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Math 113 – Calculus III Exam 3 Practice Problems Fall 2005
- Suppose the motion of a particle is given by
x = 4 cos t, y = sin t.
(a) Describe the motion of the particle, and sketch the curve along which the particle
travels.
(b) Find the velocity and acceleration vectors of the particle.
(c) Find the times t and the points on the curve where the speed of the particle is
greatest.
(d) Find the times t and the points on the curve where the magnitude of the acceler-
ation is greatest.
- Find the coordinates of the points where the line
x = t, y = 1 + t, z = 5t
intersects the surface
z = x
2
2
.
- Let
g(x, y, z) = e
−(x+y)
2
2 (x + y).
Suppose that a piece of fruit is sitting on a table in a room, and at each point (x, y, z)
in the space within the room, g(x, y, z) gives the strength of the odor of the fruit.
Furthermore, suppose that a certain bug always flies in the direction in which the
fruit odor increases fastest. Suppose also that the bug always flies with a speed of 2
feet/second.
What is the velocity vector of the bug when it is at the position (2, − 2 , 1)?
- The path of a particle in space is given by the functions x(t) = 2t, y(t) = cos(t), and
z(t) = sin(t). Suppose the temperature in this space is given by a function H(x, y, z).
Find
dH
dt
, the rate of change of the temperature at the particle’s position. (Since the
actual function H(x, y, z) is not given, your answer will be in terms of derivatives of
H.)
- Let
f (x, y) = x
3
− xy + cos(π(x + y)).
(a) Find a vector normal to the level curve f (x, y) = 1 at the point where x = 1,
y = 1.
(b) Find the equation of the line tangent to the level curve f (x, y) = 1 at the point
where x = 1, y = 1.
(c) Find a vector normal to the graph z = f (x, y) at the point x = 1, y = 1.
(d) Find the equation of the plane tangent to the graph z = f (x, y) at the point
x = 1, y = 1.
- Suppose f is a differentiable function such that
f (1, 3) = 1, f x
(1, 3) = 2, f y
f xx
(1, 3) = 2, f xy
(1, 3) = − 1 , and f yy
(a) Find gradf (1, 3).
(b) Find a vector in the plane that is perpendicular to the contour line f (x, y) = 1 at
the point (1, 3).
(c) Find a vector that is perpendicular to the surface z = f (x, y) (i.e. the graph of
f ) at the point (1, 3 , 1).
(d) At the point (1, 3), what is the rate of change of f in the direction
i +
j?
(e) Use a quadratic approximation to estimate f (1. 2 , 3 .3).
- We say that a line in 3-space is normal to a surface at a point of intersection if the
line is normal to (i.e. perpendicular to) the tangent plane of the surface at that point.
Let S be the surface defined by
x
2
2
2
= 4.
(a) Find the parametric equations of the line that is normal to the surface S at the
point (1, 1 , 1).
(b) The line found in (a) will intersect the surface S at two points. One of them is
(1, 1 , 1), by construction. Find the other point of intersection.
- Let
f (x, y) = (x − y)
3
2
− y.
(a) Find the function L(x, y) that gives the linear approximation of f near the point
(b) Find the function Q(x, y) that gives the quadratic approximation of f near the
point (1, 2)
SOLUTIONS
- (a) Note that (x/4)
2
2 = cos
2 t + sin
2 t = 1, so the path of the particle is the ellipse
(x/4)
2
2
= 1. The motion is counter-clockwise around the ellipse.
y
x
–1.
–0.
0
1
–4 –2 2 4
(b) The path is ~r(t) = (4 cos t)
i + (sin t)
j, so the velocity is
~v(t) = ~r
′
(t) = (−4 sin t)
i + (cos t)
j
and the acceleration is
~a(t) = ~r
′′
(t) = (−4 cos t)
i + (− sin t)
j.
(c) The speed is
v = ‖~v(t)‖ =
(−4 sin t)
2
2
16 sin
2
t + cos
2 t.
This has a maximum or minimum when dv/dt = 0. Now
dv
dt
15 sin t cos t
16 sin
2
t + cos
2 t
so dv/dt = 0 when sin t = 0 or cos t = 0. This means t = nπ or t = π/2 + nπ,
where n is any integer. When t = nπ, v = 1, and when t = π/2 + nπ, v = 4.
Thus the maximum speed is v = 4. It occurs when t = π/2 + nπ. When n is even,
the point in the plane where this occurs is (0, 1), and when n is odd, the point is
(d) The magnitude of the acceleration is
a = ‖~a(t)‖ =
(−4 cos t)
2
2
16 cos
2 t + sin
2
t.
This has a maximum or minimum when da/dt = 0. Now
da
dt
−15 sin t cos t
16 cos
2 t + sin
2
t
so da/dt = 0 when sin t = 0 or cos t = 0. This means t = nπ or t = π/2 + nπ,
where n is any integer. When t = nπ, a = 4, and when t = π/2 + nπ, a = 1. Thus
the maximum magnitude of the acceleration is a = 4. It occurs when t = nπ.
When n is even, the point in the plane where this occurs is (0, 4), and when n is
odd, the point is (0, −4).
- We have x = t, y = 1 + t and z = 5t, and we want z = x
2
2 , so we must solve
5 t = t
2
2
=⇒ 2 t
2
− 3 t + 1 = 0 =⇒ t =
So the line intersects the surface when t = 1/2 and when t = 1. When t = 1/2, the
point of intersection is (1/ 2 , 3 / 2 , 5 /2), and when t = 1, the point of intersection is
- Since the bug flies in the direction in which the fruit odor increases fastest, it flies in
the direction of grad g. The gradient of g is
grad g(x, y, z) =
−2(x + y)e
−(x+y)
2
2
i +
−2(x + y)e
−(x+y)
2
2
j
k,
and
grad g(2, − 2 , 1) =
i +
j.
The bug always has a speed of 2, so the velocity vector must have a magnitude of 2.
A vector with magnitude 2 and in the same direction as the gradient is
grad g(2, − 2 , 1)
‖grad g(2, − 2 , 1)‖
i +
j).
- By the chain rule,
dH
dt
∂H
∂x
dx
dt
∂H
∂y
dy
dt
∂H
∂z
dz
dt
∂H
∂x
− sin t
∂H
∂y
∂H
∂z
- (a) The gradient of f is normal to the level curve at each point. We find
grad f (x, y) = (3x
2
− y − π sin(π(x + y)))
i + (−x − π sin(π(x + y)))
j,
and
grad f (1, 1) = 2
i −
j,
so one possible answer is 2
i −
j.
(b) The line is
2(x − 1) − (y − 1) = 0, or 2 x − y = 1.
(b) We can find the points by first finding the values of t at which the line intersects
the surface x
2 +y
2 +2z
2 = 4. Plugging the parametric equations into the equation
of the surface, we have
(1 + 2t)
2
2
2
= 4
40 t
2
t(5t + 3) = 0
so t = 0 or t = − 3 /5. At t = 0, the parametric equations of the line give the
point (1, 1 , 1), which is the point we already knew. At t = − 3 /5, the parametric
equations of the line give (− 1 / 5 , − 1 / 5 , − 7 /5). This is the other point that we
want.
- (a) First get the numbers:
f (1, 2) = −1 + 4 + 1 − 2 = 2,
f x
(x, y) = 3(x − y)
2
f y
(x, y) = −3(x − y)
2
Then
L(x, y) = f (1, 2) + f x
(1, 2)(x − 1) + f y
(1, 2)(y − 2)
= 2 + 9(x − 1) − 2(y − 2).
(b) We need some more numbers:
f xx
(x, y) = 6(x − y) + 2, f xx
f xy
(x, y) = −6(x − y) + 2, f xy
f yy
(x, y) = 6(x − y), f yy
Then
Q(x, y) = L(x, y) +
f xx
(x − 1)
2
(1, 2)(x − 1)(y − 2) +
f yy
(y − 2)
2
= 2 + 9(x − 1) − 2(y − 2) − 2(x − 1)
2
- 8(x − 1)(y − 2) − 3(y − 2)
2
.
- (a) This function is not differentiable on the circle x
2
2
= 1. The graph has a
“corner” at these points.
(b) This function is not differentiable at the origin. Consider the cross section y = 0:
f (x, 0) = (x
2
)
1 / 4
=
|x|. The graph has a cusp (i.e. a point) at x = 0.
(c) This function is the composition of polynomials and the exponential function, so
it is differentiable everywhere.
(d) This function is not differentiable at points where the denominator is zero; that
is, where x
2
= y
2
. This gives the lines y = x and y = −x.
- (a)
dw
dt
= Q
x
(x, y, z)
dx
dt
+ Q
y
(x, y, z)
dy
dt
+ Q
z
(x, y, z)
dz
dt
= Qx(f (t), g(t), h(t))f
′
(t) + Qy(f (t), g(t), h(t))g
′
(t) + Qz (f (t), g(t), h(t))h
′
(t)
= Q
x
(~r(t))f
′
(t) + Q y
(~r(t))g
′
(t) + Q z
(~r(t))h
′
(t)
(b) Since
grad Q(x, y, z) = Q x
(x, y, z)
i + Q y
(x, y, z)
j + Q z
(x, y, z)
k,
we have
grad Q(~r(t)) = Q x
(~r(t))
i + Q y
(~r(t))
j + Q z
(~r(t))
k,
Also,
d~r
dt
= f
′
(t)
i + g
′
(t)
j + h
′
(t)
k
so we have
dw
dt
= Q
x
(~r(t))f
′
(t) + Q y
(~r(t))g
′
(t) + Q z
(~r(t))h
′
(t)
= (grad Q(~r(t)) ·
d~r
dt
