APR12 Solutions
The rate a t which water flows out of a pipe, in gallons per hour, is
given by a differentiable function R of time t. The table above
shows the rate as measured every 3 hours for a 24-hour period.
(a) Use a midpoint Riemann sum with 4 subdivisions of equal
length t o approximate
L24
R ( t ) dt. Using correct units, explain
the meaning of your answer in terms of water flow.
(b) Is there some time t , 0 < t < 24, such that R1(t) = O? Justify
your answer.
(c) The rate of water flow R ( t ) can be approximated by
1
Q(t) = - (768 + 23t - t2). Use Q(t) to approximate the
79
average rate of water flow during the 24-hour time period
Indicate units of measure.
t
(hours)
0
3
6
9
12
15
18
21
24
(gallons per hour)
1: R(3) + R(9) + R(15) + R(21)
1: answer
= 258.6 gallons
1: explanation
This is an approximation t o the total flow in
gallons of water from the pipe in the 24-hour
period.
(b) Yes;
Since R(0) = R(24) = 9.6, the Mean Value
Theorem guarantees that there is a t , 0 < t < 24,
such that R1(t) = 0.
(c) Average rate of flow
NN average value of Q(t)
( 1: answer
1
1: MVT or equivalent
1: limits and average value constant
1: Q(t) as integrand
( I:
(units) Gallons in part (a) and gallons/hr in
part (c) , or equivalent.
answer
1: units
Page 1 of 4
APR12 Solutions
4. Suppose that the function f has a continuous second derivative for all x , and that f ( 0 ) = 2, f ' ( 0 ) = -3,
and f l ' ( 0 ) = 0. Let g be a function whose derivative is given by g l ( x ) = e P 2 " ( 3 f( a ) 2 f 1 ( x ) )for all x .
+
(a) Write an equation of the line tangent t o the graph of f at the point where x = 0 .
(b) Is there sufficient information t o determine whether or not the graph of f has a point of inflection
when x = O? Explain your answer.
(c) Given that g ( 0 ) = 4, write an equation of the line tangent to the graph of g at the point where
x = 0.
(d) Show that g l ' ( x ) = e - 2 x ( - 6 f ( x ) - f l ( x )
your answer.
+ 2f l ' ( x ) ) . Does y have a local maximum a t x = 07 Justify
(a) Slope at x = 0 is f l ( 0 ) = -3
1: equation
Atx=O,y=2
y - 2 = -3(x-0)
(b) No. Whether f l ' ( x ) changes sign a t x = 0 is
unknown. The only given value of f " ( x ) is
f"(0) = 0.
1: answer
1: explanation
1: g l ( 0 )
1: equation
(d) y l ( x ) = e-2" ( 3f ( x )
+ 2f ' ( x ) )
g " ( x ) = (-2e-2")(3f ( x ) + 2f l ( x ) )
+ eP2"( 3f ' ( x )+ 2f I 1 ( x ) )
I
2: verify derivative
012 product or chain rule errol
< - 1> algebra errors
1: g'(0) = 0 and yl'(0)
1: answer and reasoning
Since g l ( 0 ) = 0 and gl'(0) < 0 , g does have a
local maximum at x = 0.
Page 2 of 4
APR12 Solutions
5. The graph of the function j , consisting of three line segments, is
given above. Let g ( r ) =
lx
j ( t ) dt.
(a) Compute g(4) and g(-2).
(b) Find the instantaneous rate of change of g , with respect to x , at
x = 1.
(c) Find the absolute minimum value of g on the closed interval
[-2.41. Justify your answer.
(d) The second derivative of g is not defined at x = 1 and x = 2.
How many of these values are x-coordinates of points of
inflertion of the graph of g? Justify your answer.
g(-2) =
/lP2
1
f ( t )dt = --(12) = -6
2
1: answer
(c) g is increasing on [-2,3] and decreasing on [3,4].
Therefore, g has absolute minimum at an
endpoint of [-2,1].
Since g(-2) = -6
I
( 1: interior analysis
1: endpoint analysis
1: answer
5
and g(4) = - ,
2
the absolute minimum value is -6.
(d) One; x = 1
0
1
1 (-2;
I ) , g1I(x) = f l ( x ) > 0,
On ( 1 , 2 ) , g"(x) = f l ( x ) < 0
1: choice of x = 1 only
1: show (1,g(1)) is a point of inflection
1: show (2, g(2)) is not a point of inflection
On (2,4), gl'(x) = f l ( x ) < 0
Therefore (1, g(1)) is a point of inflectio~land
(2, g(2)) is not.
Page 3 of 4
APR12 Solutions
6. 111the figure above, line P is tangent to the graph of y =
at point P, with coordinates
(wL ,
-
, where
UJ>
1
x2
-
).
p
'
0. Point
Q has coordinates (w, 0). Line crosses the x-axis a t the
point R, with coordinates (k, 0).
(a) Find the value of k when w = 3.
,
X
o
Q
Ry
(b) For all w > 0, find k in terms of w.
(c) Suppose t h a t w is increasing at the constant rate of 7 units per second When w = 5 . what is the
rate of change of k with respect t o time?
(d) Suppose that w is increasing a t the constant rate of 7 units per second. When w = 5 , what is the
rate of change of the area of APQR with respect t o time? Determine whether the area is increasing
or decreasing at this instant.
2
dx
2
--
27
and (k, 0) has slope
(b)
Line L through
and (k, 0) has slope
2
{
*I
dx
3
1: answer
2
27
-
1:
2
w3
-.
1: equation relating w and k,
using slopes
1
0 - 7
2
1
2
or 0 - - = --(k
-w)
Therefore, 2= -k-w
w3
w2
w3
1: answer
' 1: area in terms of w andlor k
1:
4
<
1:
dA
-
1
dt
implicitly
,
dt
.,=,
uslng
dw
dt
-
=
7
1: conclusion
Note: 014 if A constant
Therefore, area is decreasing.
Page 4 of 4