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Mathematics II. Homework 2
All answers have to be reasonably explained and justified.
Contents: Days 4 and 5.
- Compute the first and second order partial derivatives of:
(a) z = x^2 + 3xy + y^2. (b) z = x^2 ex^ + yex. (c) z = ln(x^2 + y). (d) z =
y^2 − x^2.
(e) z =
x^2 y
In each case, give the domain for each derivative. SOLUTION: (a) z′ x = 2x + 3y, z y′ = 3x + 2y, z′′ xx = 2, z′′ xy = z′′ yx = 3, z′′ yy = 2. Dom R^2.
(b) z′ x = 2xex^ + x^2 ex^ + yex, z′ y = ex, z xx′′ = 2ex^ + 4xex^ + x^2 ex^ + yex, z xy′′ = z yx′′ = ex, z′′ yy = 0. Dom R^2.
(c) z′ x = (^) x (^22) +xy , z y′ = (^) x (^21) +y , z′′ xx = −^2 x
(^2) +2y (x^2 +y)^2 ,^ z
′′ xy =^ z ′′ yx =^ − 2 x (x^2 +y)^2 ,^ z
′′ yy =^ − 1 (x^2 +y)^2.^ Dom {(x, y) : x^2 + y > 0 }.
(d) z′ x = −x/
y^2 − x^2 ; z y′ = y/
y^2 − x^2 ; z xx′′ = −y^2 /(y^2 − x^2 )^3 /^2 ; z′′ yy = −x^2 /(y^2 − x^2 )^3 /^2 ; z′′ xy = z yx′′ = xy/(y^2 − x^2 )^3 /^2. Dom {(x, y) : y^2 > x^2 } = {(x, y) : (y − x)(y + x) > 0 }
(e) z′ x = (^2) yx ; z′ y = −x 2 y^2 ;^ z
′′ xx =^
2 y ;^ z
′′ yy = 2
x^2 y^3 ;^ z
′′ xy =^ z ′′ yx =^ −
2 x y^2. Dom:^ {(x, y) :^ y^6 = 0}.
- Let f (x, y) = x^3 + 3y^4. Compute f (^) x′(1, 2), f (^) y′(3, −2) and the determinant of his Hessian at the point of coordinates (1, −1). SOLUTION: f (^) x′(1, 2) = 3,, f (^) y′(3, −2) = −96, |H| = −216.
- Let f (x, y) = x^3 y^4. Compute ∂^5 z ∂y^2 ∂x^3
SOLUTION:
72 y^2
- Let f (x, y) = 2x^4 − 3 x^2 y + y^2. Find the equation of the tangent plane to its surface at the point of coordinates (1, 3 , f (1, 3)). SOLUTION: z = − 10 x + 3y + 3.
- Given f (x, y) = xey/x:
(a) Show that all the tangent planes contain origin of coordinates. (Hint: find the equation of the tangent plane at a generic point (x 0 , y 0 ) of the function’s domain). (b) What can you tell about the tangent planes at the points of the domain where y = x?
SOLUTION: (a) If (a, b) belongs to f ’s domain, (a 6 = 0), the tangent plane at (a, b, aeb/a) has as general equation
z =
eb/a·(a − b) a
·x + eb/a·y
that, obviously, goes through (0, 0 , 0). (b) At the points (a, b) where a = b, the tangent plane is always the same: z = e·y
- Let f (x, y) = x^2 + 6xy − 3 y^2 − 4 y. Find, if it exists, the point on its surface where the plane 6 x − 2 y + z − 3 = 0 is the tangent plane. If such point doesn’t exist, explain why. SOLUTION: (0, − 1 , 1). We must check whether plane and surface have the tangency point in common. Indeed, f (0, −1) = 1 i (0, − 1 , 1) belongs to the plane: 6· 0 − 2 ·(−1) + 1 − 3 = 0.
- The figure below shows some level curves of a function f (x, y):
− 1 1 2 3 4 5 6 7 8 9 10 11 12 − 1
1
2
3
4
5
6
7
8
9
10
0
z = 10
z = 8 z = 6
z = 4
z = 2
P
Q
r
(a) Using the figure, discuss the signs (positive, negative or zero) of f (^) x′(P ), f (^) y′(P ), f (^) x′(Q) and f (^) y′(Q). This is, what are the signs of the partial derivatives at points P and Q? (b) On the same figure, consider the domain of f restricted to the line r. Mark the point of the domain where we find the minimum value of f (x, y). Justify your answer.
(Hint: Q(10. 000 , 625) can be obtained without the use of a calculator. Remember that 10.000 = 104 and 625 = 5^4 .)
SOLUTION: (a) Q(10. 010 , 625) ≈ 20 .015.
(b) Q(10. 000 , 623) ≈ 19 .984.
