Download 1. Recall: tan(x) = sin(x) cos(x) cot(x) = cos(x) sin(x) sec(x) = 1 cos(x

1. Recall:
tan(x) =
sin(x)
cos(x)
cot(x) =
cos(x)
sin(x)
sec(x) =
1
cos(x)
csc(x) =
1
sin(x)
(a) Sketch the graph of each of these functions.
(b) Using the di↵erentiation rules that you’ve learned, calculate the derivatives. (For
this problem, simplifying your answer really helps.)
Solution:
(a)
(b)
2. Find the equation to the tangent line of y = x + tan(x) at the point (⇡, ⇡).
Solution:
To find the slope of the tangent line, we must first take the derivative
of y = x + tan(x) and evaluate it at x = ⇡ To that end:
dy
dy
dy
(x + tan(x)) =
(x) + (tan(x))
dx
dx
dx
2
= 1 + sec (x)
And so, at x = ⇡, the slope of the tangent line is 1 + sec2 (⇡) = 1 + ( 1)2 = 2.
Given the slope of the tangent line is 2, and the point (⇡, ⇡) is on the line, we can
calculate the equation for the tangent line:
⇡ = 2(⇡) + b ) b = ⇡
y = 2x ⇡
2⇡ =
⇡ Thus, the equation for the tangent line is
2
3. A mass on a spring vibrates horizontally on a smooth level surface (see image). Its
equation of motion is x(t) = 8sin(t), where t is in seconds and x is in centimeters.
(a) Find the velocity and acceleration at time t.
(b) Find the position, velocity, and acceleration of the mass at time t =
direction is it moving in that time?
2⇡
.
3
In what
Solution:
(a) The velocity is the rate of change of distance/position with respect to time, and
acceleration is the rate of change of velocity with respect to time. Thus, formally,
we can write that:
v(t) = x0 (t)
a(t) = v 0 (t) = x00 (t)
Thus, for x(t) = 8sin(t), we have that the velocity at time t is given by:
v(t) = x0 (t) = 8cos(t)
and the acceleration at time t is given by:
a(t) = x00 (t) = v 0 (t) =
3
8sin(t)
(b) At t =
2⇡
,
3
we have that the position is given by:
✓
2⇡
x
3
◆
✓
2⇡
= 8sin
3
◆
✓p ◆
p
3
=8
=4 3
2
The velocity is given by:
✓
2⇡
v
3
◆
=x
0
✓
2⇡
3
◆
✓
2⇡
= 8cos
3
◆
=8
✓
1
2
◆
=
4
The acceleration is given by:
✓
2⇡
a
3
◆
=x
00
✓
2⇡
3
◆
Notice that the velocity at t =
negative direction.
✓
2⇡
8sin
3
=
2⇡
3
◆
=
✓p ◆
3
8
=
2
p
4 3
is negative; hence, the mass is moving in the
4
4. Find constant A and B such that the function y = Asin(x) + Bcos(x) satisfies the
di↵erential equation y 00 + y 0 2y = sin(x)
Solution:
y are.
In order to solve this problem, we first need to figure out what y 0 and
00
For y = Asin(x) + Bcos(x) we have:
y 0 = Acos(x)
Bsin(x)
And
y 00 =
Asin(x)
Bcos(x)
Thus,
y 00 + y 0
2y = ( Asin(x) Bcos(x)) + (Acos(x) Bsin(x))
= ( 3A B)sin(x) + (A 3B)cos(x)
2(Asin(x) + Bcos(x))
In order for y = Asin(x) + Bcos(x) to satisfy the di↵erential equation y 00 + y 0
sin(x), we must have:
3A B = 1
A 3B = 0
Solving this linear system of equations, we get that A =
5
3
10
and B =
1
.
10
2y =