University of Aarhus
Department of Mathematical Sciences
Calculus Quiz
week 37.2
Legend: A 4 indicates a correct response; a 8, indicates an incorrect response, in this case, the correct answer is marked with a l. Correct and follow
a 4,l for a solution. Return through n.
Fall 2004
2
Partial derivatives
1. If f (x, y) = x2 − 2y 2 + 2x − y + 5, then equation for the tangent plane
through (0, 0, 5) is:
z = x2 − 2y 2
z − 5 = x2 − 2y 2
z = 2x − y
z − 5 = 2x − y
2. The linear approximation of f (x, y) = ln(xy + 1) in (x, y) = (0, 1) is
given by
f (x, y) ≈ x
f (x, y) ≈ xy
f (x, y) ≈ xy + 1
f (x, y) ≈ ln(xy + 1)
3. The function z = f (x, y) = sin(y − x2 ) has differential
dz = −2x cos(y − x2 )dx + cos(y − x2 )dy
Yes
No
3
4. If f (x, y) = x3 − y 2 + 2, then the tangent plane through (0, 0, 2) is
horizontal.
Yes
No
5. The linear approximation of f (x, y) = sin(x + y) in (x, y) = (0, 0) is
given by
f (x, y) ≈ sin(x + y)
2
f (x, y) ≈ x + y
6. The function w = f (x, y, z) =
dw =
dw =
f (x, y) ≈ x + y
2
dy
dx
dz
x + y + z
dy
dz
− dx
x − y − z
1
x
+
1
y
+
1
z
has differential:
dy
dz
y2 − z2
dy
dz
y2 + z2
dw = − dx
x2 −
dw =
dx
x2
+
Solutions to Quizzes
4
Solutions to Quizzes
Solution to Quiz:
f (x, y) = x2 − 2y 2 + 2x − y + 5 gives
fx = 2x + 2, fy = −4y − 1
By [S] 11.4 Theorem 2 the equation for the tangent plane is
z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
I.e. in (0, 0, 5)
z − 5 = 2x − y
Solutions to Quizzes
5
Solution to Quiz:
Given f (x, y) = ln(xy + 1) and calculate
y
x
fx =
, fy =
xy + 1
xy + 1
By [S] 11.4 4 the linear approximation in (a, b) is
f (x, y) ≈ f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
I.e. for (a, b) = (0, 1)
f (x, y) ≈ x
Solutions to Quizzes
6
Solution to Quiz:
Yes, f (x, y) = sin(y − x2 ) gives:
fx = −2x cos(y − x2 ), fy = cos(y − x2 )
By [S] 11.4 4 the differential is
df = fx (x, y)dx + fy (x, y)dy
I.e.
dz = −2x cos(y − x2 )dx + cos(y − x2 )dy
Solutions to Quizzes
7
Solution to Quiz:
f (x, y) = x3 − y 2 + 2 gives
fx = 3x2 , fy = −2y
By [S] 11.4 Theorem 2 the equation for the tangent plane is
z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
I.e. horizontal with equation
z−2=0
Solutions to Quizzes
8
Solution to Quiz:
Given f (x, y) = sin(x + y) and calculate
fx = cos(x + y), fy = cos(x + y)
By [S] 11.4 4 the linear approximation is
f (x, y) ≈ f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
I.e. for (a, b) = (0, 0)
f (x, y) ≈ x + y
Solutions to Quizzes
Solution to Quiz:
w = f (x, y, z) = x1 +
9
1
y
+
1
z
gives:
fx =
−1
x2
By [S] 11.4 the differential is
dw = wx dx + wy dy + wz dz
I.e. by symmetry
dw = −
dx dy dz
− 2 − 2
x2
y
z