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Chapter 14 Review Exercises copy - Stewart Calculus ET 5e...

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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 # 0 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
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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
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x,y ( ) $ 0,0 ( ) x ! f (x,0)=0/ x 2 =0 x % 0 f (x,y) $ 0 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 $ 0 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 $ 0 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 $ 0 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 $ 0 T x (x,y+ h ) ! T x (x,y) h T xy (6,4)=lim h $ 0 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 , . Averaging these values, we have (c) , so which 3 Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; Review: Exercises
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h = & 2. T x (6,4) 3.5 T x (6,6) T x (6,2). h = & 2 T x (6,6) T (8,6) ! T (6,6) 2 = 80 ! 75 2 =2.5 T x (6,6) T (4,6) ! T (6,6) ! 2 = 68 ! 75 ! 2 =3.5 T x (6,6) 3.0 T x (6,2) T (8,2) ! T x (6,2) 2 = 90 ! 87 2 =1.5 T x (6,2) T (4,2) ! T (6,2) ! 2 = 74 ! 87 ! 2 =6.5 T x (6,2) 4.0 T xy (6,4) T xy (6,4) T x (6,6) ! T x (6,4) 2 = 3.0 ! 3.5 2 = ! 0.25 T xy (6,4) T x (6,2) ! T x (6,4) ! 2 = 4.0 ! 3.5 ! 2 = ! 0.25 T xy 6,4 ( ) ’ ! 0.25 T (6,4)=80 T x (6,4) 3.5 T y (6,4) ’ ! 3.0 T x,y ( ) T (6,4)+ T x (6,4)( x ! 6)+ T y (6,4)( y ! 4) 80+3.5( x ! 6) ! 3( y ! 4) = 3.5 x ! 3 y +71 (5,3.8) T (5,3.8) 3.5(5) ! 3(3.8)+71 77.1 ( C. f (x,y)= 2 x + y 2 , f x = 1 2 2 x + y 2 ( ) ! 1/2 (2)= 1 2 x + y 2 f y = 1 2 2 x + y 2 ( ) ! 1/2 (2 y )= y 2 x + y 2 u = e ! r sin 2 ! , u r = ! e ! r sin 2 ! u ! =2 e ! r cos 2 ! g (u,v)= u tan ! 1 v , g u =tan ! 1 v g v = u 1+ v 2 we can estimate with We have from part (a), but we will also need values for and If we use and the values given in the table, we have , . Averaging these values, we estimate . Similarly, , . Averaging these values, we estimate . Finally, we estimate : , .
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