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# Lesson20 Tangent Planes Slides+Notes

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### Lesson20 Tangent Planes Slides+Notes

1. 1. Lesson 20 (Section 15.4) Tangent Planes Math 20 November 5, 2007 Announcements Problem Set 7 on the website. Due November 7. No class November 12. Yes class November 21. OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323) Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
2. 2. Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
3. 3. Tangent Lines in one variable
4. 4. Summary Fact The tangent line to y = f (x) through the point (x0 , y0 ) has equation y = f (x0 ) + f (x0 )(x − x0 )
5. 5. Summary Fact The tangent line to y = f (x) through the point (x0 , y0 ) has equation y = f (x0 ) + f (x0 )(x − x0 ) The expression f (x0 ) is a number, not a function! This is the best linear approximation to f near x0 . This is the ﬁrst-degree Taylor polynomial for f .
6. 6. Example Example √ Find the equation for the tangent line to y = x at x = 4.
7. 7. Example Example √ Find the equation for the tangent line to y = x at x = 4. Solution We have dy 1 1 1 =√ =√= dx 4 2x 24 x=4 x=4 So the tangent line has equation y = 2 + 4 (x − 4) = 1 x + 1 1 4
8. 8. Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
9. 9. Recall Any line in Rn can be described by a point a and a direction v and given parametrically by the equation x = a + tv
10. 10. Last time we diﬀerentiated the curve x → (x, y0 , f (x, y0 )) at x = x0 and said that was a line tangent to the graph at (x0 , y0 ).
11. 11. Example Let f = f (x, y ) = 4 − x 2 − 2y 4 . Look at the point P = (1, 1, 1) on the graph of f . 0 -10 z -20 -30 -2 -2 -1 -1 0 0 y x 1 1
12. 12. There are two interesting curves going through the point P: x → (x, 1, f (x, 1)) y → (1, y , f (1, y )) Each of these is a one-variable function, so makes a curve, and has a slope!
13. 13. There are two interesting curves going through the point P: x → (x, 1, f (x, 1)) y → (1, y , f (1, y )) Each of these is a one-variable function, so makes a curve, and has a slope! d d 2 − x2 = −2x|x=1 = −2. f (x, 1) = dx dx x=1 x−1
14. 14. Last time we diﬀerentiated the curve x → (x, y0 , f (x, y0 )) at x = x0 and said that was a line tangent to the graph at (x0 , y0 ). That vector is (1, 0, f1 (x0 , y0 )).
15. 15. Last time we diﬀerentiated the curve x → (x, y0 , f (x, y0 )) at x = x0 and said that was a line tangent to the graph at (x0 , y0 ). That vector is (1, 0, f1 (x0 , y0 )). So a line tangent to the graph is given by (x, y , z) = (x0 , y0 , z0 ) + t(1, 0, f1 (x0 , y0 ))
16. 16. There are two interesting curves going through the point P: x → (x, 1, f (x, 1)) y → (1, y , f (1, y )) Each of these is a one-variable function, so makes a curve, and has a slope! d d 2 − x2 = −2x|x=1 = −2. f (x, 1) = dx dx x=1 x−1 d d 3 − 2y 4 = −8y 3 = −8. f (1, y ) = y =1 dy dx y =1 y =1
17. 17. Last time we diﬀerentiated the curve x → (x, y0 , f (x, y0 )) at x = x0 and said that was a line tangent to the graph at (x0 , y0 ). That vector is (1, 0, f1 (x0 , y0 )). So a line tangent to the graph is given by (x, y , z) = (x0 , y0 , z0 ) + t(1, 0, f1 (x0 , y0 )) Another line is (x, y , z) = (x0 , y0 , z0 ) + t(0, 1, f2 (x0 , y0 ))
18. 18. There are two interesting curves going through the point P: x → (x, 1, f (x, 1)) y → (1, y , f (1, y )) Each of these is a one-variable function, so makes a curve, and has a slope! d d 2 − x2 = −2x|x=1 = −2. f (x, 1) = dx dx x=1 x−1 d d 3 − 2y 4 = −8y 3 = −8. f (1, y ) = y =1 dy dx y =1 y =1 We see that the tangent plane is spanned by these two vectors/slopes.
19. 19. Summary Let f a function of n variables diﬀerentiable at (a1 , a2 , . . . , an ). Then the line given by (x1 , x2 , . . . , xn ) = f (a1 , a2 , . . . , an )+t(0, . . . , 1 , . . . , 0, fi (a1 , a2 , . . . , an )) i is tangent to the graph of f at (a1 , a2 , . . . , an ).
20. 20. Example Find the equations of two lines tangent to z = xy 2 at the point (2, 1, 2).
21. 21. Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
22. 22. Recall Deﬁnition A plane (in three-dimensional space) through a that is orthogonal to a vector p = 0 is the set of all points x satisfying p · (x − a) = 0.
23. 23. Recall Deﬁnition A plane (in three-dimensional space) through a that is orthogonal to a vector p = 0 is the set of all points x satisfying p · (x − a) = 0. Question Given a function and a point on the graph of the function, how do we ﬁnd the equation of the tangent plane?
24. 24. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )).
25. 25. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )). Then p must satisfy p · (1, 0, f1 (x0 , y0 )) = 0 p · (0, 1, f2 (x0 , y0 )) = 0
26. 26. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )). Then p must satisfy p · (1, 0, f1 (x0 , y0 )) = 0 p · (0, 1, f2 (x0 , y0 )) = 0 A solution is to let p1 = f1 (x0 , y0 ), p2 = f2 (x0 , y0 ), p3 = −1.
27. 27. Summary Fact (tangent planes in two variables) The tangent plane to z = f (x, y ) through (x0 , y0 , z0 = f (x0 , y0 )) has normal vector (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0
28. 28. Summary Fact (tangent planes in two variables) The tangent plane to z = f (x, y ) through (x0 , y0 , z0 = f (x0 , y0 )) has normal vector (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0 or z = f (x0 , y0 ) + f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 )
29. 29. Summary Fact (tangent planes in two variables) The tangent plane to z = f (x, y ) through (x0 , y0 , z0 = f (x0 , y0 )) has normal vector (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0 or z = f (x0 , y0 ) + f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) This is the best linear approximation to f near (x0 , y0 ). is is the ﬁrst-degree Taylor polynomial (in two variables) for f .
30. 30. Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
31. 31. Example The number of units of output per day at a factory is −1/2 1 −2 9 x + y −2 P(x, y ) = 150 , 10 10 where x denotes capital investment (in units of \$1000), and y denotes the total number of hours (in units of 10) the work force is employed per day. Suppose that currently, capital investment is \$50,000 and the total number of working hours per day is 500. Estimate the change in output if capital investment is increased by \$5000 and the number of working hours is decreased by 10 per day.
32. 32. Example The number of units of output per day at a factory is −1/2 1 −2 9 x + y −2 P(x, y ) = 150 , 10 10 where x denotes capital investment (in units of \$1000), and y denotes the total number of hours (in units of 10) the work force is employed per day. Suppose that currently, capital investment is \$50,000 and the total number of working hours per day is 500. Estimate the change in output if capital investment is increased by \$5000 and the number of working hours is decreased by 10 per day.
33. 33. Solution −3/2 −2 ∂P 1 1 −2 9 x + y −2 x −3 (x, y ) = 150 − ∂x 2 10 10 10 −3/2 1 −2 9 x + y −2 x −3 = 15 10 10 ∂P (50, 50) = 50 ∂x
34. 34. Solution −3/2 −2 ∂P 1 1 −2 9 x + y −2 x −3 (x, y ) = 150 − ∂x 2 10 10 10 −3/2 1 −2 9 x + y −2 x −3 = 15 10 10 ∂P (50, 50) = 50 ∂x −3/2 ∂P 1 1 −2 9 9 x + y −2 (−2)y −3 (x, y ) = 150 − ∂y 2 10 10 10 −3/2 1 −2 9 x + y −2 x −3 = 15 10 10 ∂P (50, 50) = 135 ∂y So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50)
35. 35. Solution, continued So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50) If ∆x = 5 and ∆y = −1, then L = 7500 + 15 · 5 + 135 · (−1) = 7440
36. 36. Solution, continued So the linear approximation is L = 7500 + 15(x − 50) + 135(y − 50) If ∆x = 5 and ∆y = −1, then L = 7500 + 15 · 5 + 135 · (−1) = 7440 The actual value is P(55, 49) ≈ 7427 13 ≈ 1.75% So we are oﬀ by 7427
37. 37. Contour plot of P 100 80 60 40 20 0 0 20 40 60 80 100
38. 38. Contour plot of L 100 80 60 40 20 0 0 20 40 60 80 100
39. 39. Contour plots, superimposed 100 80 60 40 20 0 0 20 40 60 80 100
40. 40. Animation of P and its linear approximation at (50, 50)