MATH 1225 Test 2 Free Response Solution - Spring 2020, VT

This document presents a complete solution to the MATH 1225 Test 2, a free-response exam from Spring 2020. The solution demonstrates the application of calculus concepts, including derivatives, limits, and related rates. The problems cover topics such as finding the derivative of a complex function, applying the chain rule, evaluating limits using trigonometric functions, determining the acceleration of a particle, and calculating the rate of change of volume and surface area of a balloon. The solutions are detailed, showing each step of the process and providing clear explanations. This resource is designed to help students understand the concepts tested and improve their problem-solving skills in calculus. The solution is a valuable resource for students preparing for similar exams or seeking to deepen their understanding of calculus principles.

MATH 1225 TEST 2
[DATE]

1
Problem 1
Now,

2
Problem 2
Requisite table
h(x) = 3x + eπ + f (g(x))
g(x + 4)
h′ (x) = d
dx ( 3x + eπ + f (g(x))
g(x + 4) )
h′ (x) = g(x + 4) d
dx (3x + eπ + f (g(x)) + (3x + eπ + f (g(x)) d
dx (g(x + 4)
g (x + 4)2
h′ (x) = g(x + 4) (ln(3). 3x + 0 +d
dxf (g(x)) + (3x + eπ + f (g(x)) (g(x + 4)1
g (x + 4)2
h′ (x)
= g(x + 4) (ln(3). 3x + f(x)f′ (g(x))g′ (x) − f(g(x)f′(x) + (3x + eπ + f (g(x)) (g(x + 4))
g (x + 4)2
Put x=-1
h′ (x)
= g(−1 + 4) (ln(3). 3−1 + f(−1)f′ (g(−1))g′ (−1) − f(g(−1))f′(−1) + (3−1 + eπ + f (g(−1)) (g(−1 + 4)
g (−1 + 4)2
h′ (x) = (−1) (ln(3). 3−1 + 0 ∗ f′ (3) ∗ 3 − (−2)(−1) + (3−1 + eπ + (−2) (−1)
(−1)2

3
h′ (x) = −1 (
1
3 ln(3) − 2) + (
1
3 + 2 + eπ)
1
h′ (x) = 2 −ln3
3 + 7
3 + eπ = 1
3ln(3) + 13
3 + eπ = 27.1078
Problem 3
y = f(x)
Domain = (-4, 7]
g(x) = f(2x − 2) . f′ (x + 4)
g′ (1) =?
Now,
g′ (x) = (f(2x − 2))′ . f′ (x + 4) + f(2x − 2). (f′(x + 4))′
g′ (x) = 2f′ (2x − 2). f′ (x + 4) + f(2x − 2). f′′ (x + 4)
g′ (1) = 2f′ (0). f′ (5) + f(0). f′′(5)
From the graph,
f′′ (5) = 0
g′ (1) = 2f′ (0). f′ (5)
The tangent line at x=0 is clearly horizontal.

4
f′ (0) = 0
g′ (1) = 2f′ (0)f′ (5)
g′(1) = 0
Problem 4
Limit Definition of derivative

5

6
Problem 5
lim
t → 0t3 csc2 2t cot(7t)
= lim
t → 0
t3. 1
sin2(2t) ( (cos(7t)
sin(7t) )
= lim
t → 0
. 1
sin2(2t) 1
t2
( 1
(sin 7t
t )
) (cos(7t)
= lim
t → 0
. 1
sin2(2t) 1
(4t2/4) (
1
sin 7t
7t
7 )
(cos(7t)
= lim
t → 0
. 1
4 (sin(2t)
2t )
2
1
7( 1
sin 7t
7t
) (cos(7t)
As,
lim
t → 0
.sin(ax)
ax = 1

7
= lim
t → 0
. 1
4 (sin(2t)
2t )
2
1
7( 1
sin 7t
7t
) (cos(7t) = 1
4(1)2
1
7(1) cos(7(0))
lim
t → 0t3 csc2 2t cot(7t) = 1
28cos(0) = 1
28
Problem 6
The y-coordinate at any time t of particle’s movement is given below.
y(t) = 1
3 t3 − 7
2 t2 + 10 t
The first derivative would provide the velocity of the particle.
y′ (t) = d
dt(1
3 t3 − 7
2 t2 + 10 t)
y′ (t) = 3t2
3 − 7
2 ∗ 2t + 10 = t2 − 7t + 0
y′ (t) = t2 − 7t
The second derivative would provide the acceleration of the particle.
y′ (y′ (t)) = d
dt(t2 − 7t)
y′′ (t) = 2t − 7…….. (1)
This equation would be used to determine the acceleration of the moving particle. As the task
is to find the interval at which the particle would be slowing down and hence, the equation
(1) must be negative or should say would be lesser than zero. Therefore,
2t − 7 < 0
2t − 7 + 7 < 0 + 7
2t < 7
t < 7
2

8
t < 3.5
Hence, the interval would be (0, 3.5].
Problem 7
y = xsin2(x)
The student’s answer is not correct because he has made a mistaken while differentiating the
function sin^2(x).
Problem 8
Rate of change of surface area of balloon dS
dt = 0.1cm2
min
Diameter of balloon = 10 cm
Rate of change of volume =?
Here,
Volume of spherical balloon =
4
3π r3
Surface area of spherical balloon = 4 π r2
Now,
S = 4 π r2
Rate of change of surface area

9
dS
dt = d
dt (4 π r2)
dS
dt = 4 π ∗ 2r
dr
dt
dS
dt = 8πrdr
dt
0.1 = 8πr
dr
dt
dr
dt = 0.1
8πr (eq. 1. )
Similarly,
Rate of change of volume
V =4
3π r3
dV
dt = d
dt (4
3π r3)
dV
dt = 4
3 π3r2dr
dt
dV
dt = 4π r2 dr
dt
From equation 1
dV
dt = 4π r2 0.1
8πr =0.1 r
2
Radius of balloon = 5 cm
Hence,
dV
dt = 0.1 ∗
5
2 = 0.25 cm3/min
Therefore, rate of change of volume would be 0.25 cm^3/min.