Short Solutions to Practice Material for Test #2 MATH 2421

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1 Short Solutions to Practice Material for Test # MATH 4 Kawai (#) Describe recisely the D surfaces listed here (a) + y + z z = Shere ( ) + (y ) + (z ) = 4 = The center is located at C (; ; ) and the radius is R = (b) +z = 9 Right circular cylinder erendicular to the z-lane. The generating curve is a circle of radius in the y-lane. The surface is etruded in the y-ais direction. (#) Describe the level curves for f (; y) = e y We set the two-variable function equal to a constant k e y = k ) ln e y = ln (k) ) y = ln (k) The level curves (contours) are arabolas y = + ln (k) y (#) Evaluate these artial derivatives. (a) ln + y sin () = + y sin () + y sin () (b) Product Rule. y = + y cos () + y sin () h tan (y) e =yi = tan (y) h e =yi + e =y [tan (y)] y y = tan (y) y e =y + e =y sec (y) = e =y tan (y) y + sec (y)

2 (c) Mied second artial. y y (y + ) = + (y + ) Now take the artial with resect to + (y + ) = + (#4) Find the imlicit derivative z y on the surface z + yz y 7 cos = (y + ) The left side is F (; y; z) The imlicit formula is z y = Fy z B + 7 y = sin C A + yz F z Multily to and bottom by in order to clear that one denominator. y z + 7 sin ( + yz) (#5) The lateral area of a cone (the crunchy art of an ice cream cone) is If r = and h = 4; then A is 5 sq. units. A A (r; h) = r r + h Use the total di erential to aroimate the change in A if r moves from to and h moves from 4 to 4 We need the two rst-order artials. Product Rule A r = r r r + h + r r + h + r + h r + h = = r + h r + h The other one is easier. A h = r h h r + h i = h rh = r = r + h r + h The total di erential is da = A r dr + A h dh = r + h r + h dr + rh r + h dh Based on the changes given above, we have dr = + and dh = + We substitute in r = and h = 4 (9) + (6) da = (+) + 5 () (4) 5 (+) = We eect the area to increase by aroimately this many sq. units. 46 =

3 (#6) Suose we have non-collinear oints which must lie in a unique lane P ( ; ; 5) ; Q (; ; ) ; R (4; ; ) (a) Find the arametric equations associated with the D line which asses through Q and R! QR = h; ; i [Any scalar multile will also work.] = () t + y = ( ) t + z = ( ) t + We can also use the coordinates of R (4; ; ) (b) Find P! Q and P! R; and then nd a vector n which is orthogonal to both of those vectors. Cross roduct.!! P Q = h; ; 4i P R = h6; ; 5i n =! P Q! P R = i j k = i 4 5 j k 6 = i ( 5 ) j ( 5 ( 4)) + k ( 6) = h 5; 9; 6i (c) Give the standard form for the lane which contains P; Q; and R If we use P as our base oint, then we have The other choices use Q and R 5 ( ( )) 9 (y ) 6 (z 5) = 5 ( + ) + 9 (y ) + 6 (z 5) = 5 ( ) + 9 (y ) + 6 (z ) = 5 ( 4) + 9 (y ) + 6 (z ) = All the standard forms are equivalent to the general form 5 + 9y + 6z 8 = (#7) Find the standard form for the tangent lane at (; ; ) to the function surface z = f (; y) = y + y rf = y; + y rf (; ) = h; i The standard form use n = h; ; i or h; ; i ( ) + (y ) (z ) = (y ) (z ) = y z =

4 (#8) From the deendency information in Section.5, Eercise #5, you can make a tree. Find u=z u # v. # & w y.. # # & z z t t z After all the lower variables are substituted in, we see that u is a function of t and z Thus, assuming di erentiability everywhere, we can nd formulas for u=t and u=z If z changes, there are THREE aths of change back u to u We can do some factoring. u z = u v w v w z + u v v z + u v y v y z u z = u v v w w z + v z + v y y z If we insert the actual names of the functions involved, we have u z = f g h v w z + g z + g q ; y z since u = f; v = g; w = h; = ; and y = q (#9) Section.5, Eercise #55ab is in the Student s Solution Manual! (#) Matching (i) (F) z = + y Cone that oens out onto the negative z-ais. (ii) (C) z = y Ellitic araboloid that oens out onto the negative z-ais. (iii) (A) z = y Hyerbolic araboloid. The signs of the squared terms are oosite. (iv) (B) + y z = Hyerboloid of ONE sheet. One negative sign in standard form. (#) Consider the function f (; y) = y + y (a) Evaluate rf ( ; ) rf = hf ; f y i = h y; + yi rf ( ; ) = h ( ) ( ) ; ( ) + ( )i = h ; i (b) Give the angle [in standard osition] which tells us the direction we should walk in from ( ; ) to achieve steeest ascent in f In other words, which angle gives us the greatest rate of increase if f at ( ; )? This direction is in Quadrant III. We have = 5=4 = 5 because tan vert comonent horiz comonent = tan = 4 ; but we need to adjust this for the correct quadrant. Just add 4

5 (c) Evaluate D u f ( ; ) for the unit vector u which has the same direction as h; i The associated unit vector is Dot this with the gradient. u = q h; i = h; i + ( ) D u f ( ; ) = h ; i h; i = ( + ) = In this direction, we eect f to decrease at the rate of direction of u = er unit ste in the (#) My favorite min/ma roblem. Let f (; y) = + y y We already know that (; ) is a critical oint. (a) Find the other FOUR critical oints. f = y = y = = or y = or y = Now look at f y = y y = When = ; we have y = We already know (; ) is a critical oint. When y = ; we have We have ( ; ) and (; ) Similarly, when y = We have ( ; ) and (; ) ( ) ( ) = ) = = () () = ) = = (b) Now run the Second Partials Test on the critical oint (; ) and classify it as min/ma/saddle. f = y f yy = f y = 4y d = () () = 4 > Result Local minimum. 5

6 (#) Evaluate Inner Z =6 Z =6 Z =6 sin ( + y) dy = sin ( + y) dy d = 4 [cos ( + y)]=6 = cos + cos () Outer Z =6 cos + cos () d = = = h sin + sin i =6 sin () sin sin 6! + = 4 X (#4) Evaluate Z dy =??? y Let u = y; du = dy Perfect match. Z du = sin (u) + C u The de nite integral is sin (y) y= = sin () sin () = sin () 6