MATH 1020 WORKSHEET 13.1
Vector Functions
Vector-valued functions are of the form ~r(t) = f (t) ı̂ + g(t) ̂ or ~r(t) = f (t) ı̂ + g(t) ̂ +
h(t) k̂. These can also be denoted with angle bracket notation.
Find the domain of ~r(t) =
√
4 − t2 ı̂ + t2 ̂ − 6t k̂.
Solution. To determine the domain, we first look at the domain of each component.
√
ı̂ − component :
4 − t2
defined for 4 − t2 ≥ 0 or − 2 ≤ t ≤ 2
̂ − component : t2
defined for all t
k̂ − component :
−6t
defined for
all t
The domain for the vector function is all t-values where all three components are
defined. So we have that the
Domain of ~r(t) = {t : −2 ≤ t ≤ 2} .
Find ~r(t) · ~u(t) given that ~r(t) = h3 cos t, 2 sin t, t − 2i and ~u(t) = h4 sin t, −6 cos t, t2 i.
Solution. We take the dot product of vector functions in the same way that we took
the dot product of vectors in 12.3.
~r(t) · ~u(t) = h3 cos (t), 2 sin (t), t − 2i · h4 sin (t), −6 cos (t), t2 i
= 12 cos (t) sin (t) − 12 cos (t) sin (t) + (t − 2)t2
= t3 − 2t2
Thus the dot product gives us a scalar function
~r(t) · ~u(t) = t3 − 2t2 .
Evaluate
1
t2 − 4
̂ + k̂
lim t ı̂ + 2
t→2
t − 2t
t
Solution. To determine the limit of a vector function, we take the limits of the
individual components.
ı̂ − component :
lim t = 2
t→2
0
t2 − 4
=
2
t→2 t − 2t
0
2
t −4
(t − 2)(t + 2)
lim 2
= lim
t→2 t − 2t
t→2
t(t − 2)
4
(t + 2)
= =2
lim
t→2
t
2
1
1
k̂ − component: : lim =
t→2 t
2
̂ − component :
lim
← indeterminate form
Putting the answers to each component back into a vector form we get that
1
1
t2 − 4
̂ + k̂ = h2, 2, i .
lim t ı̂ + 2
t→2
t − 2t
t
2
MATH 1020 WORKSHEET 13.2
Derivatives and Integrals of Vector Functions
When we differentiate or integrate vector-valued functions, we do so term by term or component by component. Don’t forget that when we have indefinite integration of a vector
function, each component has a different constant of integration.
Find ~r0 (t) given that ~r(t) =
1
t
ı̂ + 16t ̂ +
t2
2
k̂.
Solution. Differentiating term by term and keeping everything in vector form we have
~r 0 (t) = −
1
ı̂ + 16 ̂ + t k̂
t2
or
~r 0 (t) = h−
1
, 16, ti
t2
Find Dt [~r(t) × ~u(t)] given that ~r(t) = ht2 , sin t, cos ti and ~u(t) = h1/t2 , sin t, cos ti.
Solution. First we set up our matrix to take the cross product of the vector functions ~r(t)
and ~u(t).
ı̂
̂
k̂
2
~r × ~u = t
sin (t) cos (t) = (sin (t) cos (t) − sin (t) cos (t)) ı̂ − t2 cos (t) − 1/t2 cos (t) ̂
1/t2 sin (t) cos (t) + t2 sin (t) − 1/t2 sin (t) k̂
= h0, − cos (t) t2 − 1/t2 , sin (t) t2 − 1/t2 i
Taking the derivative of ~r × ~u component by component we get
Dt [~r(t) × ~u(t)] = h0, sin (t) t2 − 1/t2 − cos (t) 2t + 2/t3 , cos (t) t2 − 1/t2 + sin (t) 2t + 2/t3 i .
Evaluate
Z √ (2t − 1) ı̂ + 4t3 ̂ + 3 t k̂ dt
Solution. Integrating component by component we get
Z
Z
Z
Z √ √
(2t − 1) dt ı̂ +
(2t − 1) ı̂ + 4t3 ̂ + 3 t k̂ dt =
4t3 dt ̂ +
3 t dt k̂
= t2 − t + C1 ı̂ + t4 + C2 ̂ + 2t3/2 + C3 k̂
~
or ht2 − t, t4 , 2t3/2 i + C
~ = hC1 , C2 , C3 i is a constant vector.
where C
MATH 1020 WORKSHEET 13.3
Vector Functions: Arc Length and Curvature
Tangent and Normal Vectors
Find the length of ~r(t) = 2t ı̂ − 3t ̂ + t k̂ on the interval 0 ≤ t ≤ 2.
Solution. To determine the arc length we must first find the velocity vector ~v(t) or ~r 0 (t)
and its magnitude.
~r 0 (t) = h2, −3, 1i
√
0 √
~r (t) = 4 + 9 + 1 = 14
Using the formula for arc length we have that
Z 2√
2
√
√
L=
14 dt = 14 t = 2 14
0
0
Find the curvature of ~r(t) = h t2 , sin (t) − t cos (t) , cos (t) + t sin (t) i.
Solution. To find the curvature, we first find the unit tangent vector by finding ~r 0 (t) and
its magnitude.
~r 0 (t) = h2t, cos (t) − cos (t) + t sin (t), − sin (t) + sin (t) + t cos (t)i
0 q
~r (t) = 4t2 + t2 sin2 (t) + t2 cos2 (t)
√
p
√
= t2 (4 + sin2 (t) + cos2 (t)) = 5t2 = t 5
Applying the formula for the unit tangent vector we get
h2t, t sin (t), t cos (t)i
1
~
√
T(t)
=
= √ h2, sin (t), cos (t)i
t 5
5
Using the curvature formula involving the derivative the the unit Tangent vector we find
that
~ 0 (t) = √1 h0, cos (t), − sin (t)i
T
5
q
0 1
1
~
T (t) = √
cos2 (t) + sin2 (t) = √
5
5
0 1
√
~
T (t) 1
= √5 =
κ= ~r 0 (t) 5t
t 5
Thus
κ=
1
.
5t
Find the unit tangent vector T~ for ~r(t) = t ı̂+t2 ̂+t k̂ and find a set of parametric equations
for the tangent line at P (0, 0, 0).
Solution. Determining the unit Tangent vector we have
~r 0 (t) = h1, 2t, 1i
p
0 p
~r (t) = 1 + 4t2 + 1 = 4t2 + 2
h1, 2t, 1i
~
T(t)
=√
4t2 + 2
The t−value that corresponds to P (0, 0, 0) is t = 0, thus the direction numbers for the
tangent line are vbr 0 (0) = h1, 0, 1i and the parametric equations for the tangent line are
x=t
y=0
z=t
Verify that the space curves intersect at the given values and find the angle between the
tangent vectors to the curves at the point of intersection. ~r(t) = ht − 2, t2 , t/2i t = 4 and
√
~u(s) = hs/4, 2s, 3 si s = 8.
Solution. To verify that the space curves intersect, we evaluate them at the given parameter
values and see if the resulting position vector is the same, i.e. they both include the same
point.
4
~r(4) = h4 − 2, 42 , i = h2, 16, 2i
2√
3
~u(8) = h8/4, 2(8), 8i = h2, 16, 2i
Since these two position vectors are the same, the curves intersect at the point (2, 16, 2).
Next we determine the tangent vectors to each curve at the point of intersection.
1
~r 0 (t) = h1, 2t, i
2
1
1 −2/3
0
~u (s) = h , 2, s
i
4
3
1
~r 0 (4) = h1, 8, i
2
1
1
0
~u (8) = h , 2, i
4
12
Thus the angle between the tangent vectors to each curve is given by
~r 0 (t) · ~u 0 (s)
cos (θ) = ~r 0 (t) ~u 0 (s) 1
1
1
1 14 + 8(2) + 21 12
+ 16 + 24
q
q
=q
=q 4
1
1
4+64·4+1
9+4·144+1
1 + 64 + 14 16
+ 4 + 144
4
144
6+16·24+1
24
q
=q
261
4
586
144
=
√
391
24√
261 586
2
12
=√
391
√
261 586
Thus
cos (θ) = √
391
√
.
261 586
MATH 1020 WORKSHEET 13.4
Motion in Space: Velocity and Acceleration
Given the position function ~r(t) = t2 ı̂+t ̂+2t3/2 k̂, find the velocity, speed and acceleration
of the object.
Solution.
~v(t) = ~r 0 (t) = 2t ı̂ + 1 ̂ + 3t1/2 k̂
p
speed = ~v = 4t2 + 1 + 9t
3
~a(t) = ~r 00 (t) = h2, 0, √ i
2 t
Given the acceleration function ~a(t) = t ̂ + t k̂, find the velocity and position vectors when
~v(1) = 5 ̂ and ~r(1) = ~0. Evaluate the position vector at t = 2.
Solution. Integrating and using ~v(1) = 5 ̂ we find ~v(t)
Z
t2
t2
~v = ~a dt = hc1 , + c2 , + c3 i
2
2
1
1
~v(1) = hc1 , + c2 , + c3 i = h0, 5, 0i ← Solving for constants
2
2
t2 9 t2 1
we get ~v(t) = h0, + , − i
2
2 2
2
Integrating ~v(t) and using ~r(1) = ~0 we find ~r(t)
Z
t3 9
t3
~r = ~v dt = hc4 , + t + c5 , − f rac12t + c6 i
6
2
6
1 1
1 9
~r(1) = hc4 , + + c5 , − + c6 i = h0, 0, 0i ← Solving for constants
6 2
6 2
t3 9
14 t3 1
1
we get ~r(t) = h0, + t − , − t + i
6
2
3 6
2
3
Lastly, evaluating our resulting position vector when t = 2 we have that
~r(t) = h0,
23 9
14 23 1
1
4 27 14 4 3 1
17 2
+ 2− ,
− 2 + i = h0, +
− , − + i = h0, , i
6
2
3 6
2
3
3
3
3 3 3 3
3 3
Find the vector-valued functions for the acceleration, velocity and position of a projectile
launched at a height of 10 feet above the ground with an initial velocity of 88 feet per second
and at an angle of 30o above the horizontal.
Solution. Starting with the acceleration vector, we integrate and use information in our
problem to solve for the constants of integration.
~a(t) = h0, −32ift/sec
√
~v(t) = h0, −32ti + v~0 = h0, −32ti + h88 cos 30o , 88 sin 30o i = h44 3, −32t + 44i
~r(t) = h0, −16t2 i + v~0 t + r~0 =
← r~0 depends on where you place origin
√
Origin at ground : ~r(t) = h44 3t + 0, −16t2 + 44t + 10i
√
Origin at launch height : ~r(t) = h44 3t + 0, −16t2 + 44t + 0i
You can use either vector as your position vector, you just need to be consistent throughout
a given problem.
A baseball,√hit 3 feet above the ground, leaves the bat at an angle of 45o with an initial
speed of 72 2 ft/sec. How high does the ball rise above the ground? If the outfield fence
is 350 feet away and 10 feet tall, is it a homerun?
Solution. Starting with the acceleration vector, we integrate and us information in our
problem to solve for the constants of integration.
~a(t) = h0, −32ift/sec
√
√ 2
2
~v(t) = h0, −32ti + h72 2 cos 45 , 72 2 sin 45 i = h72 2
, −32t + 72 2
i = h72, −32t + 72i
2
2
~r(t) = h72t + x0 , −16t2 + 72t + y0 i
√
o
√
o
√
√
Placing the origin at the point where the ball is hit, we have that x0 = 0 and y0 = 0. To
solve for the maximum height of the ball, we find when the y component of the velocity
vector is zero
vy = 0
−32t + 72 = 0
72
9
t=
=
32
4
secs
Thus we have that the maximum height of the ball is
9
9
rx ( ) = 72
= 18 · 9 = 162 feet
4
4
To determine if the ball clears the fence and is a homerun, we set the x component of the
position vector equal to 360 feet and solve for t.
360 = 72t
→t=
360
72
→t=5
secs
Now all that remains to determine the height of the ball when it reaches the fence and
compare it to 10 − 3 = 7f eet. Note that our origin is 3 feet above the ground, so the ball
only needs to be 7 feet with respect to our origin to clear the fence.
y(5) = −16(5)2 + 72(5) = −16(25) + 360 = −400 + 360 = −40
feet
Since the answer here is negative, the ball never reaches the fence so it is not a homerun
hit.
MATH 1020 WORKSHEET 14.1
Functions of Several Variables
Find the domain and range of g(x, y) =
1
xy .
Solution. We observe that xy 6= 0 for the g(x, y) to be defined, so we have that x 6= 0 and
y 6= 0. Thus for the domain we can write:
D = {(x, y) : x 6= 0, y 6= 0}
or that the domain is the set of all points (x, y) such that x 6= 0 and y 6= 0.
For the range, we see that g(x, y) 6= 0 but can take on all other values. Thus we write the
range is −∞ < g(x, y) < 0 and 0 < g(x, y) < ∞.
Describe the level curves of the function f (x, y) = x2 + y 2 . Sketch the level curves for the
values c = 0, 2, 4, 6, 8.
Solution.
The level curves are circles of increasing radius.
c=0⇒
0 = x2 + y 2
c=2⇒
2 = x2 + y 2
c=4⇒
2
4=x +y
c=6⇒
6 = x2 + y 2
c=8⇒
8 = x2 + y 2
2
r=
√
2
r=2
√
r= 6
√
r=2 2
Describe the level surfaces of the function f (x, y, z) = x2 + 3y 2 + 5z 2 .
Solution. Setting the function f (x, yz) equal to an arbitrary constant we have
x2 + 3y 2 + 5z 2 = c
x2 y 2 z 2
+ c + c =1
c
3
5
For each value of c we have the equation of an ellipsoid. So the level surfaces of the function
f (x, y, z) are nested ellipsoids.
MATH 1020 WORKSHEET 14.3
Partial Derivatives
When taking partial derivatives, you consider all variables other than the one related to
your derivative as constants.
Find both first partial derivatives for z = x2 e2y .
Solution.
zx = 2xe2y
zy = x2 e2y (2)
or
2x2 e2y
Find the slopes of the surface z = e−x cos y in the x- and y-directions at the point P (0, 0, 1).
Solution. To find the slope in the x-direction, we find zx and evaluate the derivative when
x = 0 and y = 0.
zx = −e−x cos (y)
zx (0, 0) = −e0 cos (0) = −1(1) = −1
To find the slope in the y-direction, we find zy and evaluate the derivative when x = 0 and
y = 0.
zy = e−x (− sin (y)) = −e−x sin (y)
zy (0, 0) = −e0 sin (0) = −1(0) = 0
Find the four second partial derivatives for z = ln (x − y).
Solution. In order to find the second partial derivatives, we must first find both first partial
derivatives.
1
zx =
(1) = (x − y)−1
x−y
1
zy =
(−1) = −(x − y)−1
x−y
Now we will find the second partials that require zx :
1
(x − y)2
1
= (zx )y = −1(x − y)−2 (−1) =
(x − y)2
zxx = (zx )x = −1(x − y)−2 (1) = −
zxy
Next we will find the second partials that require zy :
1
(x − y)2
1
= (zy )y = 1(−1)(x − y)−2 (−1) = −
(x − y)2
zyx = (zy )x = −(−1)(x − y)−2 (1) =
zyy
Note that zxy = zyx .
Given f (r, s, t) = r sin (st) + st2 er , find frs , frst , and ftrs .
Solution. To find the first two partial derivatives requested, we first must find fr .
fr = sin (st) + st2 er
frs = (fr )s = t cos (st) + t2 er
frst = (frs )t = cos (st) + t (− sin (st)(s)) + 2ter
= cos (st) − st sin (st) + 2ter
Recalling that for the functions that we encounter in Calculus, mixed partials are equal
if they have the same number of derivatives for each variable, regardless of order, we can
write:
ftrs = cos (st) − st sin (st) + 2ter
or we can do the work as follow below:
ft = rs cos (st) + 2ster
ftr = s cos (st) + 2ster
ftrs = cos (st) + s (− sin (st)(t)) + 2ter
= cos (st) − st sin (st) + 2ter
We note that as expected frst = ftrs .