Practice Problems for Midterm 2

Transcription

1 Practice Problems for Midterm () For each of the following, find and sketch the domain, find the range (unless otherwise indicated), and evaluate the function at the given point P : (a) f(x, y) = + 4 y, P (, ) (b) f(x, y) = x y, P (5, 4) (c) f(x, y) = sin (x + y ), P (, ) (do not compute range) (d) f(x, y) = ln(x y), P ( e, 0) (e) f(x, y) = x y, P (, 0) (f) f(x, y, z) = ln(x + y + z), P (/3, /3, /3) (You do not need to sketch the domain) (g) f(x, y) = e xy, P (, 5) () Consider the function f(x, y) = e y x. (a) Find the domain and range of f(x, y). (b) For which values of c does the level curve f(x, y) = c have no points? (c) What do the level curves of f look like for valid values of c (namely, the ones not in (b)). (d) Sketch the level curve for c =. (3) Let g : R R be defined as g(x, y) = x y. (a) Find the domain and range of g. (b) What does the level curve of g at c = 0 look like? (c) Sketch the level curve of g going through (, ). (4) Let g : R R be defined as (a) Find the domain and range of g. g(x, y) = sin(x). y (b) What does the level curve of g at c = 0 look like? (c) Sketch the level curve of g with c =. (5) Consider the equation y = x 3 x. (a) Find a function f(x, y) with the above equation as one of its level curves. (b) Does f(x, y) = y x 3 x have the above equation as a level curve? Explain.

2 (6) Suppose the level curve to z = f(x, y) at c = is the curve y = x with a hole at (, 4) and the level curve at c = is the curve y = x + with a hole at (, 4) again. What is lim f(x, y)? (x,y) (,4) (7) Evaluate the limits of the following functions at the specified points (if they exist): (a) f(x, y) = x + e xy at (, 0) (b) f(x, y) = (c) f(x, y) = x at (3, ) x +y x at (0, 0) x +y (d) f(x, y) = ( (e) f(x, y) = sin(x) sin (f) f(x, y) = (g) f(x, y) = xy5 at (0, 0) x +y 0 x+y ) at (0, 0) (x )(y ) (x ) +(y ) at (, ) x y x +y at (0, 0) (h) f(x, y) = x + y ln(x + y ) at (0, 0) (i) f(x, y) = x +y +3z x +y at (0, 0, 0) (j) f(x, y) = xy xy+xz+yz at (,, ) (k) f(x, y) = x3 y 3 x 3 +y 3 at (0, 0) (l) f(x, y) = (x ) +(y 3) +4 (m) f(x, y) = (x ) +(y ) (x ) +(y ) +9 3 (n) f(x, y) = x y(x y) x 4 +y 4 at (0, 0) (o) f(x, y) = sin(x +y ) +x y 4 at (0, 0) (8) Consider the function at (, 3) at (, ) f(x, y) = tan (xy ). (a) What are the domain and range of f? (b) What is (c) What is lim f(x, y)? (x,y) (,) lim f(x, y)? (x,y) (0,0) (d) Show that f(x, y) is differentiable at every point in R. (e) Find the tangent planes to f(x, y) at (0, 0) and (, ) (that is, I want two different planes, one for each point).

3 (f) Find the linearization of f(x, y) at (, ), and use it to approximate f(.,.9) (you can leave you answer in terms of a decimal if you need to). (9) Calculate the partials f x, f y (and f z where appropriate) for the following functions. (a) f(x, y) = sin(x y). (b) f(x, y) = tan(xy) + e xy + ln(y) x (c) f(x, y, z) = y sin(xz) z. (d) f(x, y, z) = x yz. (0) If f(x, y) = e xy, find f xxy. That is, find 3 f y x x. () Find an equation for the set of all points in R 3 equidistant from the point (0, 0, ) and the plane z =. Classify the surface. () Suppose a surface S contains the parametric curves r (t) = + 5t, 3 t, + t t 3 and r (s) = 3s s, s + s 3 + s 4, s s + s 3. Find the equation of the tangent plane to S at (, 3, ). (3) (a) Classify the surface x + y z =. (b) Classify the surface x + y + z = 0. (c) Find and describe the intersection of the surfaces in (a) and (b). (4) Let z = f(x, y) = sin (x) y. (a) What are the domain and range of f(x, y)? (b) Sketch the level curve at c =. (c) Calculate (d) Calculate lim f(x, y). (x,y) (0,) lim f(x, y) (x,y) (0,0) along the y-axis and the line y = x. What can you conclude? (e) If we make the additional definition f(0, 0) = 0, can we say f(x, y) is continuous at (0, 0)? Explain. (f) Calculate f x, (g) Where is f(x, y) differentiable? Explain. f f, and y y x. 3

4 (h) Find the equation of the tangent plane to z = f(x, y) at the pont (π/4, ). (i) Find the linear approximation of f(x, y) at (π/4, ) and use it to approximate f(π/, ). (j) What is f(π/, )? (k) What is the derivative of f in the direction of u =, at (π/4, ). (l) Find the direction in which f(x, y) increases the most rapidly at (π/4, ). Give answer as a unit vector. (m) Find a unit vector tangent to the level curve at (π/4, ). (5) Consider the function f(x, y, z) = 3z + e x y. (a) What are the domain and range of f(x, y, z)? (b) Where is f(x, y, z) continuous? Explain. (c) Calculate lim f(x, y, z). (x,y,z) (0,0,) (d) What is the maximum rate of increase of f(x, y, z) at the point (0, 0, )? (e) Determine the set of all unit vectors u such that, at (0, 0, ), the function f(x, y, z) increases at /3 of its maximum rate (namely, /3 of the number you found in (d)) in the direction of u. (f) In three dimensional space, the set of vectors you found in (e) maps out a curve. What is the shape of that curve? (6) Consider the function (a) What is the domain of f(x, y)? f(x, y) = x y x + y. (b) By calculating the limit along different lines, show that does not exist. lim f(x, y) (x,y) (0,0) (c) If we make the additional definition f(0, 0) = 0, is f(x, y) differentiable at (0, 0)? NOTE: THE REMAINING PARTS DO NOT USE THIS EXTRA DEFINITION OF f(0, 0) = 0. (d) Explain why f(x, y) is continuous at every point of R except (0, 0). (e) Show f(x, y) is differentiable at every point of R other than (0, 0). (f) Find the equation of the tangent plane to f(x, y) at (, ). 4

5 (g) What is a vector perpendicular to the plane in (f). (h) Find the linear approximation to f(x, y) at (, ) and use it to approximate f(, ). (i) What is f(, )? How far off is the approximation in (h)? (j) At (, ), in which direction does f(x, y) increase most rapidly? How about decrease most rapidly? (k) What is the maximum rate of increase at (, )? (l) Sketch the level curves to f(x, y) at c =, 0, and. (m) Find the derivative of f(x, y) at (, ) in the direction of,. (n) Find all unit vectors u such that the directional derivative of f(x, y) at (, ) in the direction of u is 0. (o) Find a unit vector perpendicular to the level curve of f(x, y) at (, ). Sketch both the level curve and the vector. (Hint: you ve already graphed the level curve). (7) Approximate. tan(46 ). You can leave your answer in terms of π. (8) Find an equation for the surface consisting of all points P for which the distance between P and the x-axis is twice the distance from P to the yz-plane. Classify this surface. (9) Show that f(x, y) = e x sin(y) is a solution to the Laplace equation f xx + f yy = 0. (0) Show that u(x, t) = sin(x at)+ln(x+at) is a solution to the wave equation u tt = a u xx, where a is a constant. () Suppose z = x + 3xy y. Estimate the change in the function as x changes from to.05 and y changes from 3 to.96. () The base and height of a right circular cone are measured to be 0 cm and 5 cm, respectively. The possible error in measurement for each is at most 0. cm. Estimate the maximum possible error in calculated volume of the cone. (3) If z = x y + xy 4, and x = sin(t), y = cos(t), find dz dt at t = 0. (4) The temperature at a point is given by T (x, y), measured in Celsius. A bug crawls so that its position at t seconds is x = + t and y = + t, where x and y are in 3 centimeters. The temperature satisfies T x (, 3) = 4 and T y (, 3) = 3. How fast is the temperature rising on the bug s path after 3 seconds? (5) If w(x, y, z) = xy + yz + xz, x = r cos(θ), y = r sin(θ), z = rθ, find w r r = and θ = π/. w and θ when (6) Find all points at which the direction of fastest change of the function f(x, y) = x + y x 4y is i + j. (7) If f(x, y) = xy, find the gradient f(3, ) and use it to find the tangent line to the level curve f(x, y) = 6 at (3, ). Sketch the level curve, the tangent line, and the gradient vector. 5

6 (8) True/False: (a) If the angle between f(x 0, y 0 ) and u is acute, then f(x, y) is increasing in the direction of u at the point (x 0, y 0 ). (b) The gradient at a point is always tangent to the level curve passing through that point. (c) If f(x, y) is a continuous functions, then no two level curves of f(x, y) intersect. (d) To show lim (x,y) (x0,y 0 ) f(x, y) does not exist, we must exhibit two paths through (x 0, y 0 ) along which we get different limits. (e) If the function f(x, y) has different limits along the paths y = x and y = x + have different limits, then we can conclude that lim (x,y) (,0) f(x, y) does not exist. (f) The function f(x, y) at (x 0, y 0 decreases most rapidly in the direction of f(x 0, y 0 ). (g) If f(0, 0) = 3, 4, then there exists a vector u such that D u f(0, 0) = 0. (h) We can find a function f(x, y) such that f x = xy + sin(x) and f y = x + cos(y) for all points (x, y). (i) There were more true answers in this true/false section than there were in the true/false questions for the last practice midterm. 6

7 Answers: () (Sorry, sketches are not included, but type Domain of whatever function into WolframAlpha to see what they look like) (a) Domain: {(x, y) R : y }, Range: [, 3], f(, ) = (b) Domain: {(x, y) R : x y }, Range: [0, ), f(5, 4) = 3 (c) Domain: {(x, y) R : x + y 3}, f(, ) = 0 (d) Domain: {(x, y) R : y < x }, Range: R, f( e, 0) = (e) The domain is {(x, y) R : x + y }, and the range is {z R : 0 z }, f(, 0) =. (f) The domain is {(x, y, z) R 3 : x + y + z } and the range is the nonnegative reals: {z R : z 0}, f(/3, /3, /3) = 0 (g) The domain is {(x, y) R : xy 0} and the range is the positive real numbers greater or equal to {z R : z }, f(, 5) = e 0. () (a) The domain is all of R and the range is {z R : z > 0} (i.e. the positive real numbers). (b) Setting f(x, y) = c gives e y x = c. Since the range of the exponential function consists of positive real numbers, if c 0, then the level curve will have no points on it. (c) If c > 0, then the level curve is the parabola y = x + [ + ln(c)]. (d) When c =, we get ln(c) = 0, so your graph should be that of y = x +. (3) (a) The domain is {(x, y) R : y 0} (i.e. any point with y 0). The range is all of R. (b) The level curve at c = 0 looks like the vertical line x = 0 with the point (0, 0) removed (since y cannot equal 0). (c) It should be the graph of y = x with a hole at the origin (since y cannot equal 0). (4) (a) The domain is {(x, y) R : y 0} (i.e. any point with y 0). The range is all of R. (b) The level curve at c = 0 looks like a bunch of vertical lines at x = 0, ±π, ±π,..., except the point y = 0 is removed from each one. (c) It should be the graph of y = sin(x) minus the points where y = 0 (i.e. the points where graph of y = sin(x) crosses the x-axis should be removed, and there should just be a hole in the graph). (5) (a) f(x, y) = x 3 x y has the equation as a level curve at c = 0. (b) No, because (0, 0) is a solution to the equation which is not on the level curve at c =. 7

8 (6) The limit does not exist, because along the path y = x the limit is, and along y = x + the limit is. (7) (a) (b) 3/0 (c) DNE (check x = 0) (d) DNE (check x = 0 and x = y 5 ) (e) 0 (f) DNE (check x = and y = x + ) (g) 0 (h) 0 (i) DNE (j) /3 (k) DNE (l) DNE (m) 6 (n) DNE (o) 0 (8) (a) The domain of f(x, y) is all of R, and the range is the interval ( π/, π/ in R. (b) f(x, y) is continuous everywhere as g(x) = tan (x) is a continuous function in R, so to evaluate the limit we can just plug in the point, and tan () = π/4. (c) By the same logic as in (b), the limit is tan (0) = 0. (d) Observe that the partials are f x = y + x y, f 4 y = xy + x y, 4 which are both continuous everywhere as the denominator can never be 0 (the monomial x y 4 0 always). Therefore f is differentiable everywhere. (e) We calculate the partials already. We get f x (0, 0) = f y (0, 0) = 0, and since f(x, y) = 0, the plane at (0, 0) is simply z = 0. At (, ), f x (, ) =, and f y (, ) =, and since f(, ) = π/4, the plane is z π 4 = (x ) + (y ). (f) The linearization consists solely of isolating z in the previous part, so the linearization is z = L(x, y) = π 4 + (x ) + (y ). So L(.,.9) = π 4 + (. ) + (.9 ) = π

9 (9) (a) f x = xy cos(x y), f y = x cos(x y) (b) f x = y sec (xy) + y e xy ln(y) x, f y = x sec (xy) + xye xy + xy. y[zx cos(xz) sin(xz)] z. (c) f x = y cos(xz), f y = sin(xz), f z z = (d) f x = yzx yz, f y = zx yz ln(x), f z = yx yz ln(x). (0) f xxy = 4y 3 e xy + xy 5 e xy. () z = x y + 3, an elliptic paraboloid. () 8(x ) 6(y 3) + 40(z ) = 0 (3) (a) Hyperboloid of one sheet (b) Sphere (c) The intersection can be described as the union of two circles: x + y =, z = 3, x + y =, z = 3. (4) (a) The domain of f(x, y) is the set of points (x, y) with y 0. In set notation {(x, y) R y 0}. Both sin (x) and y are nonnegative, so we expect the range to be a subset of the nonnegative reals. /y can be any positive number, and sin (x) can be 0, so the range is, in fact, all nonnegative reals: {z R : z 0}. (b) At c =, we get sin (x)/y =, so y can never be 0. Multiplying, we get y = sin (x), or y = ± sin(x). So you should graph both sin(x) and sin(x), with holes at every point where y = 0. (c) The function is continuous at (0, ), so we can plug in the point to get a limit of 0. (d) Along the y-axis, the limit is 0, and along y = x it is (need to use L Hopital s rule twice). Therefore the limit does not exist. (e) No, because the limit does not exist, so the second condition for continuity fails. sin(x) cos(x) (f) f x =, and since sin(x) cos(x) = sin(x), we can write this as f y x = sin(x). From this we get y f xy = y sin(x). 3 If you left f x in its original form, you would get f xy = 4 sin(x) cos(x), y3 both answers are acceptable. We also get f y = sin (x) y 3. 9

10 (g) We find f(x, y) is defined for (x, y) with y 0. In this region, both f x and f y are continuous, and so f is differentiable on its domain (namely, the one we found in (a)). (h) We found f x and f y in the previous part. f x (π/4, ) = and f y (π/4, ) =, and z 0 = f(π/4, ) = /, so the equation is (i) The linear approximation is We get L(π/, ) = + π 4. (j) f(π/, ) =. (k) 3/ 5 (l), (m), z = (x π ) (y ). 4 L(x, y) = + (x π ) (y ). 4 (5) (a) The domain is all of R 3, the range is all of R. (b) It is continuous everywhere as all functions involved are continuous, and the sum and composition of continuous functions are continuous. (c) Since f is continuous, we can just plug in the point (0, 0, ) to get 4. (d) 3 (e) u can be any vector x, y, /3 with x + y = 8/9. (f) A circle. (6) (a) Every point in R except the point (0, 0). (b) Along the line y = mx, we get a limit of m +m. m = yields a limit of 0, m = 0 yields a limit of, therefore limit does not exist. (c) No, because it s not continuous. (d) f(x, y) is a quotient of two polynomials and the denominator is not zero for (x, y) (0, 0). (e) We know f(x, y) is defined at all such points, so show that the partials are continuous at all points other than the origin. (f) z = x y, or z = (x ) (y ) if you don t simplify. (g),, or,, (h) L(x, y) = (x ) (y ), L(, ) =. 5 5 (i) f(, ) = 0, so the approximation was off by 5 =.04. 0

11 (j), and,, respectively. (k) /, or. (l) c = 0: the lines y = ±x, with a hole at the origin. c = : the y-axis with a hole at the origin, c = : the x-axis, with a hole at the origin. (m) 3/ 5 (n) /, /, /, / (o) /, or /, is the vector you want. You should sketch this vector at (, ) to the level curve at c = 0. (7).05 + π 90 (8) 4x = y + z, a cone. (9) f xx = e x sin(y) and f yy = e x sin(y), so f xx + f yy = 0. (0) If you compute both u xx and u tt correctly, it should work out. () 0.65 () 0π cm 3 (3) 4 (4) C/s (5) w r = π, w θ = π (6) All points on the line y = x + (7) f(3, ) =, 3, x + 3y =. (8) (a) True (b) False (c) True (d) False (e) False (f) False (g) False (h) False (i) True