Chapter 14 Review Exercises copy - sin ! 1 x ! 1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sin ! 1 x ! 1 " x " 1 tan ! 1 y f (x,y)=sin ! 1 x +tan ! 1 y (x,y)| ! 1 " x " 1 { } D = (x,y,z)| z # x 2 + y 2 { } z = x 2 + y 2 z = f (x,y)=1 ! x 2 ! y 2 (0,0,1). z = f (x,y)= x 2 + y 2 ! 1 z # 1= x 2 + y 2 ! z 2 k = e ! c = e ! x 2 + y 2 ( ) ! ln k = c = x 2 + y 2 1. The domain of is while the domain of is all real numbers, so the domain of is . 2. , the points on and above the paraboloid . 3. , a paraboloid with vertex 4. , so and . Thus the graph is the upper half of a hyperboloid of one sheet. 5. Let be the level curves. Then , so we have a family of concentric circles. 1 Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; Review: Exercises k = x 2 +4 y 4( y ! k /4)= ! x 2 (0,k/4). f f 1,1 ( ) lim x,y ( ) $ 1,1 ( ) 2 xy x 2 +2 y 2 = 2(1)(1) 1 2 +2(1) 2 = 2 3 . 6. or , a family of parabolas with vertex at 7. 8. 9. is a rational function, so it is continuous on its domain. Since is defined at , we use direct substitution to evaluate the limit: 2 k = x 2 +4 y 4( y ! k /4)= ! x 2 (0,k/4). f f 1,1 ( ) lim x,y ( ) $ 1,1 ( ) 2 xy x 2 +2 y 2 = 2(1)(1) 1 2 +2(1) 2 = 2 3 . 6. or , a family of parabolas with vertex at 7. 8. 9. is a rational function, so it is continuous on its domain. Since is defined at , we use direct substitution to evaluate the limit: 2 Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; Review: Exercises x,y ( ) $ 0,0 ( ) x ! f (x,0)=0/ x 2 =0 x % f (x,y) $ f (x,x)=2 x 2 / 3 x 2 ( ) = 2 3 x,y ( ) $ 0,0 ( ) x = y f (x,y) $ 2 3 T x (6,4)=lim h $ T (6+h,4) ! T (6,4) h T x (6,4) h = & 2 T x (6,4) T (8,4) ! T (6,4) 2 = 86 ! 80 2 =3 T x (6,4) T (4,4) ! T (6,4) ! 2 = 72 ! 80 ! 2 =4 T x (6,4) 3.5 ( C/m. T y 6,4 ( ) =lim h $ T (6,4+ h ) ! T (6,4) h h = & 2 T y (6,4) T (6,6) ! T (6,4) 2 = 75 ! 80 2 = ! 2.5 T y (6,4) T (6,2) ! T (6,4) ! 2 = 87 ! 80 ! 2 = ! 3.5 T y (6,4) ! 3.0 ( C/m. u = 1 2 , 1 2 D u T (6,4)= ) T (6,4) * u = T x (6,4) 1 2 + T y (6,4) 1 2 D u T (6,4) (3.5) 1 2 +( ! 3.0) 1 2 = 1 2 2 0.35 (6,4) u 0.35 ( C/m. D u T (6,4)=lim h $ T 6+ h 1 2 ,4+ h 1 2 ! T (6,4) h h = & 2 2 . D u T (6,4) T (8,6) ! T (6,4) 2 2 = 80 ! 80 2 2 =0 D u T (6,4) T (4,2) ! T (6,4) ! 2 2 = 74 ! 80 ! 2 2 = 3 2 D u T (6,4) 3 2 2 1.1 ( C/m. T xy (x,y)= + + y T x (x,y) =lim h $ T x (x,y+ h ) ! T x (x,y) h T xy (6,4)=lim h $ T x (6,4+ h ) ! T x (6,4) h 10. As along the axis, for , so along this line. But , so as along the line , . Thus the limit doesn't exist. 11. (a) , so we can approximate by considering and using the values given in the table: , . Averaging these values, we estimate to be approximately Similarly, , which we can approximate with : , . Averaging these values, we estimate to be approximately (b) Here , so by Equation 15.6.9 [ET 14.6.9], . Using our estimates from part (a), we have . This means that as we move through the point in the direction of , the temperature increases at a rate of approximately Alternatively, we can use Definition 15.6.2 [ET 14.6.2]: , which we can estimate with Then...
View Full Document

This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.

Page1 / 19

Chapter 14 Review Exercises copy - sin ! 1 x ! 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online