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Bai giang ham so kha vi va vi phan cua ham nhieu bien

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  • 1. Chapter 8: Functions of Several Variables Section 8.4 Differentials Written by Richard Gill Associate Professor of Mathematics Tidewater Community College, Norfolk Campus, Norfolk, VA With Assistance from a VCCS LearningWare Grant
  • 2. To begin this section on differentials in three variables, we will begin with a review of differentials in two variables. Consider the function y = f(x). Now consider a generic value of x with a tangent to the curve at (x, f(x)). Compare the initial value of x to a value of x that is slightly larger. The slope of the tangent line is: ( x + ∆x, f ( x + ∆x)) dy  = f ' ( x). dx   ∆y dy  It was at this point that we first ( x, f ( x))  saw dx defined as ∆x. rise dy = x + ∆x x Since run dx and since dx is being defined as “run” then dy   becomes “rise” by definition. dx = ∆x For small values of dx, the rise of the tangent line, dy = f ' ( x) dx was used as an approximation for ∆y , the change in the function.
  • 3. Now consider the extension of the differential concept to functions of several variables. Consider an input (x,y) for this function. Its outputs z = f(x,y) create a set of points (x,y,z) that form the surface you see below. The differential dz will have two parts: one part generated by a change in x and the other part z = f(x,y) generated by a change in y. z Consider a new point in the domain generated by a small change in x. Now consider the functional image of the new point. We can use the function to ( x + ∆x, y, f ( x + ∆x, y )) calculate ∆z , the difference ( x, y, f ( x, y )) ∆x between the two z-coordinates. x By subtracting the z-coordinates: y ( x, y ) f ( x + ∆x, y ) − f ( x, y ) = ∆z ( x + ∆x , y )
  • 4. Now consider the extension of the differential concept to functions of several variables. Consider an input (x,y) for this function and its output (x,y,z). The differential dz will have two parts: one part generated by a change in x and the other part z = f(x,y) generated by a change in y. z By subtracting the z-coordinates: f ( x + ∆x, y ) − f ( x, y ) = ∆z Since the y- coordinate is constant, we can use that cross ( x + ∆x, y, f ( x + ∆x, y )) section to draw the tangent to ( x, y, f ( x, y )) ( x, y, f ( x, y )) ∆x Remember that, dz is the change in the height dz x y of the tangent line, and ( x, y ) can be used to estimate ( x + ∆x , y ) the change in z.
  • 5. Since there has been no change in y we can express the differential so far in terms of the change in x: ∂z dz = dx ∂x z = f(x,y) Now track the influence on z when a new point is generated by a change in the y direction. z ( x + ∆x, y + ∆y, f ( x + ∆x, y + ∆y )) Now that the change in z is generated by changes in x and y, ( x + ∆x, y, f ( x + ∆x, y )) we can define the total differential: ( x, y, f ( x, y )) ∆x ∂z ∂z dz = dx + dy or ∆y ∂x ∂y x y dz = f x ( x, y )dx + f y ( x, y )dy ( x + ∆x , y ) ( x, y ) ( x + ∆x, y + ∆y )
  • 6. Example 1. Suppose the graph we have been working with is generated by the equation : z = 3 x 2 + 2 y 2 . Find the total differential for z. Try this on your own first. z = f(x,y) ∂z ∂z z dz = dx + dy ∂x ∂y ( x + ∆x, y + ∆y, f ( x + ∆x, y + ∆y )) dz = 6 x dx + 4 y dy ( x + ∆x, y, f ( x + ∆x, y )) ( x, y, f ( x, y )) ∆x ∆y x y ( x + ∆x , y ) ( x, y ) ( x + ∆x, y + ∆y )
  • 7. If f ( x, y ) = y cos x, evaluate f (2,1) and f (2.1,1.08), Example 2. calculate the total differential dz and compare it to ∆z. Round to the nearest hundredth. Hint: first calculate dx and dy. Solution: dx = 2.1 – 2 = 0.1 and dy = 1.08 – 1 = 0.08 z = y cos x ∂z ∂z dz = dx + dy ∂x ∂y = − y sin x dx + cos x dy ( x, y ) = (2,1) , dx = 0.1, dy = 0.8 → dz = −1sin 2(0.1) + cos 2(0.08) = −0.0909.... − 0.0332..... = −0.124 f (2.1,1.08) = 1.08(cos(2.1)) = −0.545 f (2,1) = 1.08(cos(2.1)) = −0.416 ∆z = f (2.1,1.08) − f (2,1) = 1.08(cos(2.1)) − 1.08(cos(2.1)) dz is a good approximation for ∆z = −0.129
  • 8. Definition of Differentiability For z = f ( x, y ), the function is differentiable at (a, b) if ∆z can be expressed in the form ∆z = f x (a, b)∆x + f y (a, b)∆y + ε 1∆x + ε 2 ∆y where ε 1 and ε 2 → 0 as (∆x, ∆y ) → (0,0). Since dz = f x ( x, y )dx + f y ( x, y )dy, you can consider ε 1∆x and ε 2 ∆y to be the error when we use dz to approximate ∆z. The following theorem is presented If the partial derivatives f x and f y exist in without proof though a region containing (a, b) and are continuous you can usually find the proof in the at (a, b) then f is differentiable at (a, b). appendix of a standard Calculus textbook.
  • 9. Theorem: If a function of x and y is differentiable at (a,b) then it is continuous at (a,b). f ( x, y ) = f ( a, b) Solution: the objective is to show that lim (x, y)→ (a, b) Let z = f ( x, y ) be differentiable at (a, b). Then, by definition ∆z = ( f x (a, b) + ε 1 )∆x + ( f y (a, b) + ε 2 )∆y where ∆x = a − x, ∆y = b − y, and ε 1 → 0 and ε 2 → 0 as ( x, y ) → (a, b). We also know that ∆z = f(a,b)-f(x,y). From above, ∆z = f (a, b) − f ( x, y ) = ( f x (a, b) + ε 1 )∆x + ( f 2 (a, b) + ε 2 )∆y = ( f x (a, b) + ε 1 )(a − x) + ( f 2 (a, b) + ε 2 )(b − y ) Taking the limit as ( x, y ) → (a, b), the above expression goes to 0. f (a, b) − f ( x, y ) = 0 → f (a, b) = f ( x, y ) as ( x, y ) → (a, b) which completes the proof.
  • 10. Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2 ft. These measurements have possible errors in accuracy as laid out in the table below. Complete the table below. Comment on the relationship between dV and ∆V for the indicated errors. ∆h ∆V ∆r dV Solution: 0.1 ft. 0.1 ft. 7.54 cu ft V = f (r , h) = πr h 2 0.01 ft. 0.01 ft. ∂V ∂V .754 cu ft dV = dr + dh ∂r ∂h 0.001 ft. 0.001 ft. .075 cu ft dV = 2πrh dr + πr 2 dh ∆r = ∆h = 0.1 → dV = 2π (2)(5)(0.1) + π (2 2 )(0.1) = 2π + 0.4π = 2.4π ≈ 7.540 ft 3 ∆r = ∆h = 0.01 → dV = 2π (2)(5)(0.01) + π (2 2 )(0.01) = 0.2π + .04π = 0.24π ≈ .754 ft 3 ∆r = ∆h = 0.001 → dV = 2π (2)(5)(0.001) + π (2 2 )(0.001) = 0.024π ≈ 0.075 ft 3
  • 11. Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2 ft. These measurements have possible errors in accuracy as laid out in the table below. Complete the table below. Comment on the relationship between dV and ∆V for the indicated errors. ∆h ∆V ∆r dV Solution: 0.1 ft. 0.1 ft. 7.54 cu ft 7.826 cu ft V = f (r , h) = πr h 2 0.01 ft. 0.01 ft. ∂V ∂V .754 cu ft 0.757 cu ft dV = dr + dh ∂r ∂h 0.001 ft. 0.001 ft. .075 cu ft 0.075 cu ft dV = 2πrh dr + πr 2 dh ∆V = f (r + ∆r , h + ∆h) − f (r , h) = f (2.1, 5.1) − f (2,5) = π (2.1) 2 (5.1) − π (2) 2 (5) ≈ 7.826 ∆V = f (2.01, 5.01) − f (2,5) = π (2.01) 2 (5.01) − π (2) 2 (5) ≈ 0.757 ∆V = f (2.001, 5.001) − f (2,5) = π (2.001) 2 (5.001) − π (2) 2 (5) ≈ 0.075 The approximation of dV for ∆V gets better as ∆r and ∆h get smaller.
  • 12. Example 4. A right circular cylinder is constructed with a height of 40 cm. and a radius of 25 cm. What is the relative error and the percent error in the surface area if the possible error in the measurement of each dimension is ½ cm. Solution: If the measurements are correct the surface area will be A = 2πr 2 + 2πrh The total differential will generate an estimate for = 2π (25) 2 + 2π (25)(40) the possible error. = 1250π + 800π = 2050π A = 2πr 2 + 2πrh ∂A ∂A dh = (4πr + 2πh)dr + 2πr dh dA = dr + ∂r ∂h = (4π (25) + 2π (40))(.5) + 2π (25)(.5) = (100π + 80π )(.5) + 50π (.5) = 115π dA 115π = = 0.056 The relative error is A 2050π The percent error is 5.6%
  • 13. This concludes the material for Lesson 8.4. This also concludes the lessons on Blackboard and the material for the semester. As was the case in previous Bb lessons, we have posted three sets of exercises on Bb. Any exercises worked correctly will add to your thinkwell exercise totals. The practice exam should be up and running on Bb. Good luck preparing for the exam and congratulations on getting to the end of one of the toughest courses in the curriculum.

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