Calculus Applications--SOLUTIONS
_ FAMAT State Convention 2017
1. C—Let x be the amount of old road restored. Then the length of the new road is
9  (5  x) 2 . So, the total
cost of the plan is C ( x)  200000 x  400000 x 2  10 x  34 . The minimum cost occurs at one of the critical
points which are x  5  3 . Clearly 5  3 is not a valid answer and 5  3 is indeed a minimum.
2. A— The mouse can wait some amount of time while the table rotates and then spend the remainder of the
time moving along that ray at 1 m/s. He can reach any point between the starting point and the furthest

reachable point along the ray, (1 − θ/π) meters out. So the area is
1  

0 2 1    d  6 .
2
3. D— The volume is

2


2
2
1
1 2
1 
1 2
1 
97 2
 2
 4
 2
2
.

sin
x

dx


sin
x

sin
x

dx


sin
x
(1

cos
x
)

sin
x

dx




 


10 
5
100 
5
100 
200

 
 
2
2
2
2
4. B—The intersection points of the 2 curves are (1, 1) and (1, -1). The length of a side of the triangle is 2 y
3 x
where y  x for x  1 and y 
for x  1. So, the volume is
2
1

0
2
3
3(2 x ) 2
3  3 x 
3 3
dx  
.
 2
 dx 
4
4 
2 
2
1
36 6

x
x . Using implicit
5. B—Call the angle you want to maximize  . Note that tan   tan( A  B) 
36 6
1 
x x
2
42 x  9072
differentiation, you get  /  cos 2  
. The max occurs at x  6 6.
( x 2  216)2
6. B—The speed will cancel out, so you can assume it is 1. So,
t 1
r 2
1
1
1  5
 r 1 
r2
r t dt  2 r1 t dt  ln  r   2 ln  r  1   r  2 . Use the positive case only.
7. D—The graph is a figure 8. By symmetry, the area it encloses is four times the area under the curve that we
get from just considering 0  t 

2
. For these values of t, we can write y  2sin t cos t  2 x 1  x 2 . This can be
integrated by letting u  1  x 2 . This gives an area of
2
8
. So, the total area is 4 times that  .
3
3
8. B—The points of inflection occur when − cos x = 0, so x =
 3
,
. At

, the equation of the tangent line is
2
2 2

3
3
y   x  . At
, the equation of the tangent line is y  x  . The intersection is at x   . At this point,
2
2
2



3
. So, the height is . The x-intercepts of the tangent lines are and
both tangent lines equal
, which
2
2
2
2
2
 1   
are separated by  . So, the area of the triangle is      
.
4
 2  2 
Calculus Applications—SOLUTIONS p2 _ FAMAT State Convention 2017
9. E (20)— We want to maximize f ( x ) when sin (x) ≥ 0 and minimize f ( x ) when sin (x) < 0. We do this by
setting f ( x ) = 5 in the first case and f ( x ) = −5 in the second case. Noting that the bounds of integration cover
precisely one full period of sin, we see that the integral becomes equivalent to twice the integral of 5 sin (x) over
the half period where sin (x) ≥ 0. This results in 20.
10. A—The slope of the tangent line is 2a, so the tangent equation is y  (a 2  1)  2a( x  a). Setting
x  y  0 gives  a 2  1  2a 2  a  1.
11. A—The units of accumulation are the product of the units of the dependent and independent variables. So,
 liters 

  km  liters
 km 
12. A—The y  component of the acceleration is y // . y /   cos  t  y //   ( sin  t )   2 y.
13. B—Catch =C = x(15  x)  x . To maximize C, we find C / and set = to zero. C /  14  2 x  0@ x  7.
Checking this in the 2nd derivative shows this is a max.
14. D—At height x above his initial position, the ninja must exert a force of 5  0.2x to balance his own weight
30
and the weight of the rope below that point. So, the work, W 
 (5  0.2 x)dx  240.
0
1
1
dV
dh
dh 2
 50   h 2 . At h  5,
 .
15. D— V   r 2 h   h3 (h  r ) 
3
3
dt
dt
dt 
16. C—The change in Ethel’s position is x  x 3 . The optimal length to climb is at a critical point. The only
realistic critical point is at the solution to 1  3 x 2  0  x 
3
.
3
17. A—At time t, Lucy is at position (t, sin t) and Desi is at position (t − 5, cos t). Hence at time t, the distance,
d, between Lucy and Desi is d  (sin t  cos t ) 2  25. To find the max value of d , we solve for t such that
dd
dd
dd
(sin t  cos t )(cos t  sin t )
0.
=
. Then
= 0 ⇒ sin 2 t  cos 2 t  0  sin 2 t  cos 2 t. This happens if
2
dt
dt
dt
(sin t  cos t )  25

. To maximize d , we need to maximize (sin t  cos t )2 . This is occurs when cos t
4

3
= − sin t. Because t is a constant multiple of
, we now see that t must be a constant multiple of
. Then
4
4
(sin t  cos t )2 = 2 ⇒ d = 29.
t is any constant multiple of
18. E (75)— Use d 
1 2
1
at  vt  d 0 . After t seconds, Paul is at location 1875 + (1)( t 2 ) from prison and the
2
2
1
(−1)( t 2 ) + 100 t from prison. When the arrow hits Paul, both objects are at the same
2
1
1
distance away from the tower. Hence, 1875 + (1)( t 2 ) = (−1)( t 2 ) + 100 t and t = 25 or 75. It must be t = 25,
2
2
arrow is at location
b/c after the arrow hits Paul, he will stop running. After 25 s, the arrow is moving at v =100-25(1) = 75 ft/s .
Calculus Applications—SOLUTIONS p3 _ FAMAT State Convention 2017
19. A— The magnitude of the change in volume per unit time of the two solids is the same. The change in
volume per unit time of the cube is 1 cm 3 /s. The change in volume per unit time of the cylinder is 
(
dh
 cm 2 ,
dt
dh
dh
1
= the rate at which the water level in the cylinder is rising). Solving 
 cm 2 = 1 cm 3 /s  cm/s .
dt
dt

20. C—We wish to maximize sin x · cos x =
maximum is
1
sin 2x. But sin 2x ≤ 1, with equality holding for x = π/4, so the
2
1
.
2

4
21. B—From the problem, we can get
 (sin x  C )dx  0.
So,  cos
0

4


4
C 1  0  C 
2 2 4

.
22. D— The surface area at time t is 6 t , so the volume is t 3/ 2 . So the air is being pumped in at a rate of
3
t.
2
When the surface area is 144, t = 24, so the answer is 3 6 .
23. D—When Fred first hits the line y 
1
3
, Wilma hits the line y  . Fred is then at
2
2
 2 1
,  , and Wilma is

 2 2
 6 3
 2 , 2  . Let Fred have velocity with components vx and v y , and Wilma have velocity with components


vy
w
dy
wx and wy . Also,
 2 x,   2 and x  6. Since their midpoint moves at speed 1 along the line y  1,
dx
vx
wy
at 
1
1
(vx  wx )  1 and (v y  wy )  0. So,
2
2
2vx  v y  wy  6wx  6(2  vx ). This gives vx  3  3. Fred’s
speed is given by vx 2  v y 2  vx 2  ( 2vx ) 2  3 vx  3 3  3.
24. A—This is the correct meaning. 25. D—This is the correct meaning
26. B—This is the correct meaning. 27. A—This is the correct meaning
28. C—For 0 ≤ x ≤ 2, it is cheaper for Kanye to use green paint since x 2  2 x in the interval [0, 2]. For the
interval [2, 5], it is cheaper for Kanye to use red paint. Thus, the minimum amount of money Kanye needs to
2
spend is
5
 x dx   2 xdx 
2
0
2

29. C— dist 
1
 1 t
0
2
dt 
71
.
3

2
.
30. C—The distance to the horizon is
(13  156) 2  1562  13 132  122  65. The length of the altitude to the
hypotenuse in the 65-156-169 triangle is the same as the radius of the bug’s circular path. By similar triangles,
alt 
65 156
 60. The path length is 2 r  120 .
169