Statistics Questionnaire

Math 20C : Quiz 2 PracticeInstructions. You are allowed to consult your textbook or ebook, your notes, and the
lecture videos. Show all of your work. No credit will be given for unsupported answers,
even if correct. You are not allowed collaborate or communicate with any other
humans while working on this exam.
1
⟩ (for t ≥ 0).
t+1
(a) Find a parametrization of the tangent line to the curve at t = 1.
1. Consider the curve r(t) = ⟨t2 , t3 ,
Solution. At t = 1, the curve is at r(1) = ⟨1, 1, 1/2⟩. A tangent vector of the curve at
t = 1 is r′ (1) = ⟨2t, 3t2 , −(t + 1)−2 ⟩|t=1 = ⟨2, 3, −1/4⟩. Therefore, the tangent line can be
written as ⟨1, 1, 1/2⟩ + t⟨2, 3, −1/4⟩.
(b) Does the curve ever arrive at the point (1, 1, 1)? Explain.
Solution. The only possible time for the x-coordinate of r(t) to be 1 is when t2 = 1, i.e.
t = 1 (since t ≥ 0). Note that the z-coordinate when t = 1 is 1/2, which is not 1. Therefore,
the curve can never arrive at the point (1, 1, 1).
2. A particle is traveling on the path r2 (t), with acceleration vector ⟨6t, et , 0⟩. Initially at
t = 0, the particle is at the origin, with initial speed 4. Its initial velocity is in the positive
direction of ⟨1, 0, −1⟩ (i.e. it is a positive scalar multiple of this vector).
(a) What is the initial velocity of the particle?
√
Solution. The length of the vector ⟨1, 0, −1⟩ is 2. Thus the unit vector in the posi1
−1
tive direction of ⟨1, 0, −1⟩ is ⟨ √ , 0, √ ⟩. Therefore, the initial velocity of the particle is
2
2
−4
4
⟨ √ , 0, √ ⟩.
2
2
(b) Find r2 (t).
We first compute the velocity vector r′2 (t) via integrating the acceleration vector.
Z t
′
′
r2 (t) − r2 (0) =
a(u) du
0
r′2 (t)
4
−4
− ⟨ √ , 0, √ ⟩ = ⟨3t2 , et − 1, 0⟩
2
2
4
−4
r′2 (t) = ⟨ √ + 3t2 , et − 1, √ ⟩.
2
2
(The −1 on the y-coordinate of right side in the second line comes from plugging in the lower
bound 0 after integrating. )
Integrate the velocity to get the position vector r2 (t).
Z t
r′2 (u) du
r2 (t) − r2 (0) =
0
4
−4
r2 (t) − ⟨0, 0, 0⟩ = ⟨ √ t + t3 , et − t − 1, √ t⟩
2
2
4
−4
r2 (t) = ⟨ √ t + t3 , et − t − 1, √ t⟩.
2
2
x8 + x2 y + y 4
x6 + y 2
(a) Evaluate the limit
lim
3. Let f (x) =
f (x, y), or determine that it does not exist.
(x,y)→(−1,2)
Solution. By directly plugging in,
1 + 2 + 16
19
x8 + x2 y + y 4
=
=
.
(x,y)→(−1,2)
x6 + y 2
1+4
5
lim
(b) Evaluate the limit
lim
f (x, y), or determine that it does not exist.
(x,y)→(0,0)
Solution. On y = x3 , the limit is
x8 + x5 + x12
x2
1
x6
1
x8 + x2 x3 + (x3 )4
=
lim
=
lim
+
+
= lim ,
6
6
6
x→0
x→0 2
x→0 2x
x→0
x +x
2x
2x
2
lim
which does not exist. Therefore, the limit in question does not exist.
Math 20C worksheet: week 4
Sections covered: 13.5, 14.1, 14.2
Warmup:
i. Find the domain and range of f (x) =
p
9
x2
Solution:
Since we have a square root, we have the restriction 9 x2
0, which gives us the points x =
3, x = 3. Checking points, we find that the domain is: [ 3, 3]
9 x2 is a continuous parabola whose maximum value is 9. Since we established that the argument
p
p p
of 9 x2 is at least 0, we have that the range is: [ 0, 9] = [0.3]
ii. If r(t) = hsin t, cos ti, find r0 (⇡).
Solution:
We compute r(t) and input ⇡:
r0 (t) = hcos t,
sin ti =) r0 (⇡) = hcos(⇡),
iii. Find the limit along the line y = 0: lim(x,y)!(0,0)
sin(⇡)i = h 1, 0i
xy
x2 +y 2
Solution:
We substitute y = 0 and compute the resulting limit:
x(0)
0
=
lim
= 0
2
+0
(x,0)!(0,0) x2
lim
(x,0)!(0,0) x2
1
1. Find r(t) given that a(t) = h t+1
, e t , 6ti, v(0) = h1, 1, 2i and r(0) = h0, 0, 0i.
Solution:
We first integrate a(t) to get v(t):
v(t) =
Z
Z
1
h
, e t , 6tidt
t+1
⌧Z
Z
Z
1
t
=
dt, e dt, 6tdt
t+1
a(t)dt =
= hln(t + 1) + C1 , e
t
+ C2 , 3t2 + C3 i
We then use the initial condition v(0) = h1, 1, 2i:
v(0) = hln(0 + 1) + C1 , e
0
+ C2 , 3(0)2 + C3 i = hC1 , 1 + C2 , C3 i = h1, 1, 2i
=) hC1 , C2 , C3 i = h1, 0, 2i
=) v(t) = hln(t + 1) + 1, e t , 3t2 + 2i
We then integrate v(t) to get r(t):
1
2
r(t) =
Z
Z
hln(t + 1) + 1, e t , 3t2 + 2idt
⌧Z
Z
Z
t
=
ln(t + 1) + 1dt,
e dt, 3t2 + 2dt
v(t)dt =
Using integration by parts, we can find
1
du = t+1
and v = t
Z
R
ln(t + 1)dt. We let u = ln(t + 1) and dv = dt. This makes
Z
= t ln(t + 1)
t
dt
t+1
Z
t+1
1
dt
t+1 t+1
Z
1
1
dt
t+1
(t ln(t + 1))
= t ln(t + 1)
t + ln(t + 1) = (t + 1) ln(t + 1)
ln(t + 1)dt = t ln(t + 1)
= t ln(t + 1)
= t ln(t + 1)
t
So integrating v(t), we get:
r(t) =
Z
v(t)dt =
⌧Z
ln(t + 1) + 1dt,
= h(t + 1) ln(t + 1)
Z
t
e dt,
Z
t + t + C1 , e
= h(t + 1) ln(t + 1) + C1 , e
t
3t2 + 2dt
t
+ C2 , t3 + 2t + C3 i
+ C2 , t3 + 2t + C3 i
We then use the initial condition r(0) = h0, 0, 0i:
r(0) = h(0 + 1) ln(0 + 1) + C1 , e
0
=) hC1 , C2 , C3 i = h0, 1, 0i
+ C2 , (0)3 + 2(0) + C3 i = hC1 , 1 + C2 , C3 i = h0, 0, 0i
=) r(t) = h(t + 1) ln(t + 1), e
t
1, t3 + 2ti
2. Determine the range and sketch the domain and the graph of the function f (x, y) =
Solution:
p
x2
y2.
The range of the function is [0, 1) . Since the function is a square root function, the minimum
value would be 0, and we can vary x and fix y = 0 to achieve any non-negative number.
The domain of the function requires that the argument of the square root is non-negative. In other
words, we need:
x2
y2
0
Note that the graph of x2 y 2 = 0 is both the lines y = x and y =
intersects the origin. Pictured below:
x, which is a cross that
3
In order to find which regions of the graph are part of the domain, we can test points in each region.
(1, 0) : 12
02 = 1
( 1, 0) : ( 1)2
(0, 1) : 0
2
(0, 1) : 0
2
0
02 = 1
2
1 =
0
1