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Lesson20   Tangent Planes Slides+Notes
 

Lesson20 Tangent Planes Slides+Notes

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    Lesson20   Tangent Planes Slides+Notes Lesson20 Tangent Planes Slides+Notes Presentation Transcript

    • 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)
    • Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
    • Tangent Lines in one variable
    • Summary Fact The tangent line to y = f (x) through the point (x0 , y0 ) has equation y = f (x0 ) + f (x0 )(x − x0 )
    • 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 .
    • Example Example √ Find the equation for the tangent line to y = x at x = 4.
    • 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
    • Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
    • 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
    • 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 ).
    • 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
    • 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!
    • 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
    • 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 )).
    • 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 ))
    • 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
    • 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 ))
    • 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.
    • 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 ).
    • Example Find the equations of two lines tangent to z = xy 2 at the point (2, 1, 2).
    • Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
    • 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.
    • 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?
    • Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )).
    • 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
    • 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.
    • 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
    • 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 )
    • 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 .
    • Outline Tangent Lines in one variable Tangent lines in two or more variables Tangent planes in two variables One more example
    • 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.
    • 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.
    • 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
    • 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)
    • 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
    • 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
    • Contour plot of P 100 80 60 40 20 0 0 20 40 60 80 100
    • Contour plot of L 100 80 60 40 20 0 0 20 40 60 80 100
    • Contour plots, superimposed 100 80 60 40 20 0 0 20 40 60 80 100
    • Animation of P and its linear approximation at (50, 50)