Quiz 6
Name:
SOLUTIONS
Maths 114 - Calculus II
October 26, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1.
f (x, y) = arcsin(ex+y )
(a) On what set is the function f (x, y) continuous?
Both arcsin and the exponential function are continuous, so f (x, y) is
continuous everywhere it is defined. Therefore we need to find the domain
of f (x, y). The domain of arcsin is [−1, 1], so we want all (x, y) such that
−1 ≤ ex+y ≤ 1. The first inequality is always satisfied since 0 < ex+y , so
we only need to ensure that ex+y ≤ 1. Because ln is an increasing function,
this is the same as x + y = ln(ex+y ) ≤ ln(1) = 0. Therefore the set on
which f (x, y) is continuous is
{(x, y) ∈ R2 |x + y ≤ 0}.
(b) What is the range of f ?
As x + y ranges over all numbers ≤ 0, ex+y ranges over all numbers in the
interval (0, e0 ] = (0, 1]. Since arcsin is increasing and continuous,
f (x, y) = arcsin(ex+y ) ranges over the interval
(arcsin(0), arcsin(1)] = (0, π/2].
2. Suppose f (x, y) = x/y, x = u2 + v 2 , and y = uv. Find
u = v = 1.
∂f
∂u
and
∂f
∂v
when
We could do this by using the chain rule, or by first writing f as a function of u
2 +v 2
and v. In terms of u and v, the function is u uv
= uv −1 + vu−1 , so
∂f
∂f
= v −1 − vu−2 and ∂v = −uv −2 + u−1 . When u = v = 1, both partial
∂u
derivatives are zero.
Quiz 6
Name:
SOLUTIONS
Maths 114 - Calculus II
October 28, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1. Given that x3 + xy = xey + y 2 , use the following equation to find
dy
:
dx
dy
Fx
=− .
dx
Fy
We can rewrite the equation as x3 + xy − xey − y 2 = 0, so if we let
F (x, y) = x3 + xy − xey − y 2 , then
Fx
3x2 + y − ey
dy
=−
=−
.
dx
Fy
x − xey − 2y
2. Suppose you are climbing a hill whose shape is given by the equation
z = 10xy − y 4 − x2 ,
where x, y, and z are measured in meters, and you are standing at a point with
coordinates (2, 1, 15). The positive x-axis points east and the positive y-axis
points north. If you walk northwest, will you start to ascend or descend?
Let f (x, y) = 10xy − y 4 − x2 . Your rate of ascent with respect to horizontal
distance is given by the directional derivative of f (x, y) in the northwest
direction. A vector which points northwest is h−1, 1i, and we get a unit vector
pointing northwest by dividing it by its magnitude: u = √12 h−1, 1i.
The partial derivatives are fx (x, y) = 10y − 2x and fy (x, y) = 10x − 4y 3 , so the
gradient vector at the point (2, 1) is hfx (2, 1), fy (2, 1)i = h6, 16i. Therefore
√
1
1
Du f (2, 1) = ∇f (2, 1) · u = h6, 16i · √ h−1, 1i = √ (−6 + 16) = 5 2.
2
2
Because this is positive, you are ascending.