Math 151 - Exam 3A - solutions
Problem 1 A car has position at time t given by
s(t) = 28t − 5t2
(distance in meters, time in seconds).
(a) Find the velocity of the car at time t.
(b) Find the acceleration of the car at time t.
(c) At what time t does the car come to a stop (has velocity 0)?
(d) Will the car be able to stop before it has traveled 36 meters? (so it would
not hit an obstacle at that point in the road)?
Solution. (a) v(t) = s0 (t) = 28 − 10t (in m/s).
(b) a(t) = s00 (t) = −10 (in m/s2 ).
(c) Solve v(t) = 0 to get t = 28/10 = 2.8 (seconds).
(d) The car will only stop after having traveled s(2.8) = 39.2 (in meters), so
it would hit the obstacle first.
Problem 2 Find the absolute maximum and absolute minimum values of
the function
g(t) = t4 + 4t3 − 20t2 + 1
on the interval [0, 3]. Indicate at which t-values each of these extrema occur.
Solution. We need to calculate
g 0 (t) = 4t3 + 12t2 − 40t = 4t(t2 + 3t − 10)
and solve g 0 (t) = 0. From the factorization
g 0 (t) == 4t(t + 5)(t − 2),
g 0 (t) is zero at t = −5, 0, 2 (you could also use the quadratic formula to solve
t2 + 3t − 10 = 0). So we have only three candidates t = 0, 2, 3 in the given
interval, and the table of values of g(t) at these numbers is
2 3
t 0
g(t) 1 −31 10
and therefore the maximum of g(t) on [0, 3] is 10, occurring at t = 3. The
minimum is −31, occurring at t = 2.
Problem 3 Suppose the demand x (= number of pallets sold) of DraftEx
Styrofoamtm is linked to unit price p by
p=
20 + 5x
40 + x
(a) Find the revenue R(x) as a function of x.
(b) Find the value of x in the interval [10, 60] which gives maximal revenue.
Solution. For (a), you only need to use
R(x) = xp =
20x + 5x2
40 + x
For (b), we use the Quotient Rule:
R0 (x) =
5x2 + 400x + 800
(20 + 10x)(40 + x) − (40 + x)(20x + 5x2 )
=
.
(40 + x)2
(40 + x)2
So R0 (x) is always positive in the given interval, and therefore R(x) is increasing there. The maximal revenue will then occur at the right endpoint,
x = 60 (if you did not use that R is increasing, you would have to evaluate
R(10) and R(60) and note that R(60) is greater than R(10)).
Problem 4 Suppose the revenue R(x) for selling x pallets of DraftEx
Styrofoamtm is actually
R(x) = −90x + 6x2 − 0.1x3 .
Find the number x in [10, 60] (of pallets of DraftEx to sell) so that the revenue
is maximal.
Solution. First,
R0 (x) = −90 + 12x − 0.3x2 = −0.3(x2 − 40x + 300) = −0.3(x − 10)(x − 30)
The candidates for x-values giving maximal revenue are x = 10, 30, 60. A
table of values of R(x) says
x
10 30
60
R(x) −400 0 −5400
so maximal revenue occurs at x = 30.
Problem 5 A rectangle has dimensions x and y (in inches), and x is increasing at a rate of 5 inch per minute, y is decreasing at a rate of 4 inch per
minute. At what rate is the area of the rectangle changing when x = 20 and
y = 50?
Solution. Let A be the area of the rectangle and x, y its dimensions, so
A = xy at all times t. Therefore
A0 (t) = xy 0 + x0 y
and substituting x = 20 and x0 = 5, y = 50 and y 0 = −4, we get
A0 = −80 + 250 = 170
(in square inches per minute).
Problem 6 Suppose unemployment U and population P of a small country
are related by the equation
√
U + 0.2U P + 0.4P = 20.
Both U and P are functions of time t. At a time when U = 4 and P = 15
(in percent and millions of people, respectively), the unemployment grows
at a rate of U 0 (t) = 0.2 percent per year. Use related rates to find the
corresponding rate of change for P . Include units in your answer.
Solution. We can differentiate both sides, like in the previous problem,
using the Chain Rule.
U0
√ + 0.2(U 0 P + U P 0 ) + 0.4P 0 = 0
2 U
To solve for P 0 , collect first all terms with P 0 on the left-hand side:
U0
(0.2U + 0.4)P 0 = − √ − 0.2U 0 P
2 U
after which we can divide both sides by 0.2U + 0.4. Then we substitute
U = 4, P = 15, and U 0 = 0.2 to get
P0 = −
(in millions of people per year).
0.65
≈ −0.542
1.2
Problem 7 Two quantities x, y are related by
x2 y + y 3 = 5.
Suppose that y is a function of x, ie y = f (x) and that for x = 2, y = 1.
dy
(a) Find dx
at x = 2 using implicit differentiation.
(b) Find an equation for the tangent line of the graph of f (x) at x = 2. Use
the form y = m(x − 2) + b.
(c) What y-value do you get for the tangent line equation from part (b) at
x = 2.03?
Solution. (a) Differentiate both sides:
2xy + (x2 + 3y 2 )y 0 = 0.
then collect all terms with y 0 on the left side, factor out y 0 and solve for y 0 :
y 0 (x) = −
x2
2xy
+ 3y 2
Substitute x = 2 and y = 1 to get y 0 = −4/7. (b) Use part (a) and x = 2,
y = 1. Then the desired tangent line has slope −4/7, and passes through
(2, 1). It therefore has a point-slope equation
4
y = − (x − 2) + 1.
7
(c) Simply substitute x = 2.03 into the tangent line equation from part (b)
to get
4
y = − (2.03 − 2) + 1 ≈ 0.9829
7
Problem 8 Find the indefinite integrals (= family of all antiderivatives) of
(a) f (x) = 2x4 − x2 + 20x − 3
(b) g(t) = 3t2/5 − 4
(c) h(v) = 3v 2 (v 3 − 4)9
Solution. Make sure to keep the same variable as in the problem, and to
add on the integration constant at the end. For (c), use the substitution
u = v 3 − 4.
2
x3
f (x) dx = x5 −
+ 10x2 − 3x + C
5
3
Z
15
g(u) du = t7/5 − 4t + C
7
Z
1 3
h(t) dt = (v − 4)10 + C
10
Z
(a)
(b)
(c)
Problem 9 Suppose the marginal productivity making x units of styrofoam
cups is
8
.
P 0 (x) =
x+1
(a) Solve an indefinite integral to find P (x) up to a constant.
(b) Given that P (0) = 0, find P (x).
Solution. (a) We use our integration formulas from 7.1 to get
P (x) = 8 ln |x + 1| + C.
(b) From P (0) = 0, substituting into the preceding equation, we get
0 = 8 ln(1) + C = C
so C = 0 and
P (x) = 8 ln |x + 1|.