯β π₯ 0 ) β π π¦ π₯ 0 , π¦ 0 π¦β π¦ 0 +π§β π§ 0 =0 Rearranging terms: Equation of the tangent plane to the surface π§=π(π₯,π¦) at π( π₯ 0 , π¦ 0 , π§ 0 ) z= π§ 0 + π π₯ π₯ 0 , π¦ 0 (π₯β π₯ 0 ) + π π¦ π₯ 0 , π¦ 0 π¦β π¦ 0

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Tangent Plane Example Find an equation of the tangent plane to the paraboloid π§= 3π¦ 2 + π₯ 2 at (2,1,7) Simplifying: The paraboloid and its tangent plane at P: Zoom in Zoom in

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Linear Approximation The tangent plane to the graph of π§=π(π₯,π¦) at π( π₯ 0 , π¦ 0 , π§ 0 ) is π§=π( π₯ 0 , π¦ 0 )+ π π₯ π₯ 0 , π¦ 0 (π₯β π₯ 0 ) + π π¦ π₯ 0 , π¦ 0 π¦β π¦ 0 if π π₯ and π π¦ are continuous at P. The linear function πΏ(π₯,π¦)=π( π₯ 0 , π¦ 0 )+ π π₯ π₯ 0 , π¦ 0 (π₯β π₯ 0 ) + π π¦ π₯ 0 , π¦ 0 π¦β π¦ 0 is called the linearization of π at (π₯ 0 , π¦ 0 ) The approximation π π₯,π¦ βπΏ π₯,π¦ is called the linear approximation or tangent plane approximation of π at (π₯ 0 , π¦ 0 ) The linear approximation is a good approximation when (π₯,π¦) is near (π₯ 0 , π¦ 0 ) provided that the partial derivatives π π₯ and π π¦ exist and are continuous at (π₯ 0 , π¦ 0 ), that is, provided that the function π is differentiable.

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**Linear Approximation - Example**

Consider the function π π₯,π¦ = π₯+ π 2π¦ . Explain why the function is differentiable at 8,0 and find the linearization πΏ(π₯,π¦) Both partial derivatives are continuous at the point, so π is differentiable. πΏ(π₯,π¦)=π(8,0)+ π π₯ 8,0 (π₯β8) + π π¦ 8,0 π¦ (b) Use the linearization to approximate the function at 7.5, 0.2 . Compare with the actual value π 7.5,0.2 = π 0.4 β2.9986

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Differentials The tangent plane at π π,π, π π,π is an approximation to the function π§=π(π₯,π¦) for π₯,π¦ near (π,π) π(π₯,π¦)βπ(π,π)+ π π₯ π,π (π₯βπ) + π π¦ π,π π¦βπ π(π₯,π¦)βπ π,π β π π₯ π,π (π₯βπ) + π π¦ π,π π¦βπ βπ βπ₯ βπ¦ The quantity π π₯ π,π βπ₯+ π π¦ (π,π)βπ¦ is an approximation to βπ and it represents the change in height of the tangent plane when (π,π) changes to (π+βπ₯, π+βπ¦) Letting βπ₯ and βπ¦ approach zero, yields the following definition: Let π§=π(π₯,π¦) be a differentiable function, then the differential of the function at π,π is ππ= π π₯ π,π ππ₯+ π π¦ π,π ππ¦

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**Differentials Example**

Let π§=5 π₯ 2 + π¦ 2 . Find the differential ππ§ Use the differential to estimate the change in the function, βπ§, when (π₯,π¦) changes from (1,2) to 1.05,2.1 . (a) (b) Actual value:

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**Differentials 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: π=2π₯π¦+2π¦π§+2π₯π§ Differential: ππ= π π₯ ππ₯+ π π¦ ππ¦+ π π§ ππ§ = 2π¦+2π§ ππ₯+ 2π₯+2π§ ππ¦+ 2π¦+2π₯ ππ§ We are given βπ₯ β€0.1, βπ¦ β€0.1, and βπ§ β€0.1 To find the largest error in the surface area we use ππ₯=ππ¦=ππ§=0.1 together with π₯=70, π¦=55 and π§=30. βπβ 2(55)+2(30 )0.1+ 2(70)+2(30) (55)+2(70) 0.1=62 cm 2

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Section 2.9 Linear Approximations and Differentials Math 1231: Single-Variable Calculus.

Section 2.9 Linear Approximations and Differentials Math 1231: Single-Variable Calculus.

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