11/1/2017 Second Hourly Practice 4 Math 21a, Fall Name:

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1 //27 Second Hourly Practice 4 Math 2a, Fall 27 Name: MWF 9 Jameel Al-Aidroos MWF 9 Dennis Tseng MWF Yu-Wei Fan MWF Koji Shimizu MWF Oliver Knill MWF Chenglong Yu MWF 2 Stepan Paul TTH Matt Demers TTH Jun-Hou Fung TTH Peter Smillie TTH :3 Aukosh Jagannath TTH :3 Sebastian Vasey Start by printing your name in the above box and check your section in the box to the left. Do not detach pages from this exam packet or unstaple the packet. Please write neatly. Answers which are illegible for the grader cannot be given credit. Show your work. Except for problems - 3,8, we need to see details of your computation. All functions can be differentiated arbitrarily often unless otherwise specified. No notes, books, calculators, computers, or other electronic aids can be allowed. You have 9 minutes time to complete your work Total:

2 Problem ) True/False questions (2 points), no justifications needed ) T F For any continuous function f(x,y), we have 2 f(x,y)dxdy = f(x,y)dxdy. 2 This is not Fubini 2) T F If u is a unit vector tangent to f(x,y) = at (,) and f(,) =, then D u f(,) is zero. 3) T F Assume f is zero on x = y and x = y, then (,) is a critical point of f. 4) T F 5) T F 6) T F If(,)istheonlylocalminimumofafunctionf andf hasnolocalmaxima, then (,) is a global minimum. If (,) is a critical point for f, and f yy (,) < then (,) is not a local minimum. If f(x,y) and g(x,y) have the same non-constant linearization L(x,y) at (,) and f(,) = g(,) =, then the level sets f = and g = have the same tangent line at (,). 7) T F There are saddle points with positive discriminant D >. No. By definition not 8) T F If R is the unit disc, then R x2 y 2 dxdy is zero. Yes. This can be seen as the definition of area. 9) T F There is a nonzero function f(x,y) for which the linearization L(x,y) is equal to 2f(x,y). The condition means that the function is linear. For a linear function, the linearization is itself.

3 ) T F The directional derivative at a local minimum (, ) is positive in every direction. It is zero ) T F If r(t) is a curve on the surface g(x,y,z) =, then g( r(t)) r (t) =. Use the chain rule and the fact that g( r t)) is constant so that ( d/dtg(r t)) is zero. ( 2) T F If f(,) = 2, there is a direction in which the directional derivative at (,) is 2. Yes, we have seen this computation 3) T F If D > at (,) and f(,) = and f xx (,) < then f yy (,) <. We have seen this from D = f xx f yy f 2 xy. 4) T F x f(x,y) dydx = y f(x,y) dxdy. This is almost a correct correct switch but the dxdy have also to be switched. 5) T F The surface area of the sphere of radius L is π L2 sin(φ) dφ. The answer is 4π.

4 6) T F If f(x,y) = g(x) is a function of x only, then D = at every critical point. Indeed, then f xy,f xx f yy are both zero. 7) T F The gradient vector f(x,y ) is a vector which is perpendicular to the surface z = f(x,y). Nonsense 8) T F If f(,) = 2, then there is a unit vector v such that D v f(,) =. Intermediate value theorem. 9) T F The gradient of the function f(x,y) = y x sin(t) dt is sin(x),sin(y). Yes, this is the derivative by the fundamental theorem. 2) T F Assume f(x,y) = x 2 +y 4 and a curve r(t) satisfies r (t) = f( r(t)), then f( r(t)). d dt FTTFTTFTFFTTTTFTFTTT

5 Problem 2) ( points) No justifications needed a)(6 points) Match theregions with theintegrals. Each integral matches oneregion A F. A B C D E F Enter A-F Integral 2 y f(x,y) dxdy 2 y y f(x,y) dxdy 2 2 x 2 x f(x,y) dydx 2 2 x 2 x f(x,y) dydx 2 y 2 y f(x,y) dxdy 2 2 x f(x,y) dydx b) (4 points) Name the partial differential equations correctly. Each equation matches one name. Fill in -4 Name Laplace Wave Transport Heat Equation Number PDE g x g y = 2 g xx g yy = 3 g x g yy = 4 g xx +g yy = FCBDAE 4,2,,3

6 Problem 3) ( points) y B 3 V 2 J W O 5 H E x U 5 M 8 9 A L D 9 K C G 3 6 a) (6 points) Enter one label into each of the boxes. At which point is the length of the gradient maximal? At which point is the global maximum? At which point is f x >,f y =? At which point is D, / 2 f =, D, / 2 f <? At which point is f maximal under the constraint g(x,y) = y =? At which point does f have a local minimum? b) (4 points) Note that the zero vector is considered both parallel and perpendicular to any other vector.

7 parallel perp The gradient f is always to the surface f = c. For a Lagrange minimum, g is to f. If (,) is a min. of f then f(,) is to,. If (,) is max. of f and g = z f(x,y) then g is to,,. WBUGEO parallel perp The gradient f is always * to the surface f = c. For a Lagrange minimum, g is * to f. If (,) is a min. of f then f(,) is * * to,. If (,) is max. of f and g = z f(x,y) then g is * to,,. Problem 4) ( points) A farm costs f(x,y), where x is the number of cows and y is the number of ducks. There are cows and 2 ducks and f(,2) =. We know that f x (x,y) = 2x and f y (x,y) = y 2 for all x,y. Estimate f(2,9). Old MacDonald had a million dollar farm, E-I-E-I-O, and on that farm he had x = cows, E-I-E-I-O, and on that farm he had y = 2 ducks, E-I-E-I-O, with f x = 2x here and f y = y 2 there, and here two cows more, and there a duck less, how much does the farm cost now, E-I-E-I-O? f(,2) =. ThelinearizationisL(2,9) = f(,2)+2()(2 )+(2) 2 (9 2) = +2(2) 4 =

8 Problem 5) ( points) Find the Harvard H which has maximal area f(x,y) = 5xy +2x 2 with fixed exposed perimeter 6x+4y = 88. Find the maximum using Lagrange. The Lagrange equations are 5y +4x = 6λ 5x = 4λ 6x+4y = 88 Eliminating λ gives y = (7/)x. x =,y = 7,f(,7) = 55. Problem 6) ( points) a) (7 points) A minigolf on the cape has a hole at a local minimum of the function f(x,y) = 3x 2 +2x 3 +2y 5 5y 2. Find all the critical points and classify them. b) (3 points) A golfer hits tangent to the level curve f(x,y) = 2 through (,). Find this line. About minigolf: the first standardized minigolf course appeared in 96 in North Carolina. The world record on a round of minigolf is 8 strokes on 8 holes on eternite. No perfect round on concrete has been scored. The highest prizes reach 5 dollars only so that nobody is known to make a living by competing in minigolf.

9 - 6-6 max saddle -2 a) -6 6 saddle 8 6 min -3 b) We know the gradient at the point. x =. Problem 7) ( points) AcirculartracknearSalemisacircleofradius5which is centered at the origin (, ). A go-kart goes counterclockwise around the track r(t). The cheering intensity is given by a function f(x,y). The go-kart passes the point (3,4) at time t = with velocity 4,3. We know that f x (3,4) = 2 and f y (3,4) =. Find the rate of change d dt f( r(t)) at t =. Use the chain rule: d dt f(r(t)) = f(r(t)) r (t) At t =, we have 2, 4,3 = 8+3 = 22. Problem 8) ( points) a) (6 points) Find the integral y /5 y e x +x 7 x x 5 dxdy. b) (4 points) Integrate y 2 e x 2 +y 2 x2 +y 2 dxdy.

10 a) Change the order of integration e 7/8 b) Use polar coordinates π/2(e ). Note that the region is in the fourth quadrant so that we integrate from π/2 to. Problem 9) ( points) Find the surface area of the wormhole r(u,v) = 3v 3,v 9 cos(u),v 9 sin(u), where u 2π and v. Einstein-Rosen bridges are hypothetical topological constructions which would allow shortcuts through space-time. Tunnels connecting different parts of the universe appear frequently in science fiction. Note that r u r v dudv has a nonnegative integrand. In our case it is 9v +v 2. When we integrate this from to 2π we get 8πv +v 2. But note that we have to take either square root so that the integrand is nonnegative or just take the absolute value. In any case, we can just take twice the integral from to. The answer is not zero but 2π( 8 ). Problem ) ( points)

11 a) (5 points) We become typographer and design new mathematically defined typeface of the alphabet. The new letter e in this 2a design is given by a polar region r(t) t /7, with t 2π. Find the area of this region. b) (5 points) Integrate arccos(y) cos(x) dxdy. Remark: Computer scientist Donald Knuth once wrote an entire article about The Letter S. a) We have 2π t 2/7 /2 dt = (7/8)(2π) 9/7. b) Change the order of integration. We have π/2 y π/2 cos(x) dydx = dy = π/2.