Questions on functions of two variables

Question 1

For f(x, y) = x²y³ + 5x, find f_x and f_y.

Accepted Answer

Holding y constant, f_x = 2xy³ + 5, since y³ is a constant multiplier and the 5x differentiates to 5. Holding x constant, f_y = 3x²y², and the 5x now differentiates to 0 because it carries no y. M1 for treating the other letter as a constant, A1 for f_x, A1 for f_y.

Question 2

A surface has equation z = x² + 3y². Write down the section along x = 2, and describe the contour at z = 12.

Accepted Answer

Putting x = 2 gives the section z = 4 + 3y², a parabola opening upwards with its vertex at height 4. Putting z = 12 gives the contour x² + 3y² = 12, or x²/12 + y²/4 = 1, which is an ellipse reaching x = ±2√3 and y = ±2. B1 for the section, M1 for setting z = 12, A1 for naming the ellipse with its intercepts.

Question 3

For f(x, y) = x³ + 2x²y − y⁴, find the four second partial derivatives, and verify that the mixed ones agree.

Accepted Answer

First derivatives: f_x = 3x² + 4xy and f_y = 2x² − 4y³. Differentiating again gives f_xx = 6x + 4y, f_yy = −12y², and f_xy = 4x from f_x by y. Differentiating f_y by x gives f_yx = 4x as well, so the two agree, as the mixed derivative theorem says they should. M1 for the first derivatives, A1 for f_xx, A1 for f_yy, A1 for both mixed derivatives with the comparison stated.

Question 4

The surface S has equation z = x² + y² − 4. Write S in the implicit form g(x, y, z) = c, and state what the number z measures for a point of S lying above the point (x, y) of the horizontal plane.

Accepted Answer

Moving z across gives x² + y² − z = 4, so g(x, y, z) = x² + y² − z and c = 4. B1 for a correct implicit form with its constant. Read as a surface, z is the height above the point (x, y), so the equation z = f(x, y) assigns one height to each point of the plane and S is the graph of that assignment. B1 for the height reading.

Question 5

Find the stationary point of f(x, y) = x² + y² − 4x + 6y + 3 and determine its nature.

Accepted Answer

f_x = 2x − 4 and f_y = 2y + 6, so both vanish only at (2, −3), where f = 4 + 9 − 8 − 18 + 3 = −10. Then f_xx = 2, f_yy = 2 and f_xy = 0, so H = 2 × 2 − 0² = 4. Since H > 0 and f_xx > 0 this is a local minimum, of value −10. M1 for setting both first derivatives to zero, A1 for the point, M1 for H, A1 for the classification with a reason.

Question 6

Find the four stationary points of f(x, y) = x³ − 3x + y³ − 3y and classify each one.

Accepted Answer

f_x = 3x² − 3 gives x = ±1 and f_y = 3y² − 3 gives y = ±1, so the stationary points are (1, 1), (1, −1), (−1, 1) and (−1, −1). Here f_xx = 6x, f_yy = 6y and f_xy = 0, so H = 36xy. At (1, 1), H = 36 > 0 with f_xx = 6 > 0: a minimum, value −4. At (−1, −1), H = 36 > 0 with f_xx = −6 < 0: a maximum, value 4. At (1, −1) and (−1, 1), H = −36 < 0: saddle points, both of value 0. M1 for solving both equations, A1 for the four points, M1 for H = 36xy, A1 A1 A1 for the three kinds of verdict.

Question 7

Find the equation of the tangent plane to z = x²y − 3y² at the point where x = 2 and y = 1.

Accepted Answer

The height is f(2, 1) = 4 − 3 = 1, so the point is (2, 1, 1). The slopes are f_x = 2xy = 4 and f_y = x² − 6y = −2 there. The tangent plane is z = 1 + 4(x − 2) − 2(y − 1), which tidies to z = 4x − 2y − 5. B1 for the height, M1 for both partial derivatives evaluated at the point, M1 for the formula, A1 for the plane.

Question 8

For f(x, y) = e^3x cos y, find f_x, f_y and f_xy, and evaluate f_xy at (0, π/2).

Accepted Answer

f_x = 3e^3x cos y, with cos y a constant multiplier, and f_y = −e^3x sin y. Differentiating f_x with respect to y gives f_xy = −3e^3x sin y, and f_y differentiated with respect to x gives the same. At (0, π/2) that is −3 × 1 × 1 = −3. M1 A1 for f_x, A1 for f_y, A1 for the mixed derivative and its value.

Question 9

For f(x, y, z) = x²y + yz³ + 4z, find f_x, f_y and f_z, and evaluate f_z at (1, 2, −1).

Accepted Answer

Each derivative holds the other two letters constant. f_x = 2xy, since the last two terms carry no x. f_y = x² + z³, since the first two terms are linear in y and the 4z carries no y. f_z = 3yz² + 4. At (1, 2, −1) that is 3 × 2 × 1 + 4 = 10. M1 for holding the other letters constant, A1 for f_x, A1 for f_y, A1 for f_z with its value.

Question 10

The surface S has equation z = x² − y². Describe the section x = 1 and the section y = 0, describe the contour z = 0, and say what shape the surface has near the origin.

Accepted Answer

Putting x = 1 gives z = 1 − y², a parabola in the z-y plane opening downwards with its highest point at height 1. Putting y = 0 gives z = x², a parabola opening upwards with its lowest point at the origin. B1 B1 for the two sections. Putting z = 0 gives x² = y², that is the pair of lines y = x and y = −x through the origin. B1 for the contour. Along one section the origin is the bottom of a curve and along the other it is the top, so the surface is a saddle there. B1 for the saddle.

Question 11

For f(x, y) = ln(x² + y²), find f_x and f_y, and show that f_xx + f_yy = 0 wherever f is defined.

Accepted Answer

By the chain rule f_x = 2x/(x² + y²) and, by symmetry, f_y = 2y/(x² + y²). M1 for the chain rule, A1 for both first derivatives. Differentiating f_x with respect to x by the quotient rule gives f_xx = [2(x² + y²) − 2x(2x)]/(x² + y²)² = 2(y² − x²)/(x² + y²)², and the same working in the other letter gives f_yy = 2(x² − y²)/(x² + y²)². M1 for the quotient rule. The two numerators are negatives of each other, so the sum is 0 everywhere except the origin, where f is undefined. A1 for the sum vanishing with the origin excluded.

Question 12

Find the stationary points of f(x, y) = x⁴ + y⁴ − 4xy and classify each one.

Accepted Answer

f_x = 4x³ − 4y = 0 gives y = x³, and f_y = 4y³ − 4x = 0 gives x = y³. Substituting, x = x⁹, so x(x⁸ − 1) = 0 and x = 0, 1 or −1. The points are (0, 0), (1, 1) and (−1, −1). Now f_xx = 12x², f_yy = 12y² and f_xy = −4, so H = 144x²y² − 16. At the origin H = −16 < 0, a saddle point of value 0. At (1, 1) and at (−1, −1), H = 128 > 0 with f_xx = 12 > 0, so both are local minima of value −2. M1 for the pair of equations, M1 for eliminating to x = x⁹, A1 for the three points, M1 for H, A1 A1 for the classifications.

Question 13

Show that the Hessian test gives no verdict at the origin for f(x, y) = x⁴ + y⁴, and determine the nature of that stationary point by another argument.

Accepted Answer

f_x = 4x³ and f_y = 4y³ both vanish at the origin, so it is stationary. The second derivatives are f_xx = 12x², f_yy = 12y² and f_xy = 0, all zero at the origin, so H = 0 and the test decides nothing. Directly, though, x⁴ ≥ 0 and y⁴ ≥ 0 for real x and y, with equality only at x = y = 0, so f > f(0, 0) = 0 at every other point and the origin is a minimum. B1 for showing the point is stationary, M1 A1 for H = 0, B1 for the direct argument.

Question 14

The temperature at the point (x, y) of a flat plate is modelled by T = 100 − 2x² − y² degrees. Find the tangent plane to this surface at the point where x = 3 and y = 2, use it to estimate the temperature at (3.1, 1.9), and compare that estimate with the exact value.

Accepted Answer

At (3, 2) the temperature is T = 100 − 18 − 4 = 78, so the point is (3, 2, 78). B1 for the height. The slopes are T_x = −4x = −12 and T_y = −2y = −4 there. M1 for both partial derivatives evaluated at the point. The tangent plane is T = 78 − 12(x − 3) − 4(y − 2). A1 for the plane.
Substituting x = 3.1 and y = 1.9 gives 78 − 12(0.1) − 4(−0.1) = 78 − 1.2 + 0.4 = 77.2 degrees. M1 for the substitution, A1 for 77.2. Exactly, T = 100 − 2(9.61) − 3.61 = 77.17, so the estimate is 0.03 too high. B1 for the comparison.
The plane is the best flat model of the surface at that point, and the error it makes grows with the square of the step: doubling both increments to reach (3.2, 1.8) would multiply the gap by about four. Both signs matter in the substitution, since y has gone down while x has gone up.

Question 15

A box with a rectangular base and no lid is to hold 32 cm³. Its base measures x cm by y cm and its height is z cm. Show that the outside surface area is S = xy + 64/x + 64/y square centimetres, and find the dimensions that make S least, justifying that the value is a minimum.

Accepted Answer

The surface is one base and four sides, so S = xy + 2xz + 2yz. The volume condition xyz = 32 gives z = 32/(xy), and substituting turns 2xz into 64/y and 2yz into 64/x, so S = xy + 64/x + 64/y. B1 for the substitution eliminating z.
Then S_x = y − 64/x² and S_y = x − 64/y². Setting both to zero gives y = 64/x² and x = 64/y²; substituting the first into the second gives x = 64x⁴/4096 = x⁴/64, so x⁴ = 64x and x(x³ − 64) = 0. Since x is a length, x = 4, and then y = 4 and z = 32/16 = 2. M1 for both partial derivatives, M1 for solving them together, A1 for 4 cm by 4 cm by 2 cm, with S = 16 + 16 + 16 = 48 cm². A1 for the area.
For the nature, S_xx = 128/x³ = 2, S_yy = 128/y³ = 2 and S_xy = 1, so H = 2 × 2 − 1 = 3 > 0 with S_xx > 0, a local minimum. B1 for the Hessian test with its verdict.